Output resistance Operational amplifier !

[biolycans]

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

 

[erikl]

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

 

[xiahanzh]

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.

 

[biolycans]

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.

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

 

[xiahanzh]

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.

 

[erikl]

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 ?

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 .

 

[LvW]

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

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

 

[erikl]

... and the signal input is shorted.

That's the canonical method to measure the output impedance of an amplifier's two-port network representation with admittance 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.

 

[LvW]

That's the canonical method to measure the output impedance of an amplifier's two-port network representation with admittance 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 - 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