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aryajur
Joined: 23 Oct 2004
Posts: 675
Helped: 69
Location: Sunnyvale, USA
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 23 Oct 2004 1:00 AC Analysis in Cadence |
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Hello,
Iam fairly new to cadence and I was trying to do get a frequency response of a Common Source NMOS Amplifier in Cadence. I used a resistive load of 10M. I biased it at the gate with a DC source to set the biasing point in series with a SInusoidal signal source of 10uV peak to peak amplitude. Then I did the DC, Transient and AC analysis of the circuit. The Transient and the DC Analysis are coming out fine and the curves are perfect but in the AC response I see nothing, i.e. both the Gate node and the drain node show 0 ! I have attached the JPEG capture of the circuit setup.
What am I doing wrong. I tried doing the AC analysis of some passive filters and that was coming to be fine. Any help or suggestions would be greatly appreciated.
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cetc1525
Joined: 08 Oct 2004
Posts: 177
Helped: 3
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| 23 Oct 2004 1:37 Re: AC Analysis in Cadence |
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I think it's your source's problem!
just setup the bias voltage at the ac source.
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aryajur
Joined: 23 Oct 2004
Posts: 675
Helped: 69
Location: Sunnyvale, USA
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| 23 Oct 2004 1:40 Re: AC Analysis in Cadence |
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Hey thanks for the reply,
Actually it was the problem of the AC source. I had made the entry of 10uV at the field specifying the AC Amplitude but when I did that entry at the AC Magnitude field it worked !
Thanks again.
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devrimaksin
Joined: 13 Oct 2004
Posts: 93
Helped: 9
Location: Dallas, Texas, USA
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| 23 Oct 2004 1:48 AC Analysis in Cadence |
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The general problem with this kind of simulation
is the biasing point of the transistor.
1. Notice that the width and length of your transistor is very small. Just use a regular number
(like 2u/1u) to begin to not to be affected by the
short channel effects.
2. Set a bias current to the transistor. The way you can do that is connect in diode configuration, the
same transistor and apply a dc current to it from
vdd, The gate voltage of the diode connected transistor will be your bias voltage. You can choose the value of the current so that the IR drop at the
output will be half of the power supply to maximize the allowable voltage swing (For example, vdd=1.8V, hence quiescent output voltage 0.9V, R=1MOhm, Hence the current is 0.9/1M =900nA,
this is actually pretty low. it will result that your transistor operates in sub-threshold region, chose
a resistor value so that you can have at least 5uA
or so bias current).
Then You have to feed this bias voltage to your transistor, for that you can use a voltage controlled voltage source. And for AC analysis you have to have AC source right, for that you can use a vdc
source with 0 dc voltage and ac =1 and it should be connected in serie with the output of the voltage controlled voltage source. And that will be connected to the gate of the amplifier transistor.
I hope this can help
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