Current to voltage converter using op amp: Why the output voltage not same as equat?

[syamin0712]

Hi,

What I understand about I to V converter using op Amp is that the output voltage will equal to the resistor x current input that we have.

Something like this figure:

https://obrazki.elektroda.pl/1069454400_1442453480.png

So that the output voltage will be

Vout = -IsxRL


However,
when I simulated the circuit; in my case the current input is 2.5mA with operating frequency is 2MHz, and I'm using LT6200CS8-5 as my op amp.

https://obrazki.elektroda.pl/3328319100_1442453788.png

Noticed that, the 1k ohm resistor and 4.7pF capacitor are to stabilize the circuit only.So that, the current value is maintain.
If using the above equation; Vout= -2.5mA x 1.2kohm =-3 Vp-p
The output value was not same as the equation above. The Vp-p is around 8.64 Vp-p.

Could anybody explain it to me what is the error and how to solve it.
Or that equation will get the output in RMS value?

Thank you in advanced.

 

[BradtheRad]

operating frequency is 2MHz

What is the gain-bandwidth product of your op amp? You may be running it too fast for it to respond.

 

[syamin0712]

The GBP is 800MHz and slew rate is 210 V/us.

You may be running it too fast for it to respond.

What you mean by this?

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so, can you suggest any solution for it?

- - - Updated - - -

Because 2 MHz is the must operating frequency for my system

 

[crutschow]

The LT6200CS8-5 is designed to operate with a gain no lower than 5 for stability.
It will likely be unstable in the first circuit you referenced, which has a voltage gain of 1.

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I don't see how you get 2.5mA with an input voltage of 3.54Vrms and and input resistance of 4k ohms.

 

[syamin0712]

voltage gain of 1.

I thought the gain is represent by the RL. The resistor value that we used.
So. as long as my GBP is large than the frequency that I use I think is can be used.

Or it is actually a different gain?

The current 2.5mA is taken by considering amplitude value of voltage source which is Vamp= 3.54 x surd(2)=5V x2 =10 Vp-p of sinewave.
I=10/4K=2.5mA that represent the current input to the op amp.

Is it wrong?

 

[KlausST]

Hi,

The expected output voltage is about 3.54V x 4.6kohms/ 4.0 ohms is about 4.1V RMS.

Where the I put current is about 3.54V / 4kohms = 0.885mA RMS
The output impedance is 4k7 parallel to 4p7, estimated to be 4k6.

Now V_out = 0.885mA x 4600 = 4.1V RMS

This also means a peak output voltage of 4.1V x 1.41 = 5.8V.
But your supply is only 5.0V

So output voltage is in any case below 5V. It is distorted.

Klaus

 

[FvM]

The LT6200CS8-5 is designed to operate with a gain no lower than 5 for stability.
It will likely be unstable in the first circuit you referenced, which has a voltage gain of 1.

Gain of 5 in this specification is referring to the closed loop gain in non-inverting amplifier configuration, assuming a pure resistive feedback network. In the general case, you have to analyze the feedback loop gain and check if the phase margin is sufficient.

Due to the 4.7 pF capacitor, the feedback factor of your configuration is near to (minus) unity, similar to a +1 amplifier, creating an unstable loop with the decompensated LT6002-5. In simple words, this type isn't suitable for an I/V converter. Using a standard LT6002 would help.

Or adjust the loop gain, e.g. by placing a 20 pF capacitor between + and - input. This should work at least in simulation. In a real circuit, stability with a 800 MHz GBW OP is at risk anyway.