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Output resistance Operational amplifier !

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biolycans

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Hi,

I designed an operational amplifier two stage, and what I want to do is to measure using the simulations, the output resistance. The operational is CMOS.

Regards,

JC
 

So very often asked for. Check the Similar Threads below or use the Search function!
 

What simulation tool are you using?
General idea is putting a small signal voltage source at output node and test the small signal current.
But I know many simulation tools can measure it directly, which is easier. But It depends on which software are you using.
 

xiahanzh said:
What simulation tool are you using?
General idea is putting a small signal voltage source at output node and test the small signal current.
But I know many simulation tools can measure it directly, which is easier. But It depends on which software are you using.
Click to expand...

Hi my friend !

I am using top spice. I connected in the output a current source AC 1 and then I did an AC sweep between 1 hz and 10 Mhz. The plot I obtained represents the output voltage versus frequency. In the output I have a capacitor of 10 pf. At low frequencies the plot correspond to the output resistance (Ro) and at high frequencies the plot represents the capacitor impedance. What do you think ?

Regards,

JC
 

I never used top spice before but I think your way to simulate it is correct.
Your way to get capacitance impedance is correct but I do not understand why you need to put a capacitor here. For simulation purpose, if you want to know parasitic output capacitance from the circuit, you can have a open output so in high frequency, you can know how much capacitance your circuit has.
 

biolycans said:
I connected in the output a current source AC 1 and then I did an AC sweep between 1 hz and 10 Mhz. The plot I obtained represents the output voltage versus frequency. In the output I have a capacitor of 10 pf. At low frequencies the plot correspond to the output resistance (Ro) and at high frequencies the plot represents the capacitor impedance. What do you think ?
Click to expand...

I think you picked the best and most convenient method. Congratulations!

If you use a "unit" current source (1A) in an ac simulation, the output voltage value corresponds directly to the output impedance, e.g.
kV ≙ kΩ , MV ≙ MΩ .
 
biolycans said:
Hi my friend !

I am using top spice. I connected in the output a current source AC 1 and then I did an AC sweep between 1 hz and 10 Mhz. The plot I obtained represents the output voltage versus frequency. In the output I have a capacitor of 10 pf. At low frequencies the plot correspond to the output resistance (Ro) and at high frequencies the plot represents the capacitor impedance. What do you think ?
Regards,
JC
Click to expand...

As mentioned already by erikl - good and correct method, provided the device has the "normal" operating point and the signal input is shorted.
 
LvW said:
... and the signal input is shorted.
Click to expand...

That's the canonical method to measure the output impedance of an amplifier's two-port network representation with admittance (y) or inverse hybrid (g) parameters.

For a real amplifier, however, this is not necessary: Just use your preferred (actual) input impedance in order to get the real output impedance.

Depending on an amplifier's circuit architecture (e.g. in parallel-parallel or parallel-series feedback connection), short-circuiting its input could even result in nullifying the feedback and so change the intended closed-loop gain into an open-loop gain, thereby changing (increasing) the output impedance quite a lot.
 

erikl said:
That's the canonical method to measure the output impedance of an amplifier's two-port network representation with admittance (y) or inverse hybrid (g) parameters.
For a real amplifier, however, this is not necessary: Just use your preferred (actual) input impedance in order to get the real output impedance.
Depending on an amplifier's circuit architecture (e.g. in parallel-parallel or parallel-series feedback connection), short-circuiting its input could even result in nullifying the feedback and so change the intended closed-loop gain into an open-loop gain, thereby changing (increasing) the output impedance quite a lot.
Click to expand...

erikl - I know what you mean. Perhaps my wording was a bit "floppy".
By saying "shorted input signal" I mean, of course, that the input voltage source should be set to zero volts - and maintaining** any possible internal source resistance. The primary goal must be to establish the normal operating conditions for output impedance measurements - except input signal injection.

EDIT:** Please read retaining instead of maintaining
 
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