EEVblog

[@atmel9077]

You've got 10Vpp BUT you've still got 5VRMS. For me it's not a voltage doubler. However if you add a diode and a capacitor then you'll get 10VDC and you now have a charge pump voltage multiplier. Note that 1.5v calculators use this kind of voltage doubler to generate 3v for the digital logic and to drive the LCD. On dot-matrix LCDs the controller also uses this kind of voltage multiplier to generate voltage up to 15 volts or more in order to achieve a good contrast.

[@PhoenixRevealed]

@Integrated Electronics - Your concept is correct, but the LCD drive is not 5V RMS since this is a square wave. RMS only applies to pure sine waves. With a square AC waveform the effective voltage is the same as the peak voltage, in this case, 5V. EDIT: In retrospect I realize that RMS does apply to square AC (it applies to any waveform), its just that for square waves the RMS == Nominal so it's not normally used in reference to square waves.

[@Yotanido]

I'll just put a copy of my response to the comment on the other video, in case it helps someone: At time t0: COM is at 0V relative to circuit ground Pin A is at 5V relative to circuit ground. Pin A is therefore at 5V relative to COM. At time t1: COM is at 5V relative to circuit ground. Pin A is at 0V relative to circuit ground. Pin A is therefore at -5V relative to COM. In other words, Pin A just went from 5V to -5V relative to COM, which makes a change of 10V.

[@ahmedejaz1125]

Yes but at any given point of time the voltage difference between com and a is 5V and so voltage difference is not actually 10 but only appears to be so

[@gblargg]

@@ahmedejaz1125 The point is that you can generate a square wave of double the usual amplitude, without anything special. So you can drive your LCD with a 10V square wave instead of just 5V.

[@ArnaudMEURET]

Yes. And the “real” magic is why does this swing matter to the LCD element.

[@SimoWill75]

For those that are still struggling to understand, try this. Imagine you have a speaker, you apply 5vdc, the speaker cone moves out and stays there right? Now remove the 5v DC and the cone returns to its resting point right? Now reverse the wires and apply 5vdc again, this time the cone moves IN right? Now, how many volts AC, peak to peak, would move the speaker cone the equivalent distance max out, to max in? Have a guess.

[@michaelhawthorne8696]

Good analogy....

[@amojak]

Simo.. you are referring to the equivalent of a centre tapped transformer with the speaker analogy. the speaker is biased to be mid travel you push or pull its cone. it has nothing to do with the particularly poor video.

[@SimoWill75]

*Facepalm* Your comment only highlights your lack of understanding.

[@amojak]

i would say it highlights the lack of clarity and mis-explanation in the video actually. But if you are confident how about you explain the 10V being created to drive the LCD here ?

[@SimoWill75]

Troll on to your heart's content.

[@mcuembedded]

Here is a short note for those who still find it confusing... :) Easiest way to understand it is, let's have the two points called A and B, where probes are connected in this video. For first half cycle, A = 5V, B = 0, and for half second cycle, A = 0V, B = 5V If we call A our "ground reference", A must always be considered 0V. This makes B look like -5V in the first half cycle (because B is ALWAYS 5V less than A in first half cycle. In the second half cycle, B is 5v higher than A, which makes it +look like 5V. So you effectively have B swinging from -5V to +5V, with respect to A. Why A? Because that's where the LCD COM pin was connected to!

[@AndyHullMcPenguin]

@Pratik, yours is by far the simplest explanation. The key is that we are always measuring *relative* to A. The chip's GND pin is completely irrelevant (except for the reason below). All our measurements are relative to A, nothing to do with GND. However the gotcha is.. if we use a non isolated power source, the moment we make the mistake of grounding Pin A, while trying to measure things with a grounded oscilloscope probe by attaching the croc clip to Pin A and the probe clip to B ... fzzzt..... there goes the magic smoke, 'cos we just shorted out Pin A to GND. If we have our circuit on a bread board, and powered by a battery, then we are OK, since everything has galvanic isolation from the oscilloscope ground, so by attaching the croc clip to A, the whole breadboard will float relative to the oscilloscope GND +/- whatever offset we have at Pin A , so we can get away with measuring the difference between A and B. Power the circuit from a mains power source, where 0V is not galvanically isolated from our oscilloscope and ... fzzzttt... we blow things up. Today's bed time reading.... en.wikipedia.org/wiki/Test_probe

[@flubba86]

I had the exact same thoughts and questions when I watched the last video. But I paused the video, thought about it for a while, rewatched dave's explanation, rewatched the scope example, looked at it from a different point of view, and suddenly it became clear. I thought I was just slow, I didn't realize others would actually call you out as wrong or misleading. Its like staring at an optical illusion for a while then suddenly seeing it clearly.

[@PhoenixRevealed]

To be fair to the doubters, I've learned how LCD driving works 40 years ago and I still found Dave's explanation confusing.

[@gblargg]

@@PhoenixRevealed It is confusing. The output of the inverter is always +5V or -5V relative to the input. So feed a square wave and you get a +5V/-5V square between the input and output pins. Relative to the circuit ground it's another story, but if your device is floating this doesn't matter and it just sees a square wave going between +5V and -5V, for a total change of 10V each time.

[@russellhltn1396]

It's just like taking a stereo amp and putting it into bridge mode and wiring the speaker across the two outputs.

[@PhoenixRevealed]

Yes it is, but from the comments I don't think many people here understand amp bridging either. H-Bridge motor drivers work the same way too, but people are having trouble grasping that if you keep swapping the voltages on the ends of any component then that component sees EXACTLY the same source waveform as it would if driven by a true AC supply.

[@Marzec309]

It's simple, the scope is measuring relative to it's Gnd reference (the Gnd clip). When the Reference voltage on the Gnd clip is 0V and the probe is at +5V the difference is a +5 Volts. Now, the confusing part. The logic is changing the voltage at the Reference Point (The Gnd Clip) to +5V and at the same time the probe is now at 0V. Zero is below 5, the result is that the measured Value is now -5V. You end up with a +5v to -5v Differential swing on the scope. BUT, it is still a 5 Volt Peak to Peak signal relative to the logic circuit (no voltage doubling) the AC RMS is 2.5volts, you only end up with a "SIMULATED" -5Volts because the COM of the LCD is Floating relative to circuit com and not at a fixed reference point. Your single ended voltage is still 5 Volts no mater the polarity (of a single signal). It is simply the function of TWO different wave forms that give you the 10Volts (remember the LCD's com pin is floating). The Differential AC RMS of the to signals is still 5V. No doubling took place ;-) It's also worth mentioning that each segment of the LCD would have its own XOR gate, wired exactly as the one shown for segment A. In this way, all the driven segments will change polarity relative to the LCD's Com pin at the same time. Effectively driving the LCD with an AC Square wave. With current flowing from com to segment pin and then having the current change direction and flow from Segment pin to the Com pin (exactly as an AC Signal would).

[@alanesq1]

If you measure 5vdc with a multimeter then swap the probes over you will see -5vdc. This is basically what this circuit is doing - alternating between 5vdc and -5vdc resulting in 10v peak to peak.

[@qwertyasdf66]

Yeah that's almost exactly how I was imagining it. But I was thinking of a 9v battery hooked directly up to a scope. Then simply flip the battery and now you have -9v.

[@amojak]

except it isn't. you are just using a multimeter to look at 5V that is swapping polarity . it never is or will be 10V .

[@amojak]

with the 9V example, you still only have 9V too

[@SweatyHatMan]

amojak The claim here is it is 10V peak to peak. Which is exactly what it is doing... Vpp is the distance between the pos and neg peaks, not the value of the peak in reference to GND.

[@ranikeev]

The same trick is used in car audio amps. Due to 12V power limit they use bridge sircuit to drive loudspeekers. So it is possible to get 24 volts on the output peak to peak.

[@PilotPlater]

I understood it on the first video, but good on ya Dave for helping the community!

[@FilipMilerX]

Actually 10V voltage SWING (difference between minimum and maximum amplitude) doesn't mean 10V VOLTAGE. Actual voltage is +/-5V, but not +5V and -5V AT THE SAME TIME, so not +10 nor -10V differential. But the SWING is definitely 10V.

[@henribondar3997]

Yes your right this is an inverter and not a voltage doubler as Dave falsely suggest

[@MarkTillotson]

Its definitely doubling the AC component of the voltage across the load, just as a bridged audio amp doubles the voltage across the load. I think there's some confusion here betwen DC and AC - peak-to-peak voltage is a property only of the AC component, we are only talking here of the AC component. Another way to think of this setup is using phasors - the pre-inverter and post-inverter signals are 180 degrees out of phase, as as phasors(*) you can call them +5 and -5, the difference is 10. (*) This strictly only applies to sine waves of identical frequency, I gloss over the harmonics, but it works the same.

[@FilipMilerX]

But the amplitude of AC component (DC is zero in this case) is 5V, not 10. Swing is not amplitude.

[@alynicholls3230]

yes but there is 10v, but 10v ac.relative to virtual ground, 5v dc relative to actual ground.

[@bertbronson8395]

Ok, if true I should be able to power, say 4 red LEDs at like 2mA or more from this instead of just two right?;) No, it is only 5 volts (like he said, sort of like an H-bridge... polarities keep switching... so a motor would just see 5v ... first from one direction then from the other for example). Interestingly, RS-485 works like this also (10v noise immunity with only 5v ... ethernet based on '485).

[@yabgu79]

This should be studied in schools, after 10 years from graduating I saw this first time and this has to be something fundamental; It really changed how I look into circuits thanks really,

[@PhoenixRevealed]

It is studied in schools. This is basically the exact same technique used in bridged amplifiers and H-Bridge drivers.

[@yabgu79]

I don't think H-Bridge drivers double its voltage and I dont think this circuit will hold even if we put 1mA load there; Most likely output will become 5V with minimum amount of load. Seems like this is very special edge case to me,

[@tomasz9517]

at 5:53 You shoud not write -5 on the table, some folks could not understand it that way. I think the proper way will be draw an arrow down from 5v to 0v to show from where the magic negative voltage show up.

[@DOOMDUDEMX]

The simplest way i can think of explaining this is by taking a battery and alternate between connecting it backwards and forwards between the probes. This is essentially exactly what the circuit is doing with the inverters.

[@henribondar3997]

In common inverter technology, this technique is known as "full bridge" by comparison to the classical "half bridge" structure. It involves four switches and is equivalent to revert voltage and current in the load (then here alternating from +5V to -5V). However it is not usually called a voltage doubler as there is no way to obtain a 10V voltage relative to a common ground without using a storing capacitor somewhere. So voltage doubler: yes and no, everyone is right !

[@kkkk0freak008]

Yout absolutly right! Sadly Dave didnt mentiond it.

[@kkkk0freak008]

Oh dammed he does in the end of the vid as subtitles...

[@henribondar3997]

Yes your right, however using the word doubler for such a circuit is technically incorrect, there is never a 10V applied to the LED device only alternately a +5V and a -5V but not a 10V voltage difference between the PIN of the LED has he falsely suggests by an inappropriate understanding of what he sees on the scope (a 10Vpp signal doesn't mean that a 10V is present at any time)

[@BenjaminEsposti]

Henri BONDAR, I agree, it's a bit confusing to some I'm sure, when you say "voltage doubling" ... it's not necessarily doubling the voltage, but doubling the possible excursion. (Remember that voltage potential is only relative.)

[@martin.winkelhofer]

Hi Dave, I've agreed with every video you've made ... until now. It's not voltage doubling. It's just alternating 5 V across the load one way and then the other way. Your explanation gives an impression that there will be 10 V across the load (between those 2 test points), but in reality, there's only 5 V potential at any given moment. It's just that the polarity changes. If you hook up the scope this way, you'll see what you see, I completely agree with that, because your reference is constantly changing. Another thought experiment: if we hook up 1k resistor between those 2 test poins, what will be the power loss on that resistor? I'm saying 25 mW (that's equal to constant 5 V across 1k (0 V on one end and 5 V on the other for 50 % of the time and then the opposite for another 50 % of time)), but according to your reasoning, you'll probably say 50 mW (which would be equal to constant 10 V from your "voltage doubler"). What is correct?

[@waltercomunello121]

Although it may be seen as trivial, in *P = V x I* the voltage is *the absolute value* of a given voltage in time. Either we do have electromotive force, or we don't. The absolute value of V across the load *never surpasses 5V* , be it in a direction or in the other, the load never gets a whole 10V emf because the reference (i.e. "no charges moving because there is no force") is right in the middle. I've got to say, this channel confuses me from here and there. Good thing I want to acquire solid basics of electronics. Unfortunately, I've yet to found them here.

[@gblargg]

@@waltercomunello121 It will sound twice as loud if you drive a speaker. Something is being doubled. The waveform is jumping 10V between levels rather than just 5V.

[@mikejones-vd3fg]

@@gblargg hmm intersting, maybe because the speakers reference doesnt change it reproduces the full 10v swing as the cone moves in phsyical space, so although electronicially theres no doubling just an appearnce of it, phsyically you get the full efffect

[@gblargg]

@@mikejones-vd3fg It's similar to driving it with a 0 to 10V wave, in that you have 10V signal swings. But like the 5V case, it's only effectively continuously 5V since in the 0 to 10V case half the time it's at 0V.

[@gigglesseven]

new drinking game. every time dave says "actually" take a shot. if you can still understand the video by the end you win

[@gigglesseven]

7 miutes in, i'm drunk

[@ktos9934]

@@gigglesseven You almost killed me, now I'm back from hospital

[@PenZon]

You could think it like this: You have the two square waves ,out of phase, each on its own 5vpp. If you change the reference point to be one of these square wave outputs, you have to re-draw the voltage waveform for your new reference point, i.e., flatten it. This new flat line represents your new 0v. And since you flattened one wave form and the voltage difference did not change, your other wave form will be amplified. And since we consider our first output as the 0v (let's call it pin A), we see the inverted output (pin B) flipping from +5v relative to it (logic drives A low and B high), to -5v relative to it (logic drives A high and B low), giving us a total of 10vpp.

[@PhoenixRevealed]

I think you are missing the simplest way to explain what is happening when driving LCD segments. If you think instead of driving a DC motor with an H bridge... that is EXACTLY what is happening when driving LCD segments. The same differential drive trick can be used with VFD display heaters to prevent one end of the display being brighter than the other due to gradual voltage drop over the length of the heater (this technique makes the AVERAGE voltage the same the entire length of the heater). By switching both ends out of phase of any component between earth and Vcc (for instance) from the perspective of that component you are driving it alternately with Vcc and -Vcc, or in other words, an AC signal of 2x Vcc peak to peak.

[@BenjaminEsposti]

Yup, this is just like some audio amps that do the same thing to get more power from the same voltage supply rails. Just add another amp, with inverted signal, and presto!

[@AppliedCryogenics]

Dave, the way that really gelled it for me was the idea of driving a Piezo element with one gpio pin vs2. I think that's a pretty bare-bones example of what you're trying to demonstrate.

[@station240]

Another way to look at it, is in terms of current flow. After all that is what makes it a negative voltage. A -> B +5V B -> A -5V You can also show this differential signal by measuring the current, say by adding a resistor in the circuit and measuring the voltage across it using the scope.

[@patwicker1358]

Thanks for clarifying, I was one of the ones who was confused. I believed you I just couldn't get the numbers to add up.

[@tzubin99]

Cool! I learn something new almost every video! Thanks Dave.

[@mikewillis1592]

I think the problem people have is tied to thinking about ground and common. Much easier just to think about pin 1 and pin 2. Then its relatively simple. Two alternating options, Pin 1 2 Option 1 = 0V 5V Option 2 = 5V 0V There is only ever 5V relative to ground, but the LCD is not connected to ground and doesn't know this. As far as the LCD is concerned its got a push-pull alternating 5V alternating signal across it. This is a 10V peak to peak. This is rather useful beyond driving LCDs. For example if you are driving a loudspeaker with a 12-14V supply like in a car. The peak voltage available is ~12V. If you were to drive the speaker with a single amplifier like a TDA2003, with one side of the coil connected to ground, you could only have maximum of 12V across the coil. That's 6V peak to peak AC or 4V RMS and with a typical 4 ohm speaker only 4W of audio power, not enough if you own a Landrover. If you instead drive the speaker with two amplifiers, in antiphase with the speaker coil between them, as in the LCD example, you have up to 24 volts available, 8V RMS and 16Watts. This is called a Bridge amplifier. There are more exotic solutions, to high power audio from 12V, that this was one of the most common before switching amplifier chips became commonplace.

[@airmann90]

You made it click. Thanks!

[@chriskaye1997]

Love the back to basics whiteboard approach :)

[@ElmerFuddGun]

Not "doubling" any more than 10 Vpp is "double" 5 Vp. 10 Vpp = 5 Vp which since this is a square wave can be easily seen equal to 5 VAC RMS. The power dissipated by a resistor powered by a 10 Vpp (=5 Vp) AC square wave will be the same as if it was powered by 5 VDC. This circuit does however double what you could get using just a single logic output (tied to a 2.5 V common to provide AC to protect the LCD) as you are going from 2.5 Vp to 5 Vp. (all assuming full swing 5V logic. TTL will be different of course).

[@waltercomunello121]

Good point. Power dissipation implies that voltage is taken as *absolute value* . Resistors are polarless passive components, they don't care which way charges flow (do they?). So in this case we should keep in mind that in *P = V I* V is *always 5* and not 10.

[@capt.k8577]

thanks dave for another interesting and well explained video

[@sunebrian1423]

I suggest clarify some of the notations in the previous video, which are the voltage referencing points in the circuit i.e. the LCD.

[@TheBdd4]

It works! I just simulated it with LTSpiceXVII. Input signal = 5 volts at 1KHz, outputs labeled A and B, 10 V P-P displayed by taking V(A) - V(B) which displays as 10V P-P at 1KHz.

[@axelbostrom3606]

Took me a while to get it, but the current is changing direction and that's how you get it. First you have flow one way that we call positive, and then you have flow the other way and we get negative voltage.

[@Ma_X64]

I used this method to drive a piezo-buzzer directly from MCU. Just using two output pins that worked complementary. It's just an equivalent of an H-bridge cirquit. It not doubles your voltage but it doubles the amount of work that current does because instead of "working/not working" cycle it's doing "working forward/working backward" cycle.

[@mcflapper7591]

Dave, honestly, this is the type of topic/video I love to see on eevblog. Same for the reference to your scope gnd ref video. These sometimes not so intuitive aspects are my preferred topic. I would subscribe to this chan again if there was such a thing. :) The key to this "mystery" is probably the reference to the inverted signal as the signal levels are switching simultaneously in opposite directions, effectively doubling the voltage (relative to the LCD and w/o gnd ref). And as you already mentioned in the last video, the LCD only wants to see the difference in potential on one pin relative to another. GND is irrelevant here. Thanks, again, and thumbs up!

[@SFefy]

I think the simplest explanation is the following: when you measure the voltage with the correct polarity red (+) black(-) the scope shows +5V. When you put in some inverter into the circuit the polarity is swapping so red(-) and black(+). This case the scope shows -5V. Technically the scope (or DMM) just "substracts the voltage levels from red wire and black wire". With correct polarity: red(+5V)-black(0V)=+5V, the inverted polarity: red(0V)-black(+5V)=-5V

[@TheTruthSentMe]

How about this explanation?: Vpp is NOT A VOLTAGE. A voltage is the difference of two potentials AT THE SAME POINT IN TIME. That is not how Vpp is measured though. Vpp describes a time-dependent signal and is the difference of the highest and lowest amplitude regardless of where those appear in time.

[@Djhg2000]

+TheTruthSentMe Vpp is definitely a voltage. Ever heard of VAC?

[@TheTruthSentMe]

The unit is volts, but it's not a voltage. At no POINT in time you can measure that. It's a characteristic of a signal. For the same reason I don't consider VAC a voltage. It's a RMS value of a voltage.

[@Djhg2000]

+TheTruthSentMe Voltage is *always* a difference, either with respect to another point in DC or with respect to time in AC.

[@TheTruthSentMe]

"Voltage is always a difference" Yes. "either with respect to another point in DC or with respect to time in AC" No. You don't understand the difference between a voltage and a signal. What you call DC is a voltage. There is no AC voltage if you take it literally. There are AC signals where every point in time corresponds to a voltage. You can come up with ways to describe the AC signal in quantities that have the unit volts. That doesn't make them a voltage though. A voltage can only ever be the momentary value of the difference of two electrical potentials. Therefor you cannot call v(t1)-v(t2) a voltage where v(t) is a signal and t1 and t2 are points in time.

[@Djhg2000]

+TheTruthSentMe AC can be used for power delivery as well and you seem to be stuck in the world of DC.

[@willhelmx8388]

Nice explanation Dave, thumbs up for your channel

[@mr.amp0076]

& also their is very less current flow in the two differential points.... If you think about total power p=vi then the current flow is less... For this type of screen it operates in electrical field & less current flow... Means you can also get higher villages.... Awesome explanation dave...

[@mata7648]

Took me a while to grasp when I first used a differential op-amp.

[@QuantumFluxable]

I think a good analogy would be someone driving between three cities A, B, and C: A and C both have a distance of 50 km from B and all cities lie in a straight line on the map. If you were to drive from A to C with a short break in B, an observer in city B would never see you move further than 50 km from B, but an observer at C would see you move up to 100 km away from their location. A represents +5V, B is circuit ground, C represents -5V and the observer is the oscilloscope ground. The person driving from city to city is the output voltage of your circuit. While this isn't completely accurate it might help to illustrate your point.

[@Arnthorg]

its like connecting the leads backwards and then correctly again and again

[@yngvenystrand167]

Many comments here... maybe somebody already suggested this... but here goes: It might help to understand what Dave explained if the whole circuit with power supply, square wave source and all is floated, that is, all GND connections removed, so that there would be no "natural" reference point anymore. Then connect the GND to one of the legs of the LCD, and connect the scope to the other LCD leg, and the scope GND to the new circuit GND. The scope would then show 10 Vpp (varying between -5V and +5V), which still is 5Vrms. You wouldn't do like this in practice of course, but only "in your mind", to help visualize the whole thing. Not sure if this really helps though......

[@Jedda73]

I guess a simple way to look at it would be measuring a 1.5V battery with a multimeter. Probed one way gives +1.5V and reversing the probes gives -1.5V. The total voltage difference possible is therefore 3V from a 1.5V battery.

[@PhoenixRevealed]

I challenge you to run a component requiring 3VDC from a 1.5V battery just by continually reversing the battery. Can't be done.

[@zer0b0t]

Or simply invert the probes in a multimeter or oscilloscope and then you get a negative or positive voltage, the difference between both voltages is double the voltage you are measuring.

[@zanekaminski]

It’s the same thing as wiring up a 240V circuit (at least in the US. I don’t know how it’s done in other countries.) Two hot wires come into a building, each carrying 120V RMS AC. The two wires are called A and B and are 180 degrees out of phase, that is A = -B. For regular 120V outlets, you only connect one of the hot wires, and then the neutral provides the return. For a 240V outlet, such as the one used by your air conditioner or dryer, you use the A and B phase in conjunction. Although both are only 120V RMS, the difference is 240V, as expected.

[@PhoenixRevealed]

No, it's not the same thing. What you are describing is more akin to the output of a center-tapped transformer, with "neutral" coming into your home being the center tap. There is no "center" here, and no way to get 10V AC, no matter how you connect it up.

[@TheBdd4]

By the way, nice move Dave, all this chatter has occurred while you sneakily ( kudos to you) introduced your differential probe.

[@Shawn_White]

watched this at 1:00AM before going to bed had to run down to the shop right quick to test it and I'll be damn it actually works. Still can't quite wrap my head around it so I'll give it another go tomorrow.

[@deBug67]

You cant wrap your head around it because Dave got it wrong.

[@Shawn_White]

nope it actually worked just can't wrap my head around why. I also put kapton tape around the earth pin on my scope power cable since I'm only dealing with low voltages. This means I don't need any differential probes.

[@EE_fun]

Well this is something that you learn in first class at university in EE. It's so basic and fundamental like Ohms law is. Maybe this would get more clear, if you show it with a multimeter on a positive and negative DC voltage. You can easily demonstrate this by measuring the 5V rail against the -5V rail of your PC supply and you will read 10V on your meter. Voltage is the difference of potential: V = phi1 - phi2. 5V - (-5V) = 5V + 5V = 10V

[@vladomaimun]

Very well explained, thank you!

[@scottb721]

From the point of view of the input the output is either 5v higher or 5v lower regardless of what the input actually is. If the input is set to the zero line on the scope the output will swing between 5v higher (+5v) to 5v lower (-5v)

[@henribondar3997]

To confirm previous comments such as philips miller one below, Dave is wrong on that one, there is never a 10V voltage difference applied to the LED PINs only alternately +5V and -5V. This circuit can be called an inverter but surely not a voltage doubler !

[@djbare9]

I've used this technique to produce 9 volts from a 5 volt supply, very useful for circuit setups that require duel voltage but you don't want to add two battery types, in my case I was running from a 5 volt usb battery pack normally used for smartphones.

[@railspony]

Perhaps useful in explaining it would be to add 2 diodes and 2 capacitors to make it into a charge pump and get ~10V DC output. Your circuit in the video is the same thing, just without the storage. Lots of ICs have the circuit built in.

[@MrPinolatino]

You have great capacity of explain things! great teacher! thank you very much!

[@Keeze68]

So, you are using some kind of virtual ground, just like in many battery operated headphone amps..?

[@BenjaminEsposti]

This is how some audio amps get more power from a lower supply voltage. Instead of one speaker terminal being connected to ground, it is connected to the output of another amplifier that is set up to have an inverted signal output. So instead of being able to supply + or - 20V to the speaker, it an supply + or - 40V! (And more voltage across a given impedance means more current, and thus more power!)

[@philevans4021]

Thanks a lot for this. I was actually struggling to get my head round this properly, but now I understand it perfectly :-D

[@JohnAudioTech]

Using a couple LEDs connected anti-parallel to the outputs of two gates swapping low and high would also be a good demonstration of the current reversing in the circuit giving the 10vpp

[@fabimre]

JohnAudioTech : that only works with totempole output, not with open collector. You'd create a H bridge!

[@amojak]

these devices are totempole output. the differential voltage will only ever be 5V. it in this config cannot be anything else. which is +5V and which is gnd just alternates.

[@fabimre]

amojak, wrong. The peak to peak voltage is 10 volts, while the effective (RMS) value is 5 V (square wave). The swing of the segment is +/- 5 V. If you would rectify both sides and measure across these rectified sides, effectively making one these sides reference, you WOULD measure 10 V!

[@PhoenixRevealed]

"If you would rectify both sides and measure across these rectified sides, effectively making one these sides reference, you WOULD measure 10 V" NO YOU WOULD NOT!!!! For crying out loud, why do people keep promoting this idea? The voltage across the LCD segment when enabled is NEVER 10V, it is 5VAC. If you rectify and filter the 12VAC output of a transformer you don't get 34VDC (24 x 1.414)... you get SEVENTEEN VOLTS DC (12 x 1.414). If you don't believe me then build a simple circuit and prove it to yourself rather than just make erroneous prediction of what you THINK would happen.

[@SuperSSSSooonniicccc]

10 Vpp, not rms Martin Green. To use your transformer example, it would be 34 Vpp, not rms. At least, according to my full-bridge regulator and capacitive filter I made in Multsim it is. This is only possible when using a differential probe between the positive and negative sides of the transformer, not when only measuring the positive side with reference to ground, as seen with my standalone probe to the right of the differential probe. goo.gl/mR4cyo

[@ovalwingnut]

Back for a 2nd suck of the salve.... It's that good and clearly I'm slow. Thank you Ausi-Man

[@12325833]

simplest explanation - just flipping battery terminals at some frequency

[@Anton-hz4ev]

Hey Dave, i think the explanation makes now much more sense than before. By the way, i tried to use the coupon from the task description but got the message coupon expired :/ would you mind to check if there is a typo or something else going on? Thanks in advanced and good job with LCD series, I find it very interesting and fascinating to gain a bit of your insight ^^ Kind regards Anton

[@erikdenhouter]

My way to say what I think I know : * take an oscilloscope. * Put it on DC, with the 0V nicely set in the middle. * Connect a 9V battery to the input, the line goes up 9V. * Disconnect the battery, the line returns to 0. * Reverse the polarity of the battery. * Connect the 9V battery to the input, the line goes down 9V. * Difference between upper- and lower line is 18V. What I missed was the connection of a resistor and a capacitor that would be charged to the 10V DC. That must be possible and convincing ?

[@willishendley8924]

I think your two input waveforms to the math function highlight that the only way this makes sense is that the “reference” voltage is actually dynamically changing between 0V and 5V, which of course is irrelevant to the differential probe (and LCD apparently).

[@indrek.a]

I understood it better when inspecting the circuit with falstad.com/circuit, which showed that it is the same as if someone repeatedly swapped the polarity between the SEG and COM by swapping the leads of 0V and +5V at those points and that the direction just changes. I thought I understood it from the original video, but this video proved me wrong as I did not understand your explanations from this one. After investigating it myself, now I get why it works and can see that actually your explanations make sense too, thanks!

[@frankpiscopo]

To get your head around this, consider the simple logical inverter fed with a 1Hz square wave signal. Connect a ±5Vdc center-zero DC voltmeter as follows: 1) meter -ve to "0V" and meter +ve to the inverter output. It's obvious that the meter will swing from center-zero (inverter output low) to +5V (inverter output high). That's a 1Hz 5Vpp square wave. 2) move meter -ve to the inverter input. It's obvious that the meter will swing from -5V (inverter input high, output low) to +5V (inverter input low, output high). That's a 1Hz 10Vpp square wave. That's it. Between circuit ground and the inverter output, you have 5Vpp. Between the inverter input and the inverter output, you have 10Vpp.

[@Psi105]

You can demonstrate this with sound. Drive a raw piezo beeper or speaker connected between ground and 5V GPIO = medium volume. versus, Driving it between two GPIO that are alternating out of phase = Loud. The piezo mechanically moves X distance negative then back to ground. versus Moving X distance negative then X distance positive = A total 2 * x movement

[@KyleRepinski]

The video was confusing to me even though I actually understood what was going on. The oscilloscope demo might've been easier to picture if channels 1 and 2 (blue and yellow) were overlaid instead, forming two solid lines across the screen with alternating colors. "The difference from blue to yellow is always 5V. It's like measuring a battery's voltage with a multimeter and swapping the probes around (then demo that too)."

[@zwerko]

It's exactly the same as with multi-phase AC - two 220V lines can give 380V because they are phased by 120° from one another. You can look at this situation as integrated two-phase AC with phases removed by 180° from each other.

[@power-max]

I exploited this phenomenon when I was like young (like, 13) using 3 position DPDT switches and wiring them up in a way that I could apply power to a motor and reverse polarity. I think I also did some stuff with voltage doubling too, but might have been later on.

[@mrjazzycharon2]

I think it would be helpful to differentiate between the terms peak-to-peak and peak amplitude. Peak-to-peak is 10V, but the peak amplitude is only 5V. Try to connect a rectifier and you will get only 5V DC, because there is never a -5V reference potential available opposing the 5V.

[@mrjazzycharon2]

I think you are wrong. Imagine a LED, which would need 7V to light up. You won't be able to get it to light up. In other words, you won't be able to get a DC voltage >5V out of the bridge rectifier, which would be needed for that LED.

[@electronash]

mrjazzycharon2 Yes, but connect a rectifier across the input and output pins of the NOT gate (inverter), and you DO get the ~10V DC output, which is the point. ;) When you say to "try to connect a rectifier", that is under the assumption that one side of that rectifier would be referenced (connected) to Ground / 0V. The LCD in this case is of course essentially "floating" , as it doesn't have a connection directly to the normal Ground point of the circuit. I know what you're saying about the "-5V" thing though, as that could also confuse some people, but it's still technically accurate if you connect the oscilloscope's "0V" or "-" input to the output of the NOT gate, then connect the 'scopes "+" input to to Ground point of the circuit. Yes, you will still only see a 5V P-P waveform that way, but it demonstrates how that waveform would be the inverse of the NOT gate's input, which is why Dave had to show both signals on the 'scope at the same time. It is a tricky concept for the beginner's, especially when we're taught to always use the 0V connection in most circuits as the Ground reference, but it obviously doesn't need to be. :p It's all about the difference in voltage across any two points in a circuit of course, which is why Voltage can also be referred to as the "potential difference".

[@mrjazzycharon2]

Yes, I assumed to connect the rectifier to the two gate inverter outputs and not to any ground. The voltage across the two inverter outputs is either +5V or -5V. The rectifier will output the absolute value of that, which is then always 5V without polarity changes. But it will never be 10V.

[@mrjazzycharon2]

I think the key thing to understand is that you either have +5V or -5V. But you never have both at the same time, which would be necessary to get 10V DC.

[@electronash]

Actually, yes, you are partly right... There's still only a "static" difference of 5 Volts between the input and output pins of the NOT gate (in either of it's two states), but one of those states will effectively reverse the polarity of the voltage at those two points. Then, the reference for the oscilloscope will have moved, so... Yeah, I'm a tad confused as well now. lol It appears there must be 10 Volts across that differential probe, but I don't see how you would get 10 Volts DC if the NOT gate was not toggling? A normal charge pump would need to store the voltage on each polarity change on some caps, and then output that voltage with the caps in series? Even with the capacitance in the LCD, surely it would still only see a voltage of 5 Volts across a segment for either phase of the NOT gate?

[@uzaiyaro]

I’m really interested in that diff probe. Any reviews or videos on it explaining what it’s useful for and whatnot? I’m looking to probe around switchmode power supplies, especially measuring their noise and power quality etc. Would this fit the bill in not blowing my lowly USB oscilloscope (or the computer attached to it) to high heaven, or do I need an isolated probe instead? I want to do this safely and with minimal or no risk of letting the smoke out of anything, and I’m happy to pay for that safety. That said. Any videos on this? Cheers!

[@2oqp577]

The point is that Dave doesn't make sure people don't start to believe that there is a measurable 10V drop between two points at any given time. I thought this would have been very important to point out, albeit countless cues. teaching is a tough job.

[@ChristopherLionRoars]

There is never actually 10v across the inputs at any time. Its either 5 or -5. But it results in 10ptp ac.

[@flubba86]

correct, that is a really good way of explaining it. thanks.

[@willyarma_uk]

I said similar in a seperate comment, if youtube comments had been working I would have seen this earlier. I said there is never more than 5V difference at any given instant.

[@gotj]

Just like the AC at the wall plug.

[@MarkTillotson]

Another way to put it (which will confuse even more?) is that the original single-ended signal has an AC amplitude (and rms value) of 2.5V, whereas the differential signal has an AC amplitude/rms of 5V. For a square wave amplitude=rms. (And the DC component of the single ended signal also has an rms voltage of 2.5V as it happens, making total rms of 3.56V (2.5 x sqrt(2)), but that's going to confuse even more!) Yet another way to think about it is power into a load of fixed resistance - the original single-ended 5V square-wave signal puts 0.125W into a 100 ohm load (half of which is DC component, half AC component), whereas the differential signal puts 0.25W into 100 ohms load, 4 times the AC power from the single-ended signal).

[@markg735]

Another way to think about this is to read up on a "flying capacitor" circuit. You can actually think of the capacitance of the LCD as the capacitor in a flying capacitor power supply.

[@PhoenixRevealed]

Huh? The minuscule capacitance across the LCD segments is too small to have any appreciable effect at 100Hz. You are overthinking this. All this circuit does is continuously reverse the polarity of the 5V supplied to the LCD from the supply rails.

[@cameronsteel6147]

Hi, I'm from the Department of Redundancy Department

[@cameronsteel6147]

Adamus Alpha (Adam) Haha yeah for some reason I’ve really been noticing “LCD Display” a lot recently 😂

[@skeggjoldgunnr3167]

VIN number = Vehicle Identification Number Number. okay. Yet, everyone says it. I bother them about what they just said - every time.

[@nullunit5566]

Just imagine a LEGO block that represents 5 volts of potential. The LEGO block can't grow or shrink, but it can flip in polarity (position). So if Common is our 'base', then the 5 volt LEGO just moves above the 'base' line and below the 'base' line as the phase transitions occur. Hope that helps to visualize what is happening. Cheers.

[@notamouse5630]

Not relative to anything is also not true, each is relative to the other signal, so choose your point of reference arbitrarily between the two options. The good way to think of this is as a Fourier series. A 0-5v square wave is 2.5v dc + sum of all the components in a zero biased square wave of 5 Vpp. Thus when you take the difference between two of them 180 out of phase, the dc cancels, and the one adds the negative of the other to get double the AC voltage with no DC bias assuming perfect inputs.

[@Darieee]

Hey dave, I still think this explanation was too advanced for those who didn’t get it the first time ... I really don’t want to sound condescending, I barely know my shit ... But given that, I would’ve showed literally on a breadboard and the scope on roll mode with a few wires on the breadboard how you can get a 10VPP wave, ... if they didn’t get the arduino, yeah, maybe the gates they would, but, those still are quite abstract if you’ve never seen their internal construction with the 2 fets, how fets work ... etc, so with a breadboard and just some wires all that prior knowledge wouldn’t’ve been required Anyway, good on ya dave for the 2nd take in any case, we appreciate the effort

[@douggale5962]

5 - (-5) = 5 + 5 = 10. Easy.

[@ceneblock]

What was confusing tome at first is your initial diagram. The "diff" should have started at -5 instead of +5v Once I figured that out, then it fell into place.

[@HectorDC02]

wooow this is a very crazy thing, thanks Dave

[@lasersbee]

Great explanation and clarification...

[@2oqp577]

The point is that using terms involved with AC don't go very well with DC. It is not understood by most amateurs that there is never more than 5Vdc at any given moment.

[@10100rsn]

I thought this was magic at first but using the outputs of the two inverting gates are like using an H bridge to drive the output. You have 5 volts output on gate A and ground output on gate B but when you invert the gate signal you have ground output on gate A and 5 volts output on gate B effectively reversing the polarity, and that is where the 10Vpk-pk comes from. Peak-to-peak (pk-pk) is the difference between the maximum positive and the maximum negative (also known as trough) amplitudes of a waveform. So its not actually 10 volts but it is _10 volts peak-to-peak_, as the formula; (PeakVoltage - TroughVoltage) = (5 - -5) = (5 + 5) = 10Vpk-pk. It all works out. ;)

[@alanclarke4646]

Put a piezzo buzzer from 2nd inverter output to ground. Then put it across the inverter and notice how much louder it is.

[@gotj]

You can explain that with a DMM and a simple analogy with the AC at the wall plug.

[@manickn6819]

Made sense last video but I still watched this one. I know you tried Dave but for those who did not understand I am not sure you could convince them with this video. Either you see it or you don't.

[@EEVblog]

Yeah, probably.

[@PhoenixRevealed]

To be fair Dave, I learned how this works about 40 years ago, and I have to humbly say that both your explanations were clear as mud. Sorry mate.

[@bbrazen]

I think people are getting confused thinking that they can get a usable 10V signal. This is the same principle exploited for differential signal transmission. If you subtract inverse signals the output is amplified, ( A - (-A) = 2A) if you subtract the same signal (i.e noise) the signals are canceled ( A - A = 0).

[@alextirendi5412]

To explain it easy: you don't have 10 volts anywhere, the -5V symbolises only the current flow in the other direction than +5V.

[@alextirendi5412]

Ok to make it clear for you, one will never measure a 10V differents at any given time on any place in the circuit. You measure +5V at one time and -5V at another time and never at the same time because then you would really have a 10V differents, as mentioned before it's just the current flow that changes direction no voltage multiplier anywhere but it is legit to say the waveform is 10Vpp.

[@alextirendi5412]

Actually that's totally wrong if you would rectifiy the 5VAC signal you would get 5VDC and not 10VDC! You can test that easy by yourself .

[@alextirendi5412]

That's not just a rectifier you added a voltage multiplier and then you claim the original circuit (2 inverters) act as a voltage multiplier. You can even add a booster or a transformer the 2 inverters still don't mulitiplie voltages...

[@fabimre]

pe1dnn then you get a voltage doubler!

[@fabimre]

Alex Tirendi A 5V AC squarewave is still 10 Vpp!

[@avilik]

We can get +1.5V and -1.5V with alternating leads of one AA cell. We need two cells to have positive and negative rail at the same time.

[@mduvigneaud]

Thumbs up! And I'll have to watch this again... I'm starting to get back into doing some hobby electronics design.

[@Daveyk021]

Do you own stock in Excederine headache pills? I now understand your explanation which goes against all my normal logical thinking and training. I really now want a differential scope probe but would not use it in my day to day work. It's 10:1 but wouldn't an amplified differential probe be more useful ?

[@moristo]

Peak To Peak.. in this case when A 5v the B is 0v (as a ground),, so vice versa when A 0v (as a ground) the B is 5v, so the voltage designation should be fixed on 5 volts. But... don't forget, capacitive properties is always exist between adjacent conductors, including inside all components, this is the supporter so that's can be happen (sorry my english is bad, I'm just using google translate)

[@moristo]

additionally, and do not forget, every conductor has inductance and resistance properties

[@unlost117]

Dave, I dont see your previous video linked in like you said it would be in the first few seconds of this vid. Plonk it here some where? Cheers

[@TheCrazyInventor]

You are not *actually* doubling your voltage, that is a bit misleading and I think this is what confuses people. Only your swing has gone to 10v p-p. Basically you're just flipping the polarity constantly. Imagine you flipping the positive and negative lead of your DMM which is hooked up to a 5V supply. It'll display +5v, -5V, +5V, etc... Your peak to peak voltage swing is now 10V, but you aren't actually getting 10V anywhere in your circuit.

[@Brant92M]

It's an AC voltage, which is measured in terms of peak to peak voltage.

[@PhoenixRevealed]

@Brant Martin - Actually peak to peak is just one way of measuring an AC voltage, and not even the usual way of doing it. The much more common way is just the swing from 0V. When we describe your home AC wiring as 110V or 220V, that is not the peak to peak measurement, it is the amount the live wire swings positive and negative from the neutral wire (actually, it's not even that since sine waves are usually described as RMS and the peak voltage swing is actually 1.414 times the nominal voltage). Dave should have never mentioned the peak to peak swing since it is irrelevant to how this method works and just confused lost of viewers into looking for 10V somewhere.

[@MarkTillotson]

But the AC voltage however you define it is doubled, definitely...

[@Zonkotron]

No it isn't. The most common reference for AC is RMS and that is exactly the same. Don't try to fool a power electronics guy, we do this shit for a living......no inverter circuit doubles voltage or power. It makes AC from DC and that means it doubles the swing. but not the actual steady state voltage.

[@PhoenixRevealed]

No Mark, it isn't. When you look at 10Vp-p on a scope you are forgetting that everything below the gnd line is just a polarity reversal of what's above the line. The actual voltage never exceeds 5V.

[@Brutikus32]

Would this be similar to measuring a voltage and getting 5V and then reversing the leads and getting a measurement of -5V?

[@gamerpaddy]

yup

[@iwbnwif]

You can try this with a friend ;) Stand on a straight line, 5 metres apart facing each other. Both walk forwards 5 metres (without hitting each other). Your friend was 5 metres in front of you, now they are 5 metres behind you, so it appears as if they've moved 10 metres.

[@PhoenixRevealed]

"so it appears as if they've moved 10 metres" True, and a great analogy, but the important point is that they were NEVER actually more than 5 metres away from you.

[@iwbnwif]

Exactly, good point - I should have included that in my comment :)

[@jasonwalker4610]

Hi dave. You could have explained it with a battery, To start you have a 55v battery attached to a house light bulb. If you could swap the positive and neg leads on the battery around 50-60 times a sec then you would have 110v AC at 50-60hz, not 55v, this would be typical house voltage, just it would be in sqare wave and not sign. My question is, would the bulb draw the same power as if it was attached to mains, less/more because of the square wave. I think it would be more because of the deference in voltage on avr is more then sign wave, but I could be wrong.

[@PhoenixRevealed]

You are on the right track, but your numbers are wrong. A 55V battery being reversed 60 times peer second would produce 55VAC, not 110VAC. 110VAC swings 110V positive, then 110V negative (well 155.5V positive and negative actually, since it is RMS). Oscilloscope sine wave displays are convenient but can confuse things because visually you are looking at the voltage going up and down, but actually that voltage is completely reversing every 180 degrees.

[@philevans4021]

Another way to mathematically explain this is because the result is "A - B", When A is 0V, B is 5V and vice versa, so one cycle its A-B = 0-5 = -5V, and the next its A-B = 5-0 = +5V, proving that it does go between -5V and +5V, of course giving the 10V potential difference.

[@PileOfEmptyTapes]

Maybe it'll become clearer when asking: How much AC can you get out of a +5V supply? Use a plain single-ended output, and you can get 5 Vpp tops (0 V min, 5 V max). With a coupling capacitor to take up the DC, that translates to + or -2.5 Vp. No more. Now for our BTL output (and that's exactly what it is). Case 1, output 1 = 5 V, output 2 = 0 V -- that's +5 V across the load. Case 2, output 1 = 0 V, output 2 = 5 V -- that's -5 V across the load. So now we've gone to + or -5 Vp, or 10 Vpp for that matter. _Ta-dah._ Incidentally, the rule of thumb for BTL amplifiers is "twice the power into twice the impedance", but anything that works on tiny bits of static electricity clearly is very high impedance and as such not much of a load to begin with.

[@DAVIDGREGORYKERR]

This is like a bridge circuit so you do get the double voltage.

[@PhoenixRevealed]

It is exactly an h-bridge, but it is not doubling the voltage, it is converting the LCD drive from 5VDC to 5VAC.

[@alexanderwikstrom1829]

I would have explained it differently. Simply by looking at the difference at the input to our inverter, comparing it to the output, for both stages, and also be clear to state that it is peak to peak, not max value over the component. As the confusion is more likely a mix up between peak to peak, and the max voltage present over the device. But otherwise it were a good explanation.