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I am trying to understand the function of this input stage of a filter I am using.

The input current source is a 2uA peak-to-peak with zero offset.

(This specific stage is followed by an active bandpass with a gain of 40 and an active low pass with a gain of 30.)

Quick summary:

My first question before simulating it in Multisim was "Where is the input resistor?" and hence "How much is the gain?". Although my question still remains, I figured that the output voltage could be derived by adding the voltage drop across R2, to the non-inverting reference voltage 1.65 V. Specifically, the current passing through the feedback network is 1.65/47k =35uA (+- 1uA due to input current) and therefore Vout =1.65 + 35uA*36kΩ = 2.91 V.

However, I could not find any relationship between the input Vin(pp) of 359 nV (probe1) and Vout(pp) of 71.8 mV. In fact, Vout(pp) remains constant despite changes in R1, but is linearly proportional to a change in R2 ( if R2'=2*R2, Vout,pp'= 2* Vout,pp). So I assume there is no R2/R1 type of relationship to calculate gain. Then where does the gain of (71.8 mV)/(359 nV)~ 200,000 come from?

Another thing that I did not like is that Vout(dc) is different than the reference voltage 1.65 V. I know I just explained above how this dc output voltage V(dc) of probe2 is derived, but, up until now I thought the whole point of a reference voltage in the non-inverting pin is to perform level shifting so that the ac signal in the output will be centered around that reference voltage.

After this realization (that vout(dc) does not equal the reference voltage), I went on examining other possible level shifting circuits, like this for example non-inverting configuration with Vref for level shifting,

where I would think that Vout= Vref + (1+R1/R2)*Vin. However by doing the math (superposition), I get: Vout=(1+R1/R2)*Vin-(R1/R2)*Vref.


  1. Is there a formula for calculating the ac gain of this active low pass configuration in the first picture?

  2. If not, is the ~200,000 times gain of the ac signal intentional? Or did the the designer just aim for the low pass functionality and remained satisfied with this huge gain that is somehow provided?

  3. Please help me demystify opamp level shifting once and for all. Is the formula derived above for the opamp of the second picture correct? Vout=(1+R1/R2)*Vin-(R1/R2)*Vref. If it is, then how is level shifting achieved for the non inverting configuration?

**for an inverting configuration it can be achieved as explained by Spehro Pefhany in this post.

  • \$\begingroup\$ I'm puzzled by your supply connections to the op-amp. I expect to see the higher potential at the top connection and negative or ground potential at the bottom. \$\endgroup\$ May 5 '18 at 1:01
  • \$\begingroup\$ @RussellBorogove It simply is because of the way multisim places the supply pins on the opamp symbols. The positive supply pin is closer to the non-inverting input and the negative suppy is closer to the inverting input. \$\endgroup\$
    – Geo
    May 5 '18 at 12:38

In answer to the question posed by the title, this is a transimpedance amplifier with a resistor across the input to help set the DC level.

  1. The (ideal) gain of the transimpedance amplifier is Vo/Iin = 36K (just the value of R2). So peak-to-peak output from a 1uA peak input should be 72mV (ignoring the roll-off). R1 has virtually no effect on the gain, it just makes the output noise a bit worse.

  2. The voltage change seen at the inverting input is as a result of the finite gain of the amplifier, ideally it would be zero. It should be the output voltage change divided by the open-loop gain of the amplifier.

  3. Your formula looks correct.

If you wish to have the output voltage (at zero input current) equal the 1.65V reference in your top circuit, just lose the 47K resistor. There will be a DC 1.65V across your input current source. Of course you perhaps could connect the low side of the current source to Vref rather than ground. Or put a large capacitor in series with the input for AC coupling.

  • \$\begingroup\$ Perfect answer , thanks so much. Concerning your last paragraph, indeed, after putting a 10u coupling cap the DC offset was effectively removed. When you say I could perhaps connect the low side of the cuurent source to Vref rather than ground, should the non-inverting pin of the opamp also remain at Vref or should it be connected to ground instead? \$\endgroup\$
    – Geo
    May 5 '18 at 12:47
  • 1
    \$\begingroup\$ You need it at the desired output voltage, Vref. \$\endgroup\$ May 5 '18 at 12:48

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