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AD8137 Output when gain is 1 (unity).

In case of AD8137 differential amplifiers' simulation, I have problem to understanding the output behavior. The input and feedback resistors value is 1 KOhm. Therefore I suppose to have gain 1 and which leads the output voltage as like input voltage. Am I right? But if I see the simulation the Output voltage decreased half of the Input voltage. What is going on here? Am I missing something?

The input amplitude here is 500mV, hence 1Vp-p and R1, R2 both are 1KOhm, no offset. In my application this input may vary.

In this particular situation, the simulation result showing that input is 1Vp-p while output voltage is 500mVp-p and this is exactly half of the input (why?). Isn't this suppose to be same? No common mode voltage inserted. I have read through the data sheet but the reason I did not find. Can anyone help me to find out the reason here?

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• Hi,

Firstly, you connected supply through capacitors.

Secondly, now you see output voltages referred to ground but you should consider differential output voltage in this case. So this is difference of output voltages at VF1 and VF2 respectively.

Positive output peak = 250 mV, negative = -250 mV, therefore differential output 250 - (-250)=500 mV.

Regards,

Kirill

The capacitor is to remove the supply noise as you know. And, yes the differential output from both VF1 and VF2 is 500mV peak to peak. You see my question was the input differential supply is 500mV-(-500mV)=1V , then in my understanding the differential output(500mV) is half of the input. That is my confusing point.

From simulation result, doesn't it suggesting differential output is half of the input?

I am not getting it unfortunately.

Thank you for your time and consideration.

Regards,

Hasan

• Hi Hasan,

I'm sorry, I am sometimes inattentive:)

You drew a diagram with an error and I pointed it out. I know I'm not the first to talk about it, I read your other question. To understand the operation of a differential amplifier, I suggest a simple circuit. These are feedback loops.

Ue is error input voltage. As you know, an amplifier covered by feedback tries to keep the voltage between its inputs as close to zero as possible. Therefore, Ue should be almost zero.

Let's say the input voltage has changed but the output voltage has not yet. Then Ue is increased and the amplifier can return it to zero only by changing the output voltages accordingly.

At equality of all resistors it is possible only when between outputs there will be a voltage equal on absolute value to input voltage but with opposite polarity. To confirm my thoughts, I made a simple demonstration.

This is very simple fully differential amplifier.

The output difference voltage is indeed equal to the input voltage.

I think it's a SPICE model error. But your assumptions about the expected output level are probably correct.

Regards,

Kirill

• Hello Kirill,

Thank you for you nice explanation about this topic. I do believe that your simulation is right and it is almost like my spice model that i have uploaded. I am agree that 'the output difference voltage is indeed equal to the input voltage'. However, in the graph when you will attach port input then it is become confusing. May it is a spice error as you said. I am not sure about it.

Your response on this issue really help me to understand some points. Thank you for reply.

Regards,

Hasan

• Hi Hasan,

I did an experiment with LTC6362

Note that this is differential input and output voltages. Everything works as it should. You can also conduct an experiment to verify the incorrect behavior of the SPICE model. Simply eliminate all feedback resistors and connect the source directly to the inputs. It should be expected that the output voltage will be very large.

Regards,

Kirill

• Hi Hasan,

I did an experiment with LTC6362

Note that this is differential input and output voltages. Everything works as it should. You can also conduct an experiment to verify the incorrect behavior of the SPICE model. Simply eliminate all feedback resistors and connect the source directly to the inputs. It should be expected that the output voltage will be very large.

Regards,

Kirill

Children
• Hello Kirill,

Thank you for your continues effort. I will try this as you said and sorry I see your reply later.

I am not sure but removing the feedback components could cause problem and it should also follow that non-inverting input with OUT - and inverting input with OUT+ via feedback. But yes it is just simulation, I will try it by myself.

Best regards,

Hasan