enter image description hereenter image description here

I'm using an lm3900 quad op amp to run a smart phone audio signal to a bs170 MOSFET. When I use the inverting configuration it works well, but when I use the same resistor values (1k and 10k) with the non inverting configuration it either doesn't work or the output is too low. Is there an impedance issue that I'm not aware of between these 2 configurations or does it seem like a faulty chip? Any help would be appreciated.

  • \$\begingroup\$ Please include you entire circuit, and what you intend to do. Are you measuring the output after the mosfet or before. My guess is the driving signal to the mosfet is the problem. You have also not shown how your biasing the mosfet or the gain you have on the mosfet. Why do you have the mosfet anyway? \$\endgroup\$ – Voltage Spike Oct 19 '16 at 20:30

Is there an impedance issue that I'm not aware of

There is a difference in input impedance between the two circuits.

The input impedance of the non-inverting configuration is essentially the input impedance of the op-amp itself; that is, very high.

The input impedance of the inverting configuration is very close to \$R_{in}\$, because the summing node (negative input terminal of the op-amp) is a virtual ground. Another way to look at this is that the feedback will try to operate so that the input current is \$V_{in}/R_{in}\$ and your signal source will need to be able to supply this current.


But that is probably not your problem. Notice that the LM3900 does not have a typical op-amp input stage:

enter image description here

With those base-emitter junctions from the inputs to ground, you really need to operate this amplifier with the input terminals held near ground. Your non-inverting amplifier circuit doesn't do that, so it's not likely to work well.

  • 1
    \$\begingroup\$ I found an app note giving more info on using the LM3900 here. \$\endgroup\$ – The Photon Oct 19 '16 at 18:42
  • \$\begingroup\$ Perfect answer. Thanks for that. Makes sense to me. I'll look into the details of this chip. \$\endgroup\$ – Archaeus Oct 19 '16 at 18:48

Your Answer

