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liletian

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op amp simulation question
Apr 27^th, 2009, 8:29am

Hi Guys
I simulated a op amp feedback circuit as attached, it is wired the output has DC offset of 0.004 voltage, can anyone explain it? The input is a Vsin of 2Ghz without dc offset in the negative input, the positive input is connect to ground, I will expect there has no output DC offset. Can anyone explain it? please see the attachment. The op amp is an ideal op amp with gain of 1e10.
Thank you

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op_amp.png

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LeeX Junior Member Offline Posts: 12 Singapore	Re: op amp simulation question Reply #1 - Apr 27^th, 2009, 9:19am Is this op-amp ideal or is it some circuit you have designed? Also, for the configuration shown in the fig, what kind of task you expect the circuit to perform?
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liletian

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Re: op amp simulation question
Reply #2 - Apr 27^th, 2009, 9:37am

LeeX wrote on Apr 27^th, 2009, 9:19am:

Is this op-amp ideal or is it some circuit you have designed?
Also, for the configuration shown in the fig, what kind of task you expect the circuit to perform?

The Op-amp is ideal, it is mathmatic equation( using verilogA), I would expect this circuit to work as an amplifier with a gain. I would not like to see a DC offset in there. Very weird.
Thanks

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: op amp simulation question Reply #3 - Apr 27^th, 2009, 9:58am hi liletian, � � � � � � � i didn't understand what's your plan with this circuit?As you said if you want amplifier why did you kept that cap in the feedback and even you don't have any dc source in the circuit,how would you expect dc offset. Thanks, Rajasekhar.
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liletian

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Re: op amp simulation question
Reply #4 - Apr 27^th, 2009, 10:17am

raja.cedt wrote on Apr 27^th, 2009, 9:58am:

hi liletian,
i didn't understand what's your plan with this circuit?As you said if you want amplifier why did you kept that cap in the feedback and even you don't have any dc source in the circuit,how would you expect dc offset.

Thanks,
Rajasekhar.

I do not expect DC offset, but my simulation gives me DC offset, that's why I am so confusing. Embarrassed

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buddypoor Community Fellow Offline Posts: 529 Bremen, Germany	Re: op amp simulation question Reply #5 - Apr 27^th, 2009, 10:20am Hi liletian, as already mentioned before: what is the purpose of the feedback capacitor ? It destroys your dc operating point (no dc feedback). The circuit cannot work.
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liletian Community Member Offline Posts: 97 MD	Re: op amp simulation question Reply #6 - Apr 27^th, 2009, 10:24am buddypoor wrote on Apr 27^th, 2009, 10:20am: Hi liletian, as already mentioned before: what is the purpose of the feedback capacitor ? It destroys your dc operating point (no dc feedback). The circuit cannot work. It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose.
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salty

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Re: op amp simulation question
Reply #7 - Apr 27^th, 2009, 10:46am

My guess is numerical resolution. �The only reason this ckt "works" is because everything is ideal. �If you see 4mV on the output, that would imply 400fV on the input which is about 2^-41. �I think you are just seeing numerical issues. �If you have very much DC at all on the input of the amplifier with 10^10 gain, the amp will rail. �You have no DC path around the amplifier.

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buddypoor

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Re: op amp simulation question
Reply #8 - Apr 27^th, 2009, 1:33pm

liletian wrote on Apr 27^th, 2009, 10:24am:

It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose.

Please, excuse me. But I think its absolutely silly to use an artificial amplifier with a gain of 1E10 without a stable bias point and to worry about an offset of 4 mvolts. Simulation programs are a tool to simulate real circuits which cannot be analyzed with calculations by hand.

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LvW (buddypoor: In memory of the great late Buddy Rich)

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liletian

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Re: op amp simulation question
Reply #9 - Apr 27^th, 2009, 2:28pm

buddypoor wrote on Apr 27^th, 2009, 1:33pm:

liletian wrote on Apr 27^th, 2009, 10:24am:

It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose.

Please, excuse me. But I think its absolutely silly to use an artificial amplifier with a gain of 1E10 without a stable bias point and to worry about an offset of 4 mvolts. Simulation programs are a tool to simulate real circuits which cannot be analyzed with calculations by hand.

for my defense, actually for a gain of 500, there still has 4 mV DC offset.

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subgold

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Re: op amp simulation question
Reply #10 - Apr 27^th, 2009, 3:25pm

liletian wrote on Apr 27^th, 2009, 2:28pm:

buddypoor wrote on Apr 27^th, 2009, 1:33pm:

liletian wrote on Apr 27^th, 2009, 10:24am:

It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose.

Please, excuse me. But I think its absolutely silly to use an artificial amplifier with a gain of 1E10 without a stable bias point and to worry about an offset of 4 mvolts. Simulation programs are a tool to simulate real circuits which cannot be analyzed with calculations by hand.

for my defense, actually for a gain of 500, there still has 4 mV DC offset.

could you pls post your verilogA code?

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boe Community Fellow Offline Posts: 615	Re: op amp simulation question Reply #11 - Apr 28^th, 2009, 6:09am liletian wrote on Apr 27^th, 2009, 2:28pm: ... for my defense, actually for a gain of 500, there still has 4 mV DC offset. BTW, what is your output voltage? If the Gain is actually 500 [at DC] and you want 2 V at the output, you need 4 mV at the input. B.O.E.
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LeeX

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Re: op amp simulation question
Reply #12 - Apr 28^th, 2009, 11:02am

I think I may have found how you get the 4 mV "offset".

First of all, I do not think this configuration can work for any practical design, as no DC feedback is present.

However, for the ideal op-amp, I assume it works properly and by "offset" you refer to the difference between zero cross points for the input and output.

The transfer function between input and output is

Vout/Vin=1+1/(jw/(1/RC))

where w is the angular frequency, in your case it is 2G*2*Pi, R=10K and C=200fF, according to your circuit.

Thus

Vout/Vin=1-j0.03978

So Vout is actually shifted in phase and amplified in magnitute, comparing to Vin,

the phase shift is

-0.03978

so if Vin = 0.1*Sin(wt)

Vout is approximately

Vout = 0.1*Sin(wt-0.03978)

let t=0,

Vin = 0
Vout = -0.00397≈4 mV

Actually, by definition the word offset should refer to the voltage difference between the op-amp positive and the negative input when the output is zero (common mode).

If my guess is right, you are measuring the voltage difference between the output node and input node, non of them are the opamp input.

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liletian

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Re: op amp simulation question
Reply #13 - Apr 28^th, 2009, 1:35pm

HI
Thanks for answering, you might be right, I will think about it more. I do know there is no DC feedback, but it seems is necessary for a charge pump of PLL, not sure about it yet. Can you comment more on the DC feedback and why it should not working for any pratical circuits?
Thanks a lot
LeeX wrote on Apr 28^th, 2009, 11:02am:

I think I may have found how you get the 4 mV "offset".

First of all, I do not think this configuration can work for any practical design, as no DC feedback is present.

However, for the ideal op-amp, I assume it works properly and by "offset" you refer to the difference between zero cross points for the input and output.

The transfer function between input and output is

Vout/Vin=1+1/(jw/(1/RC))

where w is the angular frequency, in your case it is 2G*2*Pi, R=10K and C=200fF, according to your circuit.

Thus

Vout/Vin=1-j0.03978

So Vout is actually shifted in phase and amplified in magnitute, comparing to Vin,

the phase shift is

-0.03978

so if Vin = 0.1*Sin(wt)

Vout is approximately

Vout = 0.1*Sin(wt-0.03978)

let t=0,

Vin = 0
Vout = -0.00397≈4 mV

Actually, by definition the word offset should refer to the voltage difference between the op-amp positive and the negative input when the output is zero (common mode).

If my guess is right, you are measuring the voltage difference between the output node and input node, non of them are the opamp input.

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buddypoor

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Re: op amp simulation question
Reply #14 - Apr 28^th, 2009, 11:28pm

liletian wrote on Apr 28^th, 2009, 1:35pm:

HI
Thanks for answering, you might be right, I will think about it more. I do know there is no DC feedback, but it seems is necessary for a charge pump of PLL, not sure about it yet. Can you comment more on the DC feedback and why it should not working for any pratical circuits?
Thanks a lot

[/quote]

Each opamp needs dc feedback to fix the bias point.
Without dc feedback the offset drives the output in saturation.

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LvW (buddypoor: In memory of the great late Buddy Rich)

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