I'd like to buffer the output of an operational amplifier using a source follower.

Operational amplifier

The operational amplifier produces a sinusoidal wave like this.

Sinusoidal wave

I would like to buffer this signal. To do that I use a source follower like that.

source follower

I tried a few things.

mosfet add dc bias

But I see only small signal at the input of the mosfet. And mosfet doesn't enters into saturation region. I tested this circuit seperately and it works fine on his own.

Seperate version

This is the seperate version of the source follower. I'd like to replace small signal sinusoidal wave with output of operational amplifier.

How should I bias the MOSFET using LTSpice simulation software?

Update : I added a voltage divider resistor network to the non-inverting input of operational amplifier.


The input signal is a triangle wave. It's frequency is about 2.7 kHz.

op amp input output

Actual input signal, the green one in simulation looks like this:

triangle wave

Source follower is as before

source follower

This time output of source follower looks fine.

source follower

Isn't the input voltage of the source follower is quite high? Shouldn't the input voltage be very small? E.g. 500 mV etc.

Update 2:

Decoupling capacitors

At first glance I used decoupling capacitors instead of non-inverting input of the operational amplifier.

decoupling cap

As you'd see input and output waves are almost identical.

input output

Biasing from op amp non-inverting input

This time hopefully I should be able to find the non-inverting input.


There is about 3.5 volt bias voltage on the output signal.

opamp bias

This time I removed the first decoupling capacitor at the gate of the source follower to be able to operate mosfet in saturation region.

source follower

Also this time, the output of the source follower looks fine to me.


  • 1
    \$\begingroup\$ Try put a decoupling cap at the sinus node with your latest configuration. \$\endgroup\$
    – Unknown123
    Commented Apr 4, 2019 at 4:12
  • 1
    \$\begingroup\$ insert a DC_blocking capacitor in the signal chain. \$\endgroup\$ Commented Apr 4, 2019 at 4:39
  • 1
    \$\begingroup\$ You said you put it on the non-inverting input, but the schematic tells us that you put it on the inverting input. We told you to insert a capacitor in series at sinus node, but you're reverting back your circuit. \$\endgroup\$
    – Unknown123
    Commented Apr 4, 2019 at 6:20
  • \$\begingroup\$ @Erdem As you'd see input and output waves are almost identical. Isn't that what you wanted? \$\endgroup\$
    – Unknown123
    Commented Apr 5, 2019 at 0:34
  • \$\begingroup\$ Sure. I was trying to achieve this. \$\endgroup\$
    – user211748
    Commented Apr 5, 2019 at 4:14

1 Answer 1


Your resistors do nothing but make the op-amp drive a bit more current. You can either capacitively couple it, or bias the op-amp up to a few volts by putting a divider on the non-inverting input.

Vgs(th) is 0.8 to 3V for the BS170 and your op-amp circuit has a DC gain of 1.0 from the non-inverting input. So you can try 4 volts on the non-inverting input.

Bear in mind that the threshold will vary from part to part, though you will not see that in the simulation. For this reason, an NPN BJT emitter follower is considerably more predictable.


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