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

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 ?

• Are you asking for practical reasons, or just out of curiosity? I am pretty sure the op-amp is the best way to deal with it. But if the voltage does not have to be too accurate, it may be possible to do it with a few transistors. – mkeith Jan 14 '18 at 0:03
• For a known, constant load it's trivially solved just be factoring the load into the divider calculation. You use a feedback system (op-amp or otherwise) for an unkown or variable load. – Chris Stratton Jan 14 '18 at 0:08
• What is the actual question here? Not everyone is going to click the link to figure ot out. – Sean87 Jan 14 '18 at 0:08
• Yes, more or less. In contrast to the drawing which clearly shows a fixed load resistance. For the depicted fixed load you could make the top resistor 100 ohms (with appropriate power rating!) and leave the spot for the lower resistor unoccupied. But many real loads vary in use. – Chris Stratton Jan 14 '18 at 0:20
• Yeah. Depending on the ACTUAL goal, and what is available, this circuit could be replaced with many things. For example, you could replace the op-amp with a 5V voltage regulator. Technically, that is not an op-amp. – mkeith Jan 14 '18 at 0:53

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.

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

• So R2 would be the 100Ohm load which would be matched by an 100ohm at R1 and the two 10k resistors junction would go in the input ? – soundslikefiziks Jan 14 '18 at 0:52
• @soundslikefiziks It could be that $R_2$ is the actual load, I suppose. If you put something in parallel to $R_2$ then the collector current in $Q_2$ will be just that much higher. Regardless, I wouldn't get too crazy about making $R_1\approx R_2$. There will be about $60\:\text{mV}$ difference caused by a 10-fold difference in collector current. So how important all this is depends on just how closely you want to match the output to the input. Do you have a desired figure? I thought you were just doing a survey of approaches. Did I misunderstand? – jonk Jan 14 '18 at 0:56
• @soundslikefiziks "Cascaded PNP+NPN follower?" I don't know if it has a name. Hmm.... Does this mean that even through thousands of engineers have discovered, re-discovered, and learned about this circuit for many decades already... that no one has named it??? Ah hah!! I claim this much trampled, but still unnamed circuit topology and hereby and henceforth name and call it the "JK-Follower" circuit!! ;) May the name live long and prosper. – jonk Jan 14 '18 at 1:19
• @soundslikefiziks Great to hear. Now to just send this on to the textbook authors! We can add JK-Follower to the Darlington and Sziklai pedestal. – jonk Jan 14 '18 at 2:01
• @SpehroPefhany If I were the least bit serious, I might. He's been wonderful in the past and I wouldn't think twice about writing. (He's probably still horribly busy, though.) But as I'm sure you are aware, I'm just teasing. I probably first saw this idea around 1990, though, because of the 2nd edition of AofE. Poor memory prevents me from being sure. I did look in both 2nd and 3rd editions, last night, found the circuit but didn't find it named. Perhaps I should write just to see if he knows where he first heard of it? – jonk Jan 14 '18 at 17:29

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

simulate this circuit – Schematic created using CircuitLab

• Why do we need the diode here? is there another purpose for it other than slightly increasing the voltage through R3 ? – soundslikefiziks Jan 14 '18 at 0:43
• Another solution not using an Op Amp is an LDO. – Tony Stewart Sunnyskyguy EE75 Jan 14 '18 at 1:17
• The diode and the complementary emitter follower method, null the offset created when buffering some R ratio. – Tony Stewart Sunnyskyguy EE75 Jan 14 '18 at 1:23

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.

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.

• Downvoting with no comments? Speak up? – dreamcatcher Jan 14 '18 at 20:23
• Your answer embraces the "very wasteful" way of life. No one in their sane mind would ever actually implement them in reality. Your answer is barely okay for simulations. This is not an OK answer to be used in real life. – Harry Svensson Jan 14 '18 at 20:26
• two 5V Zener diodes in series can mean amps if the actual total Vz > Vcc.... = lost power and likely lost smoke. 2x10 ohms is not much better. MIddle suggestion is passable though if load is consistant. – Trevor_G Jan 14 '18 at 20:35