Does this circuit problem have a solution without an Op-Amp?

Question

I was going through the definition of a voltage follower in this site http://www.learningaboutelectronics.com/Articles/Voltage-follower

And a basic problem was presented, and an op-amp voltage follower as the solution.

Can this problem be solved without an op-amp ?

Answer 1

Oh, geez. I see a diode/BJT solution and a MOSFET solution.

No one did the obvious BJT-only solutions.

So I might as well add those too, now:

^{simulate this circuit – Schematic created using CircuitLab}

Here, I'm starting with a PNP follower and cascading it with an NPN follower (on the left.) Or, cascading an NPN follower by a PNP follower (on the right.) Either way, if you set things up so that \$R_1\approx R_2\$ then the collector currents will be similar and the \$V_{BE}\$ values therefore also similar. (It can be adjusted easily, of course, to tweak it in better.)

It's an okay way to cancel the \$V_{BE}\$ offset. And does the work of your opamp without the use of an opamp (which would be better to use because the opamp would have gigaohms of input impedance and active sink and source at the output.)

Put a resistor divider at the input, if you want.

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How did this idea get missed? I don't know.

Answer 2

Here is a slight variation on Tony's answer. If you like it, please upvote his answer, too.

In his circuit, D1 basically cancels the Vbe voltage drop of Q1.

In this circuit, M1 is supposed to cancel the Vgs(th) of M2. The idea is that by using MOS, we will load the divider even less. However, if the Vgs(th) of M1 and M2 are not closely matched, Tony's circuit may produce a better result.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

The issue with most of these BJT answers is the input impedance is low. That means whatever you feed it with must have an even lower impedance. Further the drive current is set by the pull-ups and Hfe so you are very limited on how small the load can be.

Voltage followers actually have five requirements.

1. Output voltage must follow input voltage
2. Input impedance should be high
3. Output impedance should be low
4. Output should be push-pull.
5. Output should be able to drive close, if not all the way, to both rails.

The circuit below provides a push-pull voltage follower with much higher input impedance and lower output impedance.

Finding matched-pair / dual MOSFETs with low Vgs and Rds_on is however important.

^{simulate this circuit – Schematic created using CircuitLab}

I would not recommend this one for tracking high frequency signals though.

Of course, by the time you add all that you used way more real-estate and cost than a simple Op-Amp circuit.

Answer 5

The two 10K resistors could be replaced with two 5V Zener diodes. It wouldn't be as good as using an op-amp though. Depends on how stable the 5V for the load needs to be. This has the potential to be very wasteful of power especially if the 10V source is high.

For that matter the top 10K resister could be replaced with a 5V Zener and the lower one left open.

If a resistive voltage divider is required then the 10K resistors could be replace with 10 ohm or less. Use 1 ohm for example. Very wasteful.