Is there a name for feedback amplifier topologies which include an opamp and transistor in the feed-forward path?. I need to design an amplifier with high bandwidth (almost 100 MHz) and high gain (50) and I think that I am better off getting most of my gain from a transistor.

I'm looking for literature on how to properly bias the transistor.

  • \$\begingroup\$ That doesn't sound like a good idea to me. Why don't you just cascade two op-amps? \$\endgroup\$
    – allanw
    Apr 22, 2011 at 5:56
  • 2
    \$\begingroup\$ You can probably get that from one op-amp. They make really fast op-amps (In the GHz!). If you are going to use discretes, just use only discretes. Mixing them isn't worth the effort. \$\endgroup\$ Apr 22, 2011 at 9:47
  • 3
    \$\begingroup\$ Actually, it is very common (and good) idea to place transistors on the output of operation amplifiers, still enclosed in the feedback path for the reason that the non-linearity of transistors can be (in part) compensated by the feedback, given that there is enough open-loop gain in the op amp. \$\endgroup\$
    – Gus
    Apr 22, 2011 at 16:49

2 Answers 2


What you need for this kind of job is a shunt feedback pair or whatever it's called. (some text books use other names. I stick to this one.) It's very easy to design a shunt feedback pair with discrete components. You might be able get away with some low cost 500Mhz transistors. They come in integrated circuits as well and they're called current feedback op amps.

An example of what the shunt feedback pair looks like can be found here in fig.3: http://www.douglas-self.com/ampins/discrete/2Q-VEM/2Q-VEM.htm

The ADA4861-3 might do what you need and it doesn't cost an arm and a leg. (the topology used is fairly easy to manufacture.) This device contains three 730Mhz current feedback opamps. You will need them because you have to split the required gain of 50 between two or three opamps.



What you can try is this: Connect a push-pull transistor pair to the output of an opamp and connect the feedback path to the output of the push-pull pair instead of the opamp's output. This way, you have the opamp do all the voltage amplification, and you can use the transistors for additional current amplification. The trainsistor pair's non-linearity will be compensated by the opamp because it's inside of the feedback loop.

This is an example I found after a quick search illustrating the general idea.

This application note (pdf), almost certainly written by Jim Williams (note the screenshots of his Tek 556 scope!), shows some practical schematics. Figure 3 details a circuit that does not meet the question's requirement of 100 MHz, but is capable of doing at least 10 MHz. The text says it is limited mainly by the OpAmp, and not by the discrete booster stage. It would be interesting to see what the circuit does with a faster, more modern OpAmp... Not perfect, but maybe close to what your question asks for!

Here's a later app'note (pdf), also by Jim Williams, featuring circuits with a -3 dB bandwidth beyond 100 MHz. A very good read!

  • \$\begingroup\$ That's ingenious. \$\endgroup\$
    – drxzcl
    Apr 22, 2011 at 10:35
  • 1
    \$\begingroup\$ OP wants a 5 GHz GBW product (50 * 100 MHz), which is way to much for most opamps. \$\endgroup\$
    – stevenvh
    Apr 22, 2011 at 11:41
  • \$\begingroup\$ Furthermore, the crossover distortion from that crude push-pull gets worse at higher frequencies. It's bad enough at audio frequencies... \$\endgroup\$
    – markrages
    Apr 22, 2011 at 21:52
  • \$\begingroup\$ Well, it just illustrates the basic idea. Please feel free to add base biasing or a cascode confuguration as desired. My scopes have a push-pull stage for the vertical amp, and they work beyond 100 MHz, too. I was just trying to give a starting point for further searches... If a transistor/opamp combination makes sense, it's likely something like this. \$\endgroup\$
    – zebonaut
    Apr 23, 2011 at 6:39
  • \$\begingroup\$ Getting a push-pull to work up-to 100Mhz isn't that easy. Your scope uses high freq. power transistors. A BF871 for example. Getting a stage capable of delivering some current doesn't solve the gain/bandwidth requirements. The voltage gain of a push-pull stage is a little less then 1. \$\endgroup\$
    – Hendrik
    Jun 3, 2011 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.