Differential Amplifiers, Part 1

[@joeb.fromsydneyaustralia5313]

After seaching and searching... finally GREAT to find a GREAT video which is boith clear AND Concise. Thanks Professor... you now have 'at least' one student in Sydney Australia.

[@ElectronicswithProfessorFiore]

The three Cs of a good explanation: clear, concise, ... and correct! Glad you found it useful.

[@legendgamingamusement6146]

Can you show us how to compute and design an amplifier with 70 vrail using mpsa 92 on differential stage c2073 vas c5198 a1941 driver thanks ❤

[@xila8861]

This was really clear and straightforward!

[@simonyoungglostog]

Excellent and I understood all of it (so far). :)

[@supremeleader5516]

Thanks sir Love from India

[@SMV1972]

Большое спасибо. Отличный урок

[@olvyshh147]

Key to understanding: Remark : Transistors are connected according to the common emitter circuit, in the OE circuit, the voltage at the emitter, roughly speaking, is always equal to the base voltage. Since the potential at the point of connection of the emitters MUST be the same for both transistors, someone must give in .... This small voltage difference falls directly on the PN junction of another transistor, closing it, that is, roughly speaking, if we assume that the transistor is open when the voltage across the BE is 0.7V, then with a voltage difference of 0.1V, we have a voltage drop on the other transistor on the BE 0.6V is not a completely open transition !!! This is equivalent to the fact that we stupidly applied 0.6V to the base. What will be the base current at this voltage?? Look at the CVC of the diode, the current will be almost absent, so the author talks about Ebers-Moll's formula. You can see the CVC of the diode, what current the LED will give out if its PN junction is not fully open, that is, if it is open at 2.7V, apply 2.6V to its legs and you will see that the current almost does not flow. So here, increasing the difference Like this comment if it was useful for you so that others can see it.

[@ElectronicswithProfessorFiore]

Why would you "assume that the transistor is open" when Vbe is 0.7V? Under typical conditions for silicon devices, that would imply that the transistor is conducting, not open (non-conducting). Second, are you referring to the Ebers-Moll model when you mention "Eberton Moll's formula"?

[@olvyshh147]

The PN junction conducts current below a voltage drop of 0.7V. In the general case, when the voltage drops below this value, it resists the current in a non-linear relationship. Above a value of 0.7V, the PN junction almost does not show any current resistance, then it can be considered as open.

[@ElectronicswithProfessorFiore]

@@olvyshh147 No. That is fundamentally incorrect. If a device does not show any resistance to current flow, then it cannot be considered an open. It is the exact opposite, namely a short. I suggest that you may wish to curtail any further exploration of op amps or other semiconductors and instead go to the fundamentals of DC circuit analysis. Unless the foundation is solid, the building will collapse.

[@olvyshh147]

@@ElectronicswithProfessorFiore True, if the device does not show any resistance, it is considered to be short-circuited. Perhaps you did not understand me, but I wrote that a further increase in voltage of 0.7V does not have any effect on the current. Most likely you did not understand my comment above at all, but only cling to words.

[@ElectronicswithProfessorFiore]

@@olvyshh147 The first statement directly contradicts the final sentence in your prior comment. A PN junction cannot be both open and short at the same time because those are opposite conditions. The second statement is not true in the real world, but is just an approximation using a simple model. Where did you first "learn" about electrical theory? You appear to have considerable confusion on the topic. I will suggest again that you go back to the fundamentals. I remain happy to help. For now, I suggest not placing your "explanations" in the comments. They run the risk of confusing others who are trying to learn the material.

[@Stelios.Posantzis]

Neat, tidy, concise! What more could one ask for from a basic explanation of diff pair?

[@ElectronicswithProfessorFiore]

Good question! Free cantaloupe, perhaps?

[@Stelios.Posantzis]

@@ElectronicswithProfessorFiore They're in season and quite distracting.

[@ORIofficialchannel]

so understandable

[@RixtronixLAB]

Nice video shot, keep it up, thank you for sharing :)

[@MsFireboy2]

Thanks for posting. One thought came to me if you were to put a potentiometer between both Emitters of the BJT’s you could balance out the current between the two 5k resistors. Your thoughts?

[@ElectronicswithProfessorFiore]

On some amplifiers there are pins that allow you to connect a pot (and perhaps a resistor or two) which can be used to null out any DC offset. This is done by bridging the two sides of the diff amp, so you're right in the ballpark! Theoretically, this could be done in either the emitter or collector section (source/drain for FET). The whole idea is to steer a little bit of current from the side that has too much to the side that's lacking. The need for this has gone down considerably over the years as the DC precision of op amps has increased. Many modern op amps do not even have the option any more.

[@treeski3410]

Please forgive my simple question but can you clarify what you mean by "current source" (I-tail, 1:51). I look at a circuit and think of the power supply as the current source. Why is a resistor a current source? Are Rc1 and Rc2 (as well as Rb1 and Rb2) also current sources? Also, with the I-t resistor being the "source" of current, it seems strange to me that the arrows you add representing flow of current all point to it. I assume your arrows represent the flow of conventional current (holes) so wouldn't the current source be +Vcc? Then I'm not sure why at 7:20 you indicate the current flowing into the NPN base. Thanks for any feedback.

[@ElectronicswithProfessorFiore]

First off, yes, I use conventional current flow. Thus, current flows into the base of an NPN transistor. Further, current flows into the collector, and that combination flows out of the emitter. Ie=Ic+Ib. Regarding "current sources", you seem to be confusing that with "sources of current". Yes, Vcc or Vee are sources of current, but they're modeled as voltage sources, not current sources. Going back to basics, an ideal voltage source always produces the same voltage, regardless of what is connected to it. Similarly, an ideal current source always produces the same current regardless of what is connected to it. This implies that the internal resistance of a voltage source is very small and that the internal resistance of a current source is very large. To your question, the COMBINATION of the tail resistor and the voltage supply behave like a current source. That is, it draws a specific current (Itail), and if the two sides of the circuit are identical, that means that each of the emitter currents must be half of Itail. So, no, by itself, a resistor is not a current source, but when it is combined with a voltage source (using appropriate values), the pair can be modeled as a current source.

[@0451Deus]

Do you have a textbook you would recommend for learning some of these general concepts about op-amps? I have the book from Sergio Franco and I'm not that impressed by it. So many steps are skipped and the explained examples are too generic to be of use when solving the chapter problems. I have to say though, your videos are the biggest help for learning this stuff. Thanks Professor Fiore, wish I had you as my professor for advanced analog devices!

[@ElectronicswithProfessorFiore]

Well....at the risk of being immodest, have you considered my op amp book? You can download a PDF for free or get a soft cover print copy from Amazon for about $21. Go to www.mvcc.edu/jfiore and follow the link to the "OER free books page". You'll find all of my books there (5 texts and 7 lab manuals) plus links to other useful OER materials. (direct links are also in the description of the video)

[@vespergremlin3367]

Hello Prof Fiore , about that Vc you calculated to be 10 Volts , the voltage drop across both transistors being small , isn't that 10 Volts Vc also 'seen' by the tail resistor ? I am sure there is something i am missing

[@vespergremlin3367]

I scrolled down and found your answer to Ahmad Shamim's question I need to go on a back-to-basics trip For the video and your answers in the comment section , I thank thee .

[@ORIofficialchannel]

thank you very much!

[@simonyoungglostog]

Me again... If we set the tail current to, say, 1mA using a resistor, will the circuit perform the same as if we used a transistor as a current source (sink) of 1mA? Can we substitute transistor current sources for resistors? I think I'm missing some knowledge here, Perhaps you could point me at a place in one of your texts? Thanks for all you do.

[@ElectronicswithProfessorFiore]

Using a transistor current source will provide increased performance in areas such as common mode rejection ratio (there's a video on that). The basic tail resistor form is a good place to start as it is simple and relatively easy to understand, but it does not provide the best performance.

[@simonyoungglostog]

@@ElectronicswithProfessorFiore As an example... Could we replace the Emitter resistor on a CE amplifier stage with a current source resulting in the same current?

[@ElectronicswithProfessorFiore]

@@simonyoungglostog Yep.

[@simonyoungglostog]

@@ElectronicswithProfessorFiore Excellent. I must be doing something wrong in my simulator. I'm trying different versions of amplifier circuits, Thank you!

[@pankajgope7175]

Thank you professor

[@incxxxx]

Which one on the general schematic of OpAmp is "+" and which one is " - " ??? Are they rechangable?

[@ElectronicswithProfessorFiore]

A diff amp is not an op amp. Typically, it is the first stage of an op amp. Consequently, the input that is in phase with the output will depend on the following stages and whether or not they invert. base->collector inverts, base->emitter doesn't.

[@incxxxx]

@@ElectronicswithProfessorFiore Thanks.

[@rezatayebi4731]

Very good

[@khianelouiseruego1421]

Is this the same process with just one Rc in the circuit?

[@ElectronicswithProfessorFiore]

Pretty much. Obviously, the collector voltages don't match but there is no need for that if you're using a single-ended output (i.e., just one collector).

[@ahmadshamim3289]

I understand that Vb is 0 volts and we have a typical 0.7 volt drop from base to emitter and therefore a 20volt drop from Ve to V- . But, How do we go from Vc of 10 volts to Ve of -0.7 volts? When the transistors are on (and therefore I1 + I2 can be = It), why isn't there just a nominal (say 0.2 volt) drop across the collector emitter junction(s)? A full "accounting" of voltage drops from V+ of 15 to V- of -20.7 would really be helpful. Thanks.

[@ahmadshamim3289]

Please ignore the strikethrough.

[@ElectronicswithProfessorFiore]

You would only get a few tenths of volt across CE if the transistor was in saturation. It's not, it's operating in the linear region. Basically, whatever isn't taken up by the collector and emitter resistors drops across CE.

[@ElectronicswithProfessorFiore]

I should add that more videos will be coming in the next couple of months that will discuss operational theory that precedes this video. That should help. They will be found in the Semiconductor Devices playlist. In the meantime, I suggest you download the free PDF of my Semiconductor Devices text.

[@ahmadshamim3289]

@@ElectronicswithProfessorFiore Got it, thank you very much. As a beginner, I have been doing self study on electronics over the past several months, and had forgotten about transistors operating in the linear region and the resulting drop across ce. I have been looking for a curated series on op amps, and thanks for the videos and all the texts. Really great!

[@saronna_7]

can i communication with u on Telegram?