It might be a silly question since I am new to circuits and have little knowledge about it.

I've learned that gain of a current to voltage convertor with op-amp is determined by feedback resistor R in the following figure:


Let's say I'm using an op-amp with GBWP of 10 MHz and using a feedback resistor of 47 kOhms. In this situation, will available bandwidth be 10M/47k = 212.xxx MHz ?

I am a bit confused that it is ok to regard it as a general concept of gain in other amplifying circuit.

In the figure above, gain is determined as 1 + R2 / R1 or R2 / R1 which is generally and practically 0 < gain <= 100... something. It is much smaller than a resistance, 47 k.

Is what I said is correct, or am I missing something and there are something I didn't know?


The GBWP for the setup stays constant. This means reducing low frequency open loop amplification by negative feedback increases the bandwidth by a same factor (assuming your open loop setup has a single pole, meaning gain rate of closure of 6dB/octave). Beware of proper uniting, you can't just divide bandwidth by resistance value and still have bandwidth.

To calculate closed loop gain you will need both resistance values, as you stated yourself. Suppose the open loop gain is 10^6, then there will be a pole at 10Hz (GBWP = 10 Hz * 10^6 = 10 Mhz). Reducing gain to for instance 100 shifts the pole to 100 kHz. (100 kHz * 100 = 10 MHz).

In your current to voltage transactor, it looks like there is only one resistor. There is one other, however, the output impedance of your input source. Both resistors determine voltage amplification (A = - Rsource / R). So in effect, bandwidth depends on your input source.

Also note that gain bandwidth product requires the notion of unity gain. And how exactly do you define unity gain of a current to voltage transactor?


Almost universally, OpAmps are internally compensated with a low frequency pole. This means that after some very low frequency (around 1Hz or so), open loop gain will fall by 20dB per decade of frequency. The frequency where open loop gain crosses unity is the gain bandwidth product (GBWP). This is a virtual constant for the amplifier throughout its operating frequency range due to the rate at which gain falls with frequency. GBWP is a small frequency model concept.

When thinking of OpAmps in terms of a small signal model, the gain bandwidth product is just an easy way of describing the useful bandwidth of the open loop gain of the amplifier. Signals in the small signal model are very small. Analogous to virtual work or virtual displacement in mechanics, small signals in electronics can be vanishingly small. The closer to amplifier performance boundaries operation becomes, the smaller the real signals must be for the concepts of the small signal model to be valid.

When considering real signals with amplitude, slew rate (SR) of an amplifier becomes more important than GBWP. In this case for a Trans Impedance Amplifier (TIA), converting current to voltage with 47kOhm conversion ratio, it would be better to look at the required SR. If the input current were 100uA peak, the peak output voltage of the OpAmp would have to be 4.7V. Slew rate to support a 1MHz signal of 4.7V amplitude would need to be:

SR > 2\$\pi\$ f \$V_{\text{pk}}\$ = \$\text{2$\pi $ 1MHz 4.7V}\$ or 29.5 \$\frac{V}{\text{$\mu $Sec}}\$

An OpAmp with GBWP of 10MHz will have a SR of about 20\$\frac{V}{\text{$\mu $Sec}}\$, so would not be able to support even a 1MHz, 100uA current to voltage conversion.


As far as the available bandwidth is concerned - it is the LOOP GAIN (LG) that matters only.

1.) Starting with the second circuit we have a feedback factor k=R1/(R1+R2) and the loop gain is

LG=-k*Aol (Aol=open-loop gain of the opamp).

The closed-loop gain is Acl=Aol/(1-LG). A unity gain-compensated opamp has an open-loop gain (1st order lowpass) Aol=Ao/(1+w/wo).

The closed-loop bandwidth is identical to the frequency where the loop gain magnitude is unity. Using the above given equations, setting |LG|=1, and solving for w we get the new 3dB corner frequency w1 (bandwidth) for the closed loop gain Acl: w1=wo(Ao*k-1).

Neglecting the "1" and setting 1/k=Acl (non-inverter) the final result is


(closed-loop gain times closed-loop BW = open-loop gain times open-loop BW). Note that the resistor VALUES are of no importance because only the resistor RATIO determines the feedback factor k.

2.) Regarding the first circuit, the loop gain is LG=Aol (because the internal resistance of the current source is infinite, which gives k=1). Therefore, the small-signal bandwidth is identical to the frequency where LG=Aol=0dB. Hence, the transit frequency (GBW) of the opamp is the available closed-loop bandwidth for the current-to-voltage converter.

UPDATE: The above considerations apply to the small-signal bandwidth only. Of course, in some cases the large-signal bandwidth (slew rate) will be the limiting factor (as outlined in another reply).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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