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I read that a unity gain buffer can be implemented with an op-amp like in this circuit:From UC Berkeley EE16A section 19.7 Note 19

I also read that a long-tailed pair as an op-amp (this might be the wrong part). So according to the buffer circuit above, I believe the circuit using long-tailed pair will be like this (edited: this is my own idea only of how the long-tailed pair can be turned into a unity gain buffer which I hope to be confirmed or corrected):

enter image description here

However, the problem is that this configuration implies that the non-inverting input terminal will always remains at ground (0V) while the inverting input will be at G*Vin (G: gain). Hence I can't understand how negative feedback is implemented in this case since the Vout is supposed to be G(Vin - Vout). Please correct me if there are any mistakes and demonstrate how long-tailed pair can be used as unity gain buffer (and explained with circuit schematic will be very appreciated).

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  • \$\begingroup\$ The connections in your circuit are wrong. Please read en.wikipedia.org/wiki/Differential_amplifier#Long-tailed_pair. \$\endgroup\$
    – CL.
    Jun 17 at 11:32
  • \$\begingroup\$ The second picture (and the name "long-tailed pair") are probably from a very old book. Opamps use active load and multiple gain stages for higher gain. If you have G*Vin gain, than even a small signal will achieve a big feedback signal, which counteract the change. If the non-inverting terminal voltage rises, the inverting terminal follows it through the output of the amp. Vout = G*(Vin-Vout) ==> Vout = Vin *G/(G+1). It seems correct to me. \$\endgroup\$ Jun 17 at 12:39
  • \$\begingroup\$ The second picture is indeed strange. This circuit has a differential output, while the first picture has a single ended one. Q2 is also shorted and thus it is not functional. \$\endgroup\$ Jun 17 at 12:42
  • \$\begingroup\$ @CL. Can you point out how is it wrong because I actually took the long-tailed pair image from the wikipedia link and added what I think is necessary to turn it into a unity gain buffer \$\endgroup\$
    – DivineMK
    Jun 17 at 13:16
  • \$\begingroup\$ @HorrorVacui do you mean that the long-tailed pair is not equivalent to an op-amp? about the second circuit it is actually single-ended like the first (only the non-inverting input left). Q2 is actually not shorted but the 2 terminals of Vout are connected to the two terminal of Vin- (inverting input terminal) \$\endgroup\$
    – DivineMK
    Jun 17 at 13:21

1 Answer 1

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The long tailed pair is a pretty rubbish differential amplifier, but what you propose should work. Your problem is you've wired it up wrong. Why did you short Q2's collector to emitter?

Here's my attempt:

schematic

simulate this circuit – Schematic created using CircuitLab

All I did was connect the output to the inverting input, just like a voltage follower. It does what you would expect, except with terribly high output impedance, and awful linearity, due to low open loop gain. Here's the output, with a sinusoidal input:

enter image description here

To be honest, I was surprised how well it worked.

I wasn't sure it was truly working as a voltage follower (maybe this is just a really complex emitter follower, with a saturated Q2, or something), so I plugged in a couple of resistors, for 50% negative feedback, to get a gain of two:

schematic

simulate this circuit

Here are the input and output, in a simulation:

enter image description here

Again, surprisingly good, considering it's so primitive. I love the hilariously large input offset voltage, so clearly visible here, but well done little long-tailed pair, good job.

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    \$\begingroup\$ @DivineMK Ah, homework. Well, in for a penny in for a pound. Your LTP circuit is correct, so refer to that. Ignore the Vout(-) node, because a typical op-amp only has only one, single ended, output. It's the one that goes up when the non-inverting input goes up, and that's your overall output. In the op-amp follower circuit, input is to the non-inverting input, so that's where I applied the sinusoidal test signal. Lastly, the output is fed back to the inverting input. That's all I did, connect those two nodes together, just like in the op-amp version. Look closer, you'll see it. \$\endgroup\$ Jun 17 at 14:45
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    \$\begingroup\$ @DivineMK Vout(-) can be ignored, because it's just an inverted version of Vout(+). When Vout(+) rises, Vout(-) falls by the same (or similar amount). You can still say that Vout(+) is equal to some large gain multiplied by the difference between Vin(+) and Vin(-). You could also say that Vout(-) is equal to some large negative gain multiplied by that same difference. \$\endgroup\$ Jun 17 at 14:52
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    \$\begingroup\$ @DivineMK In a nutshell, the emitter resistor is a current sink, shared by both branches. Any changes to current in one branch (via that resistor) must necessarily result in a change in the other, to keep Mr. Kirchhoff happy (I'm referring to Kirchhoff's Current Law). That's why changes in Vin on the left branch can influence Vout on the right. That's such a hand-wavy explanation, I feel dirty. \$\endgroup\$ Jun 17 at 15:50
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    \$\begingroup\$ @DivineMK Ideally, yes, which is why you almost always see a constant current sink there in real-life opamps. \$\endgroup\$ Jun 17 at 16:37
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    \$\begingroup\$ I think the "input offset" is more of an increase in output impedance at higher voltages. It's visible in the unity gain case (mainly because you run that one up to 10V), but becomes more pronounced when you load the output. \$\endgroup\$ Jun 17 at 20:30

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