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Closed loop or open loop gain of the transimpedance amp?

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BillQ

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Hi,

I am designing a transimpedance amplifier to convert an input current to an output voltage.

I have used an iprobe to simulate this amplifier, then I runned stb analysis in Cadence and got the bode diagramm for the GAIN(dB) and phase margin.

Is this gain value(for example 80dB) the closed loop gain of this amplifier?
If so, then the phase margin cann't be deduced from this bode diagramm, as for the phase margin, a plot of the open loop gain is needed?

Moreover, the transimpedance amplifier should have a gain with a unit of Ohm(V/A), this can't be converted into dB. So how can I describe the technical parameter GAIN of this amplifier, V/A or the simulated dB value?

Thanks!
 

Usually when simulating a Transimpedance amplifier the simulator will report the V/A as dB. That means 80dB is 10000V/A so I guess you have a 10k feedback resistor.

Keith
 

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    BillQ

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BillQ said:
I have used an iprobe to simulate this amplifier, then I runned stb analysis in Cadence and got the bode diagramm for the GAIN(dB) and phase margin.

Is this gain value(for example 80dB) the closed loop gain of this amplifier?
Click to expand...

stb analysis determines loop gain. From data on loop gain phase and gain margin are calculated.

To obtain closed loop gain (transimpedance gain) you should use AC analysis: set AC mag of input current source to 1, run AC analysis, AC value of the output voltage is the transimpedance gain.

BillQ said:
Moreover, the transimpedance amplifier should have a gain with a unit of Ohm(V/A), this can't be converted into dB.
Click to expand...

V/A can be converted to dB. This unit is called dBOhm: <value_in_dbOhms>= 20*log(<value_in_Ohms> / 1 Ohm ).

Transimpedance can be specified in either units.
 

    B

    BillQ

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Thanks very much

so this 80dB from std analysis is the open loop gain, and correspondingly I can read phase margin from this simulation results

But actually this amplifier is used for DC operation, I mean the input signal is just DC current. From the large signal conversion curve(DC sim) I can easily find Vout/Iin. Is this large signal Vout/Iin the closed loop gain of this circuit?




dedalus said:
BillQ said:
I have used an iprobe to simulate this amplifier, then I runned stb analysis in Cadence and got the bode diagramm for the GAIN(dB) and phase margin.

Is this gain value(for example 80dB) the closed loop gain of this amplifier?
Click to expand...

stb analysis determines loop gain. From data on loop gain phase and gain margin are calculated.

To obtain closed loop gain (transimpedance gain) you should use AC analysis: set AC mag of input current source to 1, run AC analysis, AC value of the output voltage is the transimpedance gain.

BillQ said:
Moreover, the transimpedance amplifier should have a gain with a unit of Ohm(V/A), this can't be converted into dB.
Click to expand...

V/A can be converted to dB. This unit is called dBOhm: <value_in_dbOhms>= 20*log(<value_in_Ohms> / 1 Ohm ).

Transimpedance can be specified in either units.
Click to expand...
 

No, the 80dB is closed loop, not open. The phase is closed loop not open. 80dB is simply the effect of converting 10k ohms or 10000V/A into dB.

Keith

I should add that this is based on my guess at what you are simulating. It would be easier to judge if I saw a schematic and simulation results. I don't use Cadence but have been designing and simulating transimpedance amplifiers with Spice for over 20 years.

Keith
 

    B

    BillQ

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BillQ said:
so this 80dB from std analysis is the open loop gain, and correspondingly I can read phase margin from this simulation results
Click to expand...

This 80 dB is loop gain (loop gain = open loop gain * feedback factor). stb analysis automatically calculates phase margin from loop gain, so you haven't to calculate anything, just add getData("phaseMargin" ?result "stb_margin") to expressions in your analog design environment and you'll get phase margin.

BillQ said:
From the large signal conversion curve(DC sim) I can easily find Vout/Iin. Is this large signal Vout/Iin the closed loop gain of this circuit?
Click to expand...

Yes, this is DC closed loop gain.
 

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    BillQ

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so is my following understanding correct?

the gain normally what the data sheet talks about refers to the closed loop gain, which can only be simulated by AC analysis

with stb simulation, only the open loop gain is shown, we can use it to analysis the phase margin and the stability problem

i

Added after 2 minutes:

by the way, although my project has not specified the slew rate and CMRR and noise such parameters

I am wondering how important are the slew rate, CMRR for an analog circuit? is it a must to explore them in each circuit?
 

BillQ said:
so is my following understanding correct?
the gain normally what the data sheet talks about refers to the closed loop gain, which can only be simulated by AC analysis
with stb simulation, only the open loop gain is shown, we can use it to analysis the phase margin and the stability problem
...........
Click to expand...

BillQ, lets first clarify the terms.
*The data sheet can only give information on the opamp itself, that means:
opamp gain when there is no feedback (open loop gain).
* You have used the opamp (which in fact is a VCVS) as a transimpedance amplifier, which means: You have established neg. feedback and you are driving this input with a current.
* As KEITH has mentioned, if you then have a voltage-to-current ratio of 10k (80 dBohm) there will b a 10k resistor in the feedback path.
* This "gain" (better: transimpedance) is a closed-loop-parameter!
* Then, the loop gain is voltage-to-voltage and is identical to the opamps specified open loop gain (see first point).

Added after 3 minutes:

The foregoing assumes that your transimpedance amplifier consists of a voltage opamp which is wired as a transimpedance amp (with resistive feedback).
The situation is completely different in case you speak about a current feedback amp, which sometimes also is called "transimpedance amplifier (TIA).
 

I am not sure that 'open loop' characteristics exist for a transimpedance amplifier. To make a transimpedance you have effectively closed the loop. If you take a stable opamp and put it in a transimpedance amplifier configuration with parasitic input and feedback capacitances it can easily end up unstable.

Keith.
 

keith1200rs said:
I am not sure that 'open loop' characteristics exist for a transimpedance amplifier. To make a transimpedance you have effectively closed the loop. If you take a stable opamp and put it in a transimpedance amplifier configuration with parasitic input and feedback capacitances it can easily end up unstable.

Keith.
Click to expand...

Hi KEITH, I see no problem to define "open loop" for such an amplifier. It has 100% voltage feedback - and when the device is not unity gain stable it will become unstable in this configuration. Disconnecting the feedback resistor from the output means: Opening the loop. And this leads to a loop gain (neglecting parasitic influences and opamp input current) which is identical to the opamps gain. This situation is identical to the classical buffer case (voltage follower).
LvW
 

BillQ said:
Thanks very much

so this 80dB from std analysis is the open loop gain, and correspondingly I can read phase margin from this simulation results

But actually this amplifier is used for DC operation, I mean the input signal is just DC current. From the large signal conversion curve(DC sim) I can easily find Vout/Iin. Is this large signal Vout/Iin the closed loop gain of this circuit?
Click to expand...



dear Keith

I having problem with my AC analysis for transimpedance gain
I have simulate out transimpedance gain (Vout/Iin) ===wf0
the unit is in dB....but why is 0.846 db only (305/360)
and If i want to convert to the correct unit for transimpedance gain, what should i do ?


 

Could you post your circuit including test stimulus? You results don't make much sense at the moment. You should add or subtract dB - never divide them.

Keith
 

but transimpedance is (Vout/Iin ) rite ?

---------- Post added at 00:51 ---------- Previous post was at 00:49 ----------



---------- Post added at 01:18 ---------- Previous post was at 00:51 ----------

From what i analysis, what you mean by subtrate is it I already have my Log Vout and Log Iin ?
to get my tranismpedance gain , I just subtrate them ?

after i subtrate, I got 55dB
is it correct for my circuit ?1K feedback resistor
 

Yes, you would subtract dB to get the result. However I would like to understand why you get such ridiculous dB values. I cannot read the values on the current source - what are they? I am guessing you don't have AC=1 which would be the most sensible value to use.

Keith
 

ricopt has posted a vast number of circuit variants and simulations of his basic (one and three transistor) transimpedance amplifier. More data for the same variant, also a transient analysis are shown here:

https://www.edaboard.com/threads/192334/

They indicate, that the circuit has a finite transimpedance if biased correctly (around 1k, as expectable). So the AC analysis gain value seems simply to be a case of erratic simulator usage.

The course up to now raises my doubts, that you can acquire missing circuit analysis basics this way, though several contributors have seriously tried to teach some fundamental ideas.
 

keith1200rs said:
Yes, you would subtract dB to get the result. However I would like to understand why you get such ridiculous dB values. I cannot read the values on the current source - what are they? I am guessing you don't have AC=1 which would be the most sensible value to use.

Keith
Click to expand...

Ya, I am using AC=1
But why is ridiculous for the 55db value ?
Is too small or too large ?
What should i do to make it correct ?
 

i just substract it, i got 55dB ^^
thx bro
 

If you apply an AC current source of value 1 to a transimpedance amplifier the output should be the transimpedance ie 55dB not -305dB.

Keith
 

Did you set the source to 1a? Seems like your tool understands the current specification as 1 attoA (10e-18 A).
 

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