Open loop gain operational amplifier

[biolycans]

Hi,

Why in the simulation of the open loop gain of an operational amplifier I have to use a voltage source AC 1V ?

I know that AC 1 means that the output will plot correctly in db as gain. If I plot volts in the output versus volt in the input, I will have really big values of the gain in the op amp, but my question is related to the AC sweep.

I know that the AC sweep is a small signal analysis but in this case why I am using a signal AC of 1 volt amplitude instead of a really small signal (10 mV for example or even less) for the small signal analysis ?

Regards,

Joaquin

 

[Audioguru]

Why do you have more than one thread about the opamp you designed?

Most opamps have an open-loop voltage gain of 200,000 or more so an input of 75uV peak causes the output to saturate.
Then why are you asking about BLASTING the input with 1V?

 

[D.A.(Tony)Stewart]

Who said you must use 1V for open loop.


THere are practical ways suggested here. https://www.arrl.org/files/file/QEX_Next_Issue/Mar-Apr_2011/QEX_3_11_Post.pdf

 

[Venkadesh_M]

Before that you don know the AC sweep and the gain of the circuit before you inspect, Its just a conventional way of doing it.. you can use any voltage range without output saturation...

 

[crutschow]

Why in the simulation of the open loop gain of an operational amplifier I have to use a voltage source AC 1V ?
........................

As has been noted by others, you can use any source value you want. A Spice AC analysis doesn't really use the actual voltage in it's calculations. It uses a linear model of the circuit and calculates the gain ratio between input and output. So whether you use 1pv, 1V, or 1kV the gain ratio remains unchanged due to the linear model. 1(V) is used for convenience because the output "voltage" then is equal to the gain ratio, either in numerical units or in dB.

The Spice transient simulation is different as that uses the non-linear models of the devices along with real voltage levels so the input voltage will be reflected as real output voltages as determined by circuit non-linearities and/or supply voltages.

 

[LvW]

In addition to crutschow´s explanation: AC small-signal analysis means that ALL voltages (even kilo-volts) are treated as small signals (because of linearization of all non-linear characteristics).
As a consequence, non-linear parts (transistors, opamps,..) must be operated with the wanted quiescent DC voltage and/or current.

 

[rocky56hk1]

ac analysis measure the frequency response of a system. freq response usually use 0db(1v) as reference. eg , gain of amplifer, filter or system usually in express in db. use 0db is easy for reference

 

[biolycans]

As has been noted by others, you can use any source value you want. A Spice AC analysis doesn't really use the actual voltage in it's calculations. It uses a linear model of the circuit and calculates the gain ratio between input and output. So whether you use 1pv, 1V, or 1kV the gain ratio remains unchanged due to the linear model. 1(V) is used for convenience because the output "voltage" then is equal to the gain ratio, either in numerical units or in dB.

The Spice transient simulation is different as that uses the non-linear models of the devices along with real voltage levels so the input voltage will be reflected as real output voltages as determined by circuit non-linearities and/or supply voltages.

Hi,

If for example I apply a voltage source of 100 mV instead of 1V as I did, then the gain in dB is less than in the case of 1V. I understand that what Spice does is to derivate the small signal model of my schematic, then doing this I can apply a signal of 1V 10 V, etc because the model is lineal, but related to what you told here, the gain ratio doesn´t remains unchanged in my case.

regards,

JC

 

[LvW]

Hi,

If for example I apply a voltage source of 100 mV instead of 1V as I did, then the gain in dB is less than in the case of 1V. I understand that what Spice does is to derivate the small signal model of my schematic, then doing this I can apply a signal of 1V 10 V, etc because the model is lineal, but related to what you told here, the gain ratio doesn´t remains unchanged in my case.
regards,
JC

It seems that there is a misunderstanding on your side.
The gain is always Vout/Vin or Vout(dB)-Vin(dB).
Of course, the (linearized) small-signal gain does not depend on the input level.
The only point is: For Vin=1V (0 dB) the output voltage is numerically identical to the measured/simulated gain value.

 

[biolycans]

It seems that there is a misunderstanding on your side.
The gain is always Vout/Vin or Vout(dB)-Vin(dB).
Of course, the (linearized) small-signal gain does not depend on the input level.
The only point is: For Vin=1V (0 dB) the output voltage is numerically identical to the measured/simulated gain value.

Hi,

Let me see if I understand.

for example, if my differential input is 100 mV. the gain ratio is the same but in this case my reference is different as in the case of 1V (0 dB. Is this correct ? That is the reason that in the bode of my circuit, for a differential signal of 100 mV the gain at 1 Hz is 57.58 dB, my reference in this case is -20 dB = 20 log (100 mV). So in this case from -20 dB to 57.58 dB the gain is the same as in the case of 1 V, approximately 77.58 dB.

Is this correct ?

Regards,

Joaquin

 

[LvW]

Hi,
Let me see if I understand.
for example, if my differential input is 100 mV. the gain ratio is the same but in this case my reference is different as in the case of 1V (0 dB. Is this correct ? That is the reason that in the bode of my circuit, for a differential signal of 100 mV the gain at 1 Hz is 57.58 dB, my reference in this case is -20 dB = 20 log (100 mV). So in this case from -20 dB to 57.58 dB the gain is the same as in the case of 1 V, approximately 77.58 dB.

Is this correct ?

Yes - I think, now you understood.
However, one single correction: "the gain at 1 Hz is 57.58 dB"

It is not the "gain" - it is the output level which is 57.58 dBV.
Hence, the gain is 57.58 dBV-(-20 dBV)=77.58 dB.
(Note: the difference dBV minus dBV gives dB).

 

[biolycans]

Yes - I think, now you understood.
However, one single correction: "the gain at 1 Hz is 57.58 dB"

It is not the "gain" - it is the output level which is 57.58 dBV.
Hence, the gain is 57.58 dBV-(-20 dBV)=77.58 dB.
(Note: the difference dBV minus dBV gives dB).

Yes, I understood. !

Thank you !

Joak