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Open loop settings??

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pekachoo007

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Is this setting valid for finding the Open loop gain of the opamp.

If yes then hows its done in cadence spectre.
 

There must be no OP capacitive output load, remove one capacitor.

Use high L and C values, e.g. 1000 H / 1000F or higher, depending on the frequency range of interest.
 

FvM said:
There must be no OP capacitive output load, remove one capacitor.
Click to expand...

The opamp is actually an error amplifier part of a linear voltage regulator. The output capacitance represents the gate capacitance of pass transistor.

FvM said:
Use high L and C values, e.g. 1000 H / 1000F or higher, depending on the frequency range of interest.
Click to expand...

Using 10^12H and 10^12F for L and C values for the frequency of operation of 10MHz.
 

A small question. Why do you need inductance in feed back? Can't you directly connect the -ve terminal of the opamp to the required DC potential and measure the open loop gain? Correct me if I am wrong.
 

The opamp is actually an error amplifier part of a linear voltage regulator. The output capacitance represents the gate capacitance of pass transistor.
Click to expand...
^
O.K., in this case the regular output C should be used.
 

What settings should I give to Cadence Spectre for simulation results of gain and phase margin
 

Run an ac simulation for your frequency range of interest, then display gain (dB20 scale) and phase.

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AMS012 said:
Why do you need inductance in feed back? Can't you directly connect the -ve terminal of the opamp to the required DC potential and measure the open loop gain?
Click to expand...

Because you need the DC control loop.

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pekachoo007 said:
Using 10^12H and 10^12F for L and C values for the frequency of operation of 10MHz.
Click to expand...

Calculate its cutoff frequency resp. time constant! DC control will take much too long for your simulation period: don't forget the output impedance of the opAmp.
 

erikl said:
Run an ac simulation for your frequency range of interest, then display gain (dB20 scale) and phase.
Click to expand...

I have done this but not satisfied with results.


erikl said:
Calculate its cutoff frequency resp. time constant! DC control will take much too long for your simulation period: don't forget the output impedance of the opAmp.
Click to expand...

What would be use of these extra calculation, how they would be helpful. As you suggested, could you provide the formulas also.
 

eares.PNG

Set AC source with AC Magnitude=1 V, DC Voltage=1.1 V, f=10 MHz

db20 = 44 db
What would be phase Margin, is it 180-67= 113??
 

pekachoo007 said:
What would be use of these extra calculation, how they would be helpful. As you suggested, could you provide the formulas also.
Click to expand...

Just use the known or estimated value of the opAmp's DC output resistance R and calculate the cutoff frequency of this low-pass filter RLC time constant; its inverse is the time constant of your DC input control voltage, i.e. before this time you can't expect to get the correct input bias.

At this cutoff frequency you can neglect the value of your chosen L=10^12H impedance against the opAmp's DC output resistance R value - let's say this is 100Ω - so the cutoff frequency - considering your chosen C=10^12F value - fco = 1/(2πRC) =0.16*10^-14 Hz, its time constant is τ = 2πRC = 6.28*10^14 s ≈ 20 million years. After this time you can expect that the input bias has reached about 63% of its correct value, more than 99% after 5τ ≈ 100 million years.
 
before this time you can't expect to get the correct input bias
Click to expand...
SPICE achieves a correct bias point in "no time" because it disables all Ls and Cs during initial transient solution.
 
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    erikl

    erikl

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pekachoo007, is the max. gain of 40 dB the value you have expected?
Why a dc voltage of 1.1 V ?
 

erikl said:
At this cutoff frequency you can neglect the value of your chosen L=10^12H impedance against the opAmp's DC output resistance R value - let's say this is 100Ω - so the cutoff frequency - considering your chosen C=10^12F value - fco = 1/(2πRC) =0.16*10^-14 Hz, its time constant is τ = 2πRC = 6.28*10^14 s ≈ 20 million years. After this time you can expect that the input bias has reached about 63% of its correct value, more than 99% after 5τ ≈ 100 million years.
Click to expand...

Why to fret about million years, the values are for the test bench.

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LvW said:
pekachoo007, is the max. gain of 40 dB the value you have expected?
Click to expand...
Around 70.

LvW said:
Why a dc voltage of 1.1 V ?
Click to expand...

What it should be then.
 

pekachoo007 said:
The purpose might be certain, with the certainity principle lying somewhere.
Click to expand...
So - you dont know why?
This is not a big problem - however, in this case you simply should ask instead of (blindly) using a dc component without knowing its purpose.
 

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