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concept quesiton: open loop gain of opamp for simulation

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ffsher100

Junior Member level 3
Hi all,

I am new to analog world and survey some opamp's simulation document for a while.

I got question about "open loop gain"

While open_loop_gain state, my understand is that all mos in opamp should be satisfied with Saturation.
first, i use dc sweep offset voltage between inverter(vip) non-inverter(vin)

vip more than vin

Thus, i put Vos(offset voltage) -1.4mv on vip such that vin=vdd/2, vip=(vdd/2)-1.4mv and all mos in saturation.
.TF V(vout) vip also let me obtain open loop gain in hspice report.

is this right way to find open loop gain??

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I'm not sure exactly what you've done, but in my understanding the open loop gain of an OPAMP is the voltage gain when the OPAMP has no feedback loop like for instance an inverting amplifier, etc.

yes, my configuration. opamp's out is left nothing to do and no feedback from out to input(vin/vip).

............
While open_loop_gain state, my understand is that all mos in opamp should be satisfied with Saturation.
first, i use dc sweep offset voltage between inverter(vip) non-inverter(vin)

Thus, i put Vos(offset voltage) -1.4mv on vip such that vin=vdd/2, vip=(vdd/2)-1.4mv and all mos in saturation.
.TF V(vout) vip also let me obtain open loop gain in hspice report.

is this right way to find open loop gain??

I am not quite sure what you really did, but - as far as I can see - at first you have placed a small voltage source between both inputs? That's not correct because this allows no stable operating point.
Recommendation: Ground the inverting input and connect the dc source between the non-inv. input and ground.
The dc sweep gives an output voltage that crosses the 0 V line. At this point you can read the offset voltage on the horizontal axis. This dc voltage must be applied to the non-inv. input node in order to allow linear operation of the device.

Remark: Dc sweep, for example, between -2mV and +2mV with a very good resolution (voltage steps) in the µV range.

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eklikeroomys

eklikeroomys

Points: 2
You have determined the OP's DC transfer curve. It's slope will in fact give the open loop DC gain. You'll mostly find it specified as a Large Signal Voltage Gain among OP DC parameters. Open loop gain as a frequency dependant parameter is determined in an AC measurement and needs an auxilary bias circuit to operate the OP in it's linear voltage range, or at least an exactly adjusted input bias cancelling the OP's offset voltage.

eklikeroomys

eklikeroomys

Points: 2
I am not quite sure what you really did, but - as far as I can see - at first you have placed a small voltage source between both inputs? That's not correct because this allows no stable operating point.

Hi LvW, as you see i put small voltage source between input only for dc sweep. for example vdd=3.3V, vip(non-invert) range1.65v-2mv)~(1.65v+1mv) while vin(invert) keep 1.65v.
your base voltage is ground and my is 1.65v.
i think there is no different from you mention vip=dc sweep crosses 0v line and vin is ground.

as below figure, is Vos(offset voltage) and Aol(open loop gain area) right?

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eklikeroomys

eklikeroomys

Points: 2
Quote: ....while vin(invert) keep 1.65v

As far as I can read your drawing (isn't it possible to select a white background?) the inv. input is at ground potential.
New information for me: Single supply. Therefor a good operating point is app. at Vdd/2=1.75 volts. This is in full agreement with your plot. Thus: Offset voltage app. +60 µV (if the voltage on the horizontal axis is 1.65+Vdc, cannot be recognized on the plot)
Use this dc bias to perform an ac analysis.

hi Lvw,
as a result, the right biasing is vip(non-invert) less than vin(invert) 60uV since DC sweep over vip at above figure.
is it right?

hi Lvw,
as a result, the right biasing is vip(non-invert) less than vin(invert) 60uV since DC sweep over vip at above figure.
is it right?

Sorry, I do not understand the meaning of the above.
Why don't you answer my questions?
* Inv. input is grounded? (it must be at app. 1.65 volts)
* What is on the x-axis of the drawing? (Drop the question, it is Vout).

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Open loop gain as a frequency dependant parameter is determined in an AC measurement
hi FvM,
my understand, when simulating and looking for "unity gain frequency", i think this is what you mean "frequency dependant".

and needs an auxilary bias circuit to operate the OP in it's linear voltage range, or at least an exactly adjusted input bias cancelling the OP's offset voltage.
does the "linear voltage range" you say is the area in flatten curve(ex:vdd/ground) without area in DC transfer curve?

---------- Post added at 21:01 ---------- Previous post was at 20:50 ----------

* Inv. input is grounded? (it must be at app. 1.65 volts)
yes, the second configuration circuit is inv in ground.
the first configration circuit is inv in vdd/2(1.65v).

* What is on the x-axis of the drawing? (Drop the question, it is Vout).
x-axis is DC sweep over vip within first and second configuration.
For my understanding:
if first circuit(vin=1.65v), the right biasing vip=1.65V-60uV
if second circuit(vin=0:ground), the right biasing vip= 0-60uV

---------- Post added at 21:05 ---------- Previous post was at 21:01 ----------

* Inv. input is grounded? (it must be at app. 1.65 volts)
yes, the second configuration circuit is inv in ground.
the first configration circuit is inv in vdd/2(1.65v).

* What is on the x-axis of the drawing? (Drop the question, it is Vout).
x-axis is DC sweep over vip within first and second configuration.
For my understanding:
if first circuit(vin=1.65v), the right biasing vip=1.65V-60uV
if second circuit(vin=0:ground), the right biasing vip= 0-60uV

ffsher100,

I am sorry, but my very first answer was not correct - and your circuit as shown in the first posting was correct!
The reason: I did not recognize immediately the fact that you were going to use a single power supply because, normally, the offset voltage is defined and measured in a double-supply configuration.
In most - if not in all - cases the offset influence plays no role if a single supply is used.
Now, since this point is clarified - what is your problem now?

You would want to perform the AC measurement with the output biased to about half supply voltage. You didn't mention yet the specified input voltage range, it's not clear, if 0 V is a suitable common mode input voltage for the amplifier. Unless otherwise requested, I would also expect a common mode voltage equal to half supply voltage for the AC measurements.

hi Lvw,
after reference textbook, i result that:
(1)the fully differential opamp(two output) is vin in ground while DC sweep on vip
(2)the single output opamp is vin in vdd/2 voltage while vip in (vdd/2 + dc sweep)
am i right?

for my first circuit and DC transfer curve, x-axis is DC sweep over vip.
the right biasing is vin=1.65v and vip=1.65v-1.45mv(offset voltage: 1.45mv approximatley)
is vip/vin biasing setup right ?
if the biasing is all right, ac analysis(unit gain frequency, phase margin) is next setp.
am i right?

for my first circuit and DC transfer curve, x-axis is DC sweep over vip.
the right biasing is vin=1.65v and vip=1.65v-1.45mv(offset voltage: 1.45mv approximatley)
is vip/vin biasing setup right ?
if the biasing is all right, ac analysis(unit gain frequency, phase margin) is next setp.
am i right?

Yes, I think so. However, the offset is negative, isn' it? (-1.45 mvolts app.)
You can apply the ac source directly in series with the offset correcting dc source.
(By the way, there are other methods to simulate the ac gain vs. frequency without compensating the offset voltage before.
However, your method - with offset correction - is also OK).

Hi Fvm,

Open loop gain as a frequency dependant parameter is determined in an AC measurement.
i think parameter unity gain frequency, phase margin, gain margin are frequency dependant in AC measurement.
am i right??

and needs an auxilary bias circuit to operate the OP in it's linear voltage range, or at least an exactly adjusted input bias cancelling the OP's offset voltage.
does "linear voltage range(L.V.R)" mean slop curve(-1.45mv~-1.35mv) without vdd/ground flatten curve as following figure??

for my first circuit(opamp: single supply voltage, single output), if i want to get unity gain frequency, phase margin, gain margin,
my correct biasing is vin:1.65v, vip:1.65v-1.45mv(offset voltage: -1.45mv approximately) for AC analysis.
am i right??

---------- Post added at 02:24 ---------- Previous post was at 01:49 ----------

hi Lvw,

Yes, I think so. However, the offset is negative, isn' it? (-1.45 mvolts app.)
yes, offset voltage is -1.45mv

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ffsher100, I have detected another source of error:

Your inverting amplifier has again of +5 if referenced to the pos. terminal.
Therefore the dc voltage for correct biasing at Vdd/2 must be (Vdd/2)*0.2=0.1*Vdd.

As another alternative - if you like to stay at Vdd/2 - put a large Capacitor in series with the resistor at the inv. terminal.
Thus, the dc gain is for the pos. input is unity and for the oscillating frequency you still have app. "4".

Your inverting amplifier has again of +5 if referenced to the pos. terminal.
I have no idea what you mean???

Therefore the dc voltage for correct biasing at Vdd/2 must be (Vdd/2)*0.2=0.1*Vdd.
Do you mean actual biasing: vin=0.33v, vip= 0.33v-1.45mvolts while vdd=3.3v???

As another alternative - if you like to stay at Vdd/2 - put a large Capacitor in series with the resistor at the inv. terminal.
yes, the method is good i have ever used it.

another question: since offset voltage is -1.45mvolts we talk before.
i try ac analysis by biasing condition as following(vdd=3.3v):
case 1: vip=(vdd/2)-1.40mVolts
case 2: vip=(vdd/2)-1.45mVolts
case 3: vip=(vdd/2)-1.5mVolts
all three cases on vin keep vdd/2.
the three cases got different result on gain and unity gain frequency.
If offset voltage vary, the ac analysis result looks like different.
If i not sure pick up true offset voltage, how could guarantee correct ac analysis result???

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Hi ffsher100, please forget my last posting (yesterday 19:09)

I was in a hurry - and this answer was related to another thread. More than that, yesterday evening we had still more than 30 deg in germany.
LvW

ffsher100, I have detected another source of error:

Your inverting amplifier has again of +5 if referenced to the pos. terminal.
Therefore the dc voltage for correct biasing at Vdd/2 must be (Vdd/2)*0.2=0.1*Vdd.

As another alternative - if you like to stay at Vdd/2 - put a large Capacitor in series with the resistor at the inv. terminal.
Thus, the dc gain is for the pos. input is unity and for the oscillating frequency you still have app. "4".

hi LvM,
above quote sentence, you want me to discard?
if yes, i am ok.

By the way,
Basing on our talk, i simulate ac analysis and got another question:
since offset voltage is -1.45mvolts we talk before.
i try ac analysis by biasing condition as following(vdd=3.3v):
case 1: vip=(vdd/2)-1.40mVolts
case 2: vip=(vdd/2)-1.45mVolts
case 3: vip=(vdd/2)-1.5mVolts
all three cases on vin keep vdd/2.
after ac analysis, the three cases got different result on (open loop)gain and unity gain frequency.
Once offset voltage vary, the ac analysis result looks like different.
If i not sure pick up whether offset voltage is true(maybe: -1.4mV, -1.45mV, -1.5mV), how could guarantee the final right ac analysis result???

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i try ac analysis by biasing condition as following(vdd=3.3v):
case 1: vip=(vdd/2)-1.40mVolts
case 2: vip=(vdd/2)-1.45mVolts
case 3: vip=(vdd/2)-1.5mVolts
all three cases on vin keep vdd/2.
after ac analysis, the three cases got different result on (open loop)gain and unity gain frequency.

ffsher, if you evaluate the transfer curve that you have posted yesterday (18:24) you will notice that the above mentioned offset voltages result in 3 different operating points, which exhibit different slopes (in particular: -1.5 mV).
Thus, it is clear that all ac analyses show different results since they are based on the slope of this curve.

For comparing purposes it would be wise to use the alternative method with dc feedback:
Place a (large) resistor between output and inv. input (to realize 100% dc feedback) and place a large capacitor between this input and ground. Thus the feedback is restricted to very low frequencies only and above the cut-off frequency you can simulate the open-loop gain vs. frequency.
Example: Rf=100 kohms, C=100 Farad result in a lower cut-off frequency of app. 16 mHz.

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