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Question is based on this document.

Quoting from there:

Voltage Follower Biasing: This method is exactly the same as the voltage divider biasing, except it uses an op-amp (or transistor) to buffer the bias voltage, so choosing small resistor values is no longer necessary.

The schematics for op-amp voltage follower for biasing is provided:
enter image description here

Then we feed the bias voltage instead of ground in negative feedback loop of non-inverting amplifier:
enter image description here

But there is no schematic in the document for voltage follower using transistor.

I've tried to find circuit on internet, but it seems most of the time it is different type of voltage follower, e.g. like this one:
enter image description here

From my understanding this is not going to work for biasing non-inverting op-amp, since we take voltage from resistor i.e. we introduce it to negative feedback loop, so we can not use large value resistors, basically making this circuit senseless - we can't save power on it, it will be almost the same as just using voltage divider(split resistor biasing).

The I tried to design my own circuit to mimic op-amp voltage follower, I came up with something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The Vbias as with op-amp circuit goes to negative loop instead ground.
But this does not work, the behavior is somewhat random, signal jumps around, etc. The logic I was using is that we need to have capacitor on the output to prevent gain from changing(because we don't introduce resistors to the loop). I tried different caps and resistors values, but nothing changes, so basically my circuit is wrong fundamentally.

So the question is - how to build a Voltage Follower for biasing the negative feedback loop of op-amp? Or better said - how to do what the first circuit in this post does, but with MOSFET? Mimicking means: 1) Possibility to use large value resistors to decrease power consumption 2) Do not change the gain of non-inverting amplifier

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  • \$\begingroup\$ 1. In your first diagram, the capacitive load on the op-amp is likely to degrade the circuit's stability or lead to oscillation. 2. Your third diagram is not a follower, but a common-source amplifier, which is an inverting configuration. \$\endgroup\$ – The Photon Aug 30 '16 at 22:27
  • \$\begingroup\$ 3. Your fourth diagram drives the amplifier input outside the power supply rail, so it is not likely to work very well. \$\endgroup\$ – The Photon Aug 30 '16 at 22:30
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Your third image is not a (source) follower, rather it is a common-source amplifier and generally is configured to provide gain > 1.

A source follower looks like this:

enter image description here

Note that there is no drain resistor at the top and that the output is taken from the source terminal rather than the drain.

\$V_{out}\$ will follow \$V_{in}\$ with a drop of roughly the gate threshold voltage of the MOSFET, \$V_{th}\$. This may be somewhere around 5V depending on the device, so one might reasonable choose a BJT emitter follower instead:

enter image description here

You can see the circuit is essentially identical excepting the transistor type. \$V_{out}\$ will follow \$V_{in}\$ with a drop of \$V_{BE}\$, typically only 0.7 V.

The input impedance of an emitter follower is relatively high, and it's output impedance is relatively low. So placed between a resistive voltage divider and the rest of the circuit (such as an amplifier input) has the effect of stabilizing the bias voltage developed across the divider against variation due to changes in current drawn from the divider.

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  • \$\begingroup\$ But the resistor Re will now be basically in the negative feedback loop, so we can't make it large value in order to save power. Or do I understand something incorrectly? I mean in the first circuit with op amp follower I have 0% influence on the gain of amplifier because I only have cap at the output. \$\endgroup\$ – ScienceSamovar Aug 30 '16 at 23:03
  • \$\begingroup\$ The output impedance of the emitter follower is going to be quite low, on the order of tens of ohms, depending on the rest of the circuit. So in series with your R1 it will not change current in the feedback much by resistance contribution. However, current will flow through the BJT, so depending on your choice for \$R_E\$, there could be additional power consumption. You'd need to analyze and simulate to get an idea and strike a balance. But consider that the emitter follower is active, and should be thought of as its Thevenin equivalent, not just adding \$R_E\$ in series with \$R_1\$. \$\endgroup\$ – scanny Aug 30 '16 at 23:24
  • \$\begingroup\$ but in the end of the day with this circuit I am forced to use quite small resistance Re, if I plug lets say 220k as Re the gain will be affected(even though it is active emitter follower). And if I use small value resistor Re - I do not save power, so I might as well use simple voltage divider or op-amp buffer, transistor follower(this one) is least favorable way. I will check calculations with this follower to get precise data, but at a glance there is not much sense to use it here(this particular voltage follower) \$\endgroup\$ – ScienceSamovar Aug 30 '16 at 23:40
  • \$\begingroup\$ You'll have to be the one to decide which circuit configuration to go with. However, keep in mind that an op amp will also draw current and therefore consume power. Also, note that like The Photon mentioned, a capacitor on the op amp output is going to degrade its stability and perhaps cause it to oscillate. So such a circuit may require more components than you've shown so far. \$\endgroup\$ – scanny Aug 30 '16 at 23:51
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    \$\begingroup\$ @ScienceSamovar, the emitter of the BJT is effectively in parallel with Re, so you can use high Re and still have a low output impedance from this design. 220 kohms might be too high, but you could use maybe 10 kohms and still have an output impedance below 5 ohms. With the divider solution, you'd need the two resistors in the same order of magnitude (~10 ohms each for a divide-by 2) to achieve 5 ohms output impedance. \$\endgroup\$ – The Photon Aug 31 '16 at 2:17
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The above answer is completely correct... but if the ~0.7V drop is nagging at you just a bit, you can fix that:

schematic

simulate this circuit – Schematic created using CircuitLab

This works by essentially doing another voltage follower after the first one, but with the opposite "type". In this way the voltage is cancelled out. Just be sure to use complementary PNP and NPN transistors with the same Vbe.

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