OPA140: Why is the gain not the same as the calculation?

Hi Hai,

this is no surprise. What is 0V x 392.8?

The OPA140 looks good:

1122.hai_opa140.TSC

Kai

kai klaas69 said:

Hi Hai,

this is no surprise. What is 0V x 392.8?

The OPA140 looks good:

(Please visit the site to view this file)

Kai

Dear Kai,

I do not understand what you mean. Can you explain in more detail?
I use a pulse source with a pulse width of 2us in this case.

Vin =-1mV

Vout = -Vin*G = (-1mV)*(-392.8) = 392.8mV

Thanks,

Hai

Hi Hai,

Something is weird about the Vin, until I changed DC Level to -1mV and Start of Edge with a delay of 20usec at the same time, then the Tina simulation is simulated your correct answer, which is same as calculated answer. 392.8mV is the correct answer, if Vos and other errors are not taken into account. The simulation is shown at approx. 403mV. 

1122.hai_opa140 modified.TSC

regards,

Raymond 

Raymond Zhang1 said:

Hi Hai,

Something is weird about the Vin, until I changed DC Level to -1mV and Start of Edge with a delay of 20usec at the same time, then the Tina simulation is simulated your correct answer, which is same as calculated answer. 392.8mV is the correct answer, if Vos and other errors are not taken into account. The simulation is shown at approx. 403mV. 

1122.hai_opa140 modified.TSC

regards,

Raymond 

Hi Raymond,

My signal is that pulse is not unit step.

And DC Level =0.

Thanks.

Hai5488.1122.hai_opa140 modified.TSC

Hi Hai,

it's normal that an OPAmp becomes slower when you increase the gain. You would need to decrease the gain of OPA140 to be able to capture your narrow signal peak. You can counterpoise the loss of gain by putting several OPA140 in series:

hai_opa140_1.TSC

As an alternative you could choose a much faster OPAmp, of course. But it might be difficult to find a much faster OPAmp which shows as high DC precision as the OPA140. So, a series circuit of several OPA140 can absolutely make sense.

Kai

Dear Kai,

Your opinion is quite reasonable. However, I am looking for a reasonable explanation for the my circuit.

Hai,

Hi Hai,

what exactly is your question?

Kai

Dear Kai,

Is the output voltage not right with the formula due to the opamp's response to the frequency?

Hai

Dear Kai,

How to examine the parameters as in the file you gave.

5381.1122.hai_opa140.TSC

I searched in TINA but only a few.

Thanks.

Hai

Hi Hai,

one error source is the input offset voltage of OPA140 which is also amplified with the programmed gain of OPAmp. And concerning the dynamic behaviour of OPA140, you just must wait enough and allow the output signal of OPA140 to settle to the final value which is determined by the input signal times gain.

Kai

Dear Kai,

I have found a circuit quite similar to mine.

http://physicsopenlab.org/2016/04/21/pmt-pulse-processing/

But the way the voltage is calculated is quite strange.

Vout = R2* Isignal.

So what does R1 do?

Hai

Hi Lai,

I guess the photomultiplier is assumed to be a current source and R1=50R is considered to have no significant influence on the current signal generation. In any case the -input of OPA354 sits on virtual ground and R1 in combination with the complex source impedance of photomultiplier forms sort of a pulse shaper.

But, very probably, R1 also plays another role: It isolates the -input of OPA354 from stray capacitance (cable capacitance and detector capacitance of the photomultiplier itself). This can considerably help to stabilize the OPAmp by preventing the phase margin from being drastically eroded.

Kai

Dear Kai,

R1 does not affect the gain?

Following is the signal from my circuit:

Channel 1 is the signal before entering the amplifier circuit.

Channel 2 is the signal behind the amplifier circuit.

Is there a mathematical formula to represent the relationship between Channel 1 and Channel 2? What I was looking for was that.

Hai

Dear Everyone,

I have expanded my circuit and have a few strange things:

At 1khz frequency I get the tissue sample equal to the result from the gain curve.

db= 20*log(Vout/Vin) => Vout =Vin*10^(db/20) = 1u*10^(105.85/20) = 196.6 mV

but at 1MHz, it's not true.

Hai

Hi Hai,

yes, because of C3. C3 is in parallel to RF and decreases the parallel impedance at high frequencies. The corner frequency of this process is 1 / (2 x pi x RF x C3) = 17.3kHz.

Kai