BJT Constant Current Driver

[@bobdurk5180]

Between your video and another one with similar theory, I was able to achieve the required results. I tested by varying the input voltage for the range I require, and all worked very stable. Thank you for the tutorial.

[@robertneill3057]

You could also use another transistor to control the current on the base of the first transistor. The second transistor has its base-emitter junction connected across RE and its collector connected to RB1/first load transistor's base in the place of RB2. The varying voltage drop across RE will control the first load transistor's base current by diverting the current away. This final circuit can have a tendency to cyclically oscillate however.

[@AdityaPrasad007]

2:41 explained (correct me if I am wrong) voltage drop is almost equal both sides - Re(Ie) = Rb2(Ib2) Re(beta)(Ib) = (Ib2)(Rb2) If we want Ib to be less compared to Ib2, say Ib/Ib2 < 0.1 then that is the same as Rb2/(beta)(Re) < 0.1 Or Rb2 < 0.1*Re*beta

[@klave8511]

You didn’t mention effect of temperature on the current controllers you show. The multiple diode version is going to be terribly temperature sensitive. The transistor is also temperature sensitive, the Vbe will change as the transistor warms up, thus the current will vary. The transistor is going to get hot when regulating from a high voltage or regulating a powerful led, Vbe drops and current goes up, transistor gets hotter etc. The emitter resistor reduces the effect if large enough.

[@kevinwolf1056]

Hi David, excellent refresher, thank you for the time you put into this video. Can you explain why you chose a V_e of 2V?

[@akshaykumar-lf3es]

same doubt here!

[@anirudhshandilya4032]

I think it was arbitrary and the solution made sense at Ve = 2 Volts.

[@gubelly]

I think it may be the value to drive the LED ie RED approx 2v@@anirudhshandilya4032

[@ceto150]

Transistor has to be in active mode. We will need at least 0.7v to turn on diode. If ve is 2v, that leaves 2.3v across the transistor. Rule of thumb, approximately half supply across vce.

[@sauravti]

One word for your video that is AWESOME

[@gc5643]

Thank you very much, until now I had no idea how to do this, thank you for sharing your knowledge, really appreciate it.

[@WistrelChianti]

Thanks this really helped, at least in terms of the maths and design. Still not sure I know why it's limiting the current though. I always thought current through the collector/emitter was a factor of current at the base... and why would that change because the load volage changed? Why did RB2 need to be less than/equal to your corrected equation etc?

[@ElectronXLab]

Your two questions are actually related. We wanted to pick an RB2 value so that the current flowing into the base is very small compared to what flows through RB2 - then RB1 and RB2 can be treated as having approximately the same current flowing through them. Then, since we're assuming the base current is small enough to ignore, we can use the approximation that VE is (VB-0.7) and then use VE/RE to calculate IE. We could have done a more complicated analysis and thevenized the VCC/RB1/RB2 part of the circuit then use that to calculate IB, and use beta to calculate IE. This would give a more accurate result, but is more work.

[@JacklapottTv]

Best explanation ever

[@polemos74]

Excuse me, but what is the difference with not using a transistor to supply constant current with predefined series resistor? When i change to another color LED the current changes.

[@redenmontesclaros3348]

How did you get Rb1=0.85

[@DannySebahar]

I somehow missed where you got the equation (Beta)(Emitter Resistance)(.1) especially the .1 because you didn't mention anything about that. Also, if R1 is 850 Ohm that makes the current through the diode 270mA as 5*(850/1850)=2.3 & 2.3V/850R=2.70mA so if the base current is multiplied by beta than the collector current would be 2.7mA*100=270mA which surely isn't what you intended was it? EDIT: I see now how I was mistaken. The current through RB1 contains both the Base current and the current through RB2. However I still don't see where you get that RB2=(beta)(Re)(.1). In your calculation with the fuor 1N4148 Diodes with A forward Voltage Drop of 0.7 Volts and 20.7V the Forward current is 5mA then wouldn't you need to have the resistor set such that the resistance is equal to the 2.2 Voltage drop over 4 x 5mA or 20mA so that Rb = 100 Ohms?

[@buffme98]

I think R2 is suppose to be 850 ohms and r1 is 1000

[@cyklop1977]

I calculate like this beta=100 , led red =2v and 0.02A ----- Re=3v/0.02A =150ohm , Rb2=3.7v/0.002A (1/10 beta) =1850ohm , Rb1=1.3v/0.0022A (1/10 beta+extra 1/100beta = 0.002A+0.0002)=590ohm .

[@kalamatagames2631]

if you use zener diode you can set voltage at any point you can find zener diode. but what sorce you will use battery or wall shocket

[@dieutran9066]

Done it. It works great but notice that Rb2 < Re*10

[@ElectronXLab]

Oops, that was a bit of an oversight - thanks for the comment.

[@artiomoable]

would it be not simpler to use only LED and series resistor in series to set a current as VCC is constant? LED voltage is also constant...

[@magnuswootton6181]

you can do this with a pure RC circuit as well. its a voltage divider with the capacitor on the "load" but the r2 resistor is where the load goes.

[@ElectronXLab]

I think that you need an active circuit to have a constant current driver; I'm not sure how you'd do it with just an RC circuit. With just RC, the capacitor will help hold a steady voltage for a short period of time...depending on its capacitance and the load.

[@arielcampos2008]

It helped me a lot, thank you.

[@jeeplvr2000]

I'm trying to come up with a reliable LED circuit for a bug zapper board I have made. I'm using three .5 watt LED's. I have a 110 v ac source and would like to use that to power the LED's so I don't have to use a buck converter to get dc for the LED's. I want everything to be on the board I have made and not have a buck converter to worry about. Do you have a circuit that would show me a simple and reliable way to do this? I assume that I will need a transformer to reduce the voltage but I cant find anything that will reduce the voltage and still fit on a small PCB.

[@codebeard]

What are the advantages of the diode version here?

[@kiranjotsingh4040]

constant drop i.e. 2.8V

[@HaiTran-wd6nu]

Dear Sir The Beta we could choose one from the HFE range of specified transistor? Thanks

[@theengineer9910]

Im getting the collector current i want but the multimeter voltages on the base and emitter do not match the calculations

[@EnfermeiroPensador]

What's about the efficiency?

[@arduinocowboy3960]

Hi, David. Like most videos out here on KZbin, you're only dealing with a single LED running a fv of 5v or less. What about a load of 40 LED's? 4 in series X 10 in parallel, powered by a car battery of between 12v and 14v? I've watched this video 5 times, but I can't seem to get my head around that much voltage and a CC of .8mA. Although it works fine on my bench power supply where I have a constant voltage, I can't get it to work on my car once the Alternator kicks in and the voltage goes up. How would you build a CC driver for my situation?

[@kennmossman8701]

There is a voltage drop across the CC, and with the Vf of White LEDs there is a limit of 3 in series

[@RevCorner]

When Vce changes (as shown around 7:30), shouldn't Ic (i.e I load) change too? Would be really helpful if someone could help me out. Thanks in advance.

[@krisea3807]

Can we use this in Power Supply Circuit as Power Indicator, just before regulator IC? My Bridge Rectifier gives 12v or 24v, current 5A.

[@kennmossman8701]

yes use the 2 BJT version

[@peacebewu]

When determining Re, shouldn't our eq be: (5V-2V)/Iload?

[@futuhcoklu1396]

No. Correct calculation should be: Re = (5V - VLed - Vce) / ILoad. VLed takes different values according to the LED type.

[@darshanm.b.7005]

If I have a voltage of 5 V at Vcc, but the operating voltage of laser is 2.2 V, will it damage the laser?? It would be really helpful if you can clear my doubt. Thanks

[@kennmossman8701]

No problem as long as the current is under the LEDs max current

[@Peter_A1466]

Will the current stay constant if Vcc changes?

[@futuhcoklu1396]

If you use resistive voltage divider to supply base current, can not keep the current fix when Vcc changed. But, if use Zener diode, you can keep it fix.

[@kennmossman8701]

@@futuhcoklu1396 are the changes significant

[@kalamatagames2631]

very nice video thanks you give me idea for control current with potesiometer without change voltage 😁

[@reofpv6665]

How many 200 watt led drivers am I able to run off one 15 amp circuit

[@abbeed1]

thanks

[@shazmiah]

Hi David can you please tell me how you arrived Rb1= 0.85*Rb2 at 4:35 , and also at 7:22, why you think that at saturation why we won't get the 20mA?

[@ElectronXLab]

At 3:33, I showed that VB needs to be 2.7V, then at 4:15, showed that the ratio of VRB2/VRB1 is 1.17. VRB2/VRB1 = RB2/RB1 = 1.17, so with a little algebra we can see RB1 = 0.85RB2.

[@ElectronXLab]

AT 6:58 in the case where the forward voltage of the LED is 3.3V, the voltage at the emitter cannot be 2V, because 3.3V + 2V = 5.3V which is more than VDD

[@shazmiah]

@@ElectronXLab : I'm trying to learn here. You mentioned that led 3.3v will saturate the transistor. I agree. My understanding of saturation is that the transistor won't pass any more current via the collector. The maximum current will be that what is set by Re(100 ohms) resistor, which is 20mA? If I am wrong would you clarify further? Thanks.

[@ElectronXLab]

@@shazmiah let's approach this a different way. In the last example, I said that the LED turned on at 3.3V and that it needed 20mA to reach that voltage. If 20mA goes through the 100 ohm resistor, the voltage across it will be 2V. The power supply is at 5V, so that means after you take off the 2V across the 100 ohm resistor, you only have 3V left. If you have 3V across the LED, then it will not have 20mA through it, so the actual current must be something less than 20mA.

[@shazmiah]

@@ElectronXLab : Thanks for your explanation and quick reply.

[@michaeld9682]

Why right Vcc and Vdd?

[@joliynill2430]

Sir is it Linear system?

[@kennmossman8701]

yesssssssssssssssssssssssssssssssssssss

[@sham1234567891]

Excellent stuff David. I need your immediate help with the following. I must wire a 3Watt (700mA) Luxeon LED to a car battery. Could you please help me design a simple driver for same? thanks & Regards. Sham.

[@kalamatagames2631]

shamrai 4.2v 700mA 3Watt 3Ω that is you LED

[@leblancmeneses3650]

Can you help me design a 24 volt 7 amp constant power source to use as a charger. I have the I have the 24 vac 10 amp transformer and 10amp bridge rectifier but I dont know how to maintain constant 7 amp delivery.

[@MegaInformazione]

Buy a LM317T

[@tianapyre2837]

It won't provide constant current.

[@kennmossman8701]

@@tianapyre2837 Yes it will. IF used in CC configuration. Read the DATA SHEET

[@Enigma758]

Problem is that you need a big wattage low value resistor and it dissipates a bunch of heat.

[@ВоваТеремок]

На русском как будет?

[@ElectronXLab]

Извини. Я не знаю

[@trongod2000]

I have to admit, I'm totally baffled by this explanation. Current is the flow of electrons right? Electrons have a negative charge, right? Doesn't that mean that your electromotive source of Vcc can be explained as a lack of electrons rather than a excess of them? How can you say that current (electrons) are going to flow from a region of few electrons (+) to a region of excess electrons (-). The way your talking about this circuit feels like your confusing voltage and current directions. Another way of explaining what I'm saying is that current flows against the arrows in diodes and transistor gates. If that is not the case, why do they call the leg of the transistor opposite the leg with the arrow on it, the Collector. Doesn't it get that name from it's job of Collecting electrons emitted by the opposite leg?

[@the1aboveall483]

Get your basics right. *Current flows in the opposite direction of electrons*

[@trongod2000]

@@the1aboveall483 You said it, "CURRENT FLOWS". What is flowing? water? air? heat? How about Current is the flow of ELECTRONS. Unless ELECTRONS are swimming like salmon and going against the "flow" so as to go upstream. And if you use that analogy then current "flow" is water? and electrons are salmon. I'll accept that if you can explain what is "flowing" in electric current if it's not electrons. kzbin.info/www/bejne/b3nPdWdmiLGqqJI Here is a basic discription of how a battery works. Weird that it states that the energy in the battery goes from the - pole around and back into the + pole of the battery. The battery is an energy storage device. It has a chemical reaction taking place inside it that wants to push electrons OUT the - pole while the + pole has a hole trying to attract an electron back in to take that one's place. I'll help you out a bit. Electromotive force (volts) exist in the battery and in any circuit hooked up to that battery. Electrons moved from the negative pole to the positive pole until there were no electrons to drive the chemical action in the battery of sufficient quantity to move current through the circuit attached to the battery. So whatever current is, it's not moving in the opposite direction of the electrons flowing in a circuit. This stuff was a whole lot easier in the old days. Everyone could see the filaments of an electron tube glow a pretty bright orange and it made perfect sense that the glow came from electrons moving through the filament heated it up. Gee wonder what heats up light bulbs and makes them glow? I bet it's electrons. Electrons that are being PUSHED by VOLTS NOT PULLED by VOLTS. There in lies the explanation . Electrons passing through a resistance generates heat. Proof that there are actually electrons flowing. In regards to the video above.. current flows up through the transistor, NOT down through it.

[@An-wd9kk]

@@trongod2000 In short. You're right. Electrons flow in the opposite directions. But scientists in the old days thought otherwise and it's not until lately that we discover current flows in the electron flow's direction. The problem is everybody is used to the old way (AKA CONVENTIONAL CURRENT) that nobody bothers to change. So calculation and explainantions all use the opposite direction of electron flow.

[@trongod2000]

@@An-wd9kk One way to show people how current flow "works" is to show them how current flowing through a conductor can pass energy to do work in something completely detached from the wire. Show them an electro-magnet for instance. Perhaps seeing a solenoid in action will help also. Both these examples however do not show the "direction" or "flow" of the energy within the conductor. At least not usually. It might open their eyes to the idea that there is "direction" if you reverse the leads going to a solenoid and see that the piston portion of the solenoid is pushed in the opposite direction. That alone though does not visually show which way electrons are flowing and even less enlightening on the subject of current flow. Current is a very hard concept to show visually and direction of current flow is even harder. One of the best ways to show the effects of current flow (electrons flow) is to take an uncharged capacitor and measure the voltage on it. Since it not charged there won't be an excess of electrons on either of it's plates. Hook the capacitor up to a source of electrons and then disconnect it and see what happened to the capacitor. Surprisingly, almost nothing. In order for electrons to build up on a plate the opposite plate needs to shed electrons. This adding and shedding electrons is pretty instructive in determining direction of flow.

[@bimbo-yw6ny]

Beta is too variable, unreliable as a fixed value

[@kennmossman8701]

think again