Differential Amplifiers, Part 2

[@DuffyDuck117]

Absolute legend! Very well explained. This (and rest of series) helps a lot for building an early intuition for what the datasheet is trying to convey. You've helped an Aussie guy out a lot here.

[@ElectronicswithProfessorFiore]

Glad to be of service!

[@simonyoungglostog]

That makes sense this morning. The twisted pair/phase-splitter is similar to a dark-frame subtraction on camera sensors.

[@rudygomez6996]

Thank you!

[@brewman467]

Hello Mr. Fiore, very well explained, thx. At 19:06 you say that the signal with gain of one is 1mV. But before you said that sin-(-sin) =2sin. Shouldn't in that case, even with a gain of 1, output be 2mV?

[@ElectronicswithProfessorFiore]

The "output signal" depends on what you use for the output. Most times, the output is "single-ended", meaning that you take the signal from one of the collectors. You can, however, use a differential output, meaning from one collector to the other. The same is true for the input (you can use a single-ended input or differential). Thus, diff-in/diff-out gain is twice the diff-in/single-out (or single-in/diff-out) gain, and four times the single-in/single-out gain. The bit about sin-(-sin) = 2 sin is where the above comes from, and also shows (when considering an inverted vs. not inverted signal), that signals can add or completely cancel. Of course, that's the theoretical ideal. In reality, cancellation is never perfect. I will be exploring this more in some upcoming videos that will focus on CMRR (common mode rejection ratio).

[@chrisw.3077]

Did you end up making a 3rd part to this video? Turns out this old circuit I'm studying is a bridged differential amplifier. The outputs are driven in opposite phase by the four output transistors.

[@ElectronicswithProfessorFiore]

? Any standard diff amp can have a differential output (i.e., outputs of opposite phase); you just use the output of both collectors instead of just one. That's one way to make a phase splitter.

[@simonyoungglostog]

Would there be any benefits to adding swamping resistors to the emitters? I was wondering if they would help mitigate differences in Beta and increase stability? (You answered that later in the video.

[@Pooja-zz9tu]

Hi, Professor Fiore amazing explanation !!! .......just have a small question if the signal has a noise before entering the phase splitter will it be amplified as well.

[@ElectronicswithProfessorFiore]

Yep. A diff amp only gets rid of common mode noise, that is, noise that has entered the system after the phase splitter (i.e., in the lines). Any noise that exists prior to the phase splitter will be presented as a differential signal by the phase splitter, and thus, amplified by the diff amp.

[@richwiskin4860]

Good vid. I would maybe explain a bit more about Re and I think you also wrote RE. Also 2k for the base resistors is not really practical in the real world.

[@ElectronicswithProfessorFiore]

The circuits are generally designed to show how things work, not necessarily be production models. FWIW, many diff amps don't even need base resistors as the base bias current comes from the surrounding circuitry (i.e., assuming DC return), but that would just confuse the explanation above.

[@rudygomez6996]

At 8:00 im confused of how output is in phase. The direction of arrow of current would suggest negative voltage/anti-phase on both collectors, wouldnt it?

[@ElectronicswithProfessorFiore]

I can see how that would be confusing. The current sources are pointing in the direction expected for the transistor model (i.e., the total current which reverse biases the CB junction and establishes the value of r'e). The AC current rides on top of that. The AC current on the right/lower side is flowing left to right for a positive input, producing an in-phase signal at that collector. A "tell" that these are not the instantaneous directions for the AC current is that the sources would be feeding into each other. The reason I didn't draw the arrow the other way is that some students would then ask why the second transistor model appears backwards to what they're used to seeing. It's sort of a no win situation for me as I see problems either way I draw it. I recommend that you ignore that detail and concentrate on the following: For the lower right output, the first stage acts as a follower which is non inverting. The second stage is a common base configuration which is also non inverting. The end result is an output which is in phase with the original input. The upper output, in contrast, appears as a standard inverting common emitter configuration. Hence the two outputs are in anti-phase.

[@rudygomez6996]

@@ElectronicswithProfessorFiore i think i mighta sorta actually understand lol at least the gist of what you said. thank you!

[@robertthompson5908]

Very nice explanation.

[@simonyoungglostog]

At 6:10, does the second transistor's Rb get ignored because the second power supply is grounded? I'm a bit lost at 7:40. I can see the signal coming in and through the first r'e, then through the 2nd r'e (2 x r'e in series), I don't see why that would reduce the signal by 2. Guidance, much appreciated. I've paused at 12:00, my poor brain! :)

[@tepiaxeria6971]

Regarding the second transistor's Rb: yes, I think you have to short Vin2 to ground, as per superposition theorem.

[@te9781]

What I'm trying to understand is does the output signal of diff amp has a DC component ? I really understand the dual supply bias that Q point is in middle of BJT curve so we need a dc blocking capacitor at the output.. but when I read about opamps I find that dual supply opams produce the ac signal without DC component that consume me alot can you please answer me ?

[@te9781]

Confuse* me

[@ElectronicswithProfessorFiore]

That depends on the how the diff amp (or op amp for that matter) is designed. With a bipolar supply, it is possible to set the collector voltage at 0VDC. Further, if you use a differential output (i.e., from the first collector to the second), the DC values will cancel leaving you with no DC (this is true up to the level of matching between the two sides). Op amps often have DC level shifting stages which produce the 0VDC output you mentioned, but this is not true of all op amps. Some are designed to work off of single polarity DC supplies and will have a DC offset, typically equal to half of the total DC supply. For details on how the whole 0VDC output thing is achieved, I suggest you take a look at the BJT biasing section of my Semiconductor Devices text and associated videos (also a free OER title, just like the op amp text). Side note, it is not required that the Q point be centered on the DC (or even the AC) load line in order to create 0VDC at the output. A Q point centered on the AC load line is important if you want to maximize the output compliance (and thus the maximum unclipped power for a given load).

[@te9781]

@@ElectronicswithProfessorFiore Thank you 🙏

[@LL-ue3ek]

the equivalent resistive circuit is a bit confusing。

[@ElectronicswithProfessorFiore]

There is a bit more detail in the text. Perhaps that will help clarify. It's a free download, see the links in the video's description.

[@vassilpopov2559]

Could you please write down the equations on how the second transistor impedance is just r`e (@5:15)? Thank you!

[@ElectronicswithProfessorFiore]

Z is just v/i. The voltage at the emitter is Ve and the entering current is Ie (AC values, of course). The ratio of Ve to Ie is r'e. This assumes that all of Ve drops across that resistance. In reality, there is also a small drop across the associated base resistor, but the base current is much, much smaller than the emitter current, and the base biasing resistor is effectively in parallel with the internal impedance of the second source which is quite small, typically. Those two factors (small base current and small effective base resistance) combine to make a very small voltage drop across the base resistor, meaning that the original assumption is valid for the typical case. The computation is essentially the same as finding the output impedance of an emitter follower (voltage follower). See p 177 of the Semiconductor Devices text for that, or the section on common base amplifiers.

[@incxxxx]

I understand how OpAmp amplifies AC but how it also amplifies DC voltage, its mystery. The same scheme that works for AC and DC. What is the key which forces to work the both way?

[@ElectronicswithProfessorFiore]

The key is that the circuit contains no lead networks (i.e., coupling capacitors). It's directly coupled, and thus, has no lower frequency limit.

[@incxxxx]

@@ElectronicswithProfessorFiore But that scheme you have showed can not amplify DC voltage. So, there is trigger that triggers the real OpAmp scheme to the correct one?

[@ElectronicswithProfessorFiore]

@@incxxxx No, a direct coupled amplifier does amplify DC (that's DC as a signal, of course).

[@JojoLearning-t5r]

Hi all. Can anyone explain to me why is r'e equals 26mV/ Ie = 26 Ohms. I don't understand where is 26mV from.

[@ElectronicswithProfessorFiore]

You will find the derivation of that equation in my Semiconductor Devices textbook, free to download (see video description for details). I might also suggest that you go through the Semi Dev playlist from the beginning. Everything builds on what came before.

[@torikenyon]

Huh?