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current to voltage converter using ua741 op amp PSPICE

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For both the AC and DC analysis, the independent source used for the simulation must be a CURRENT source, not a VOLTAGE source.
 
Thank you for the instant reply but i don't understand....
Can you show me both analysis?
Because for both analysis i used a current source(in the figure the current source is called I4 and in the plot is called I(Iin), but I4=I(Iin) ): in the DC i used an I(DC) (x axis) and for the AC analysis i used an I(AC) (the current source that i named before).
I never used a voltage source for this circuit...VOUT is the output tension...
 
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You are right; I was confused by the symbol that is shown on the schematic.

Your currents seem to be much less than the input bias current of a 741 opamp. The 741 is not suitable for detecting such small currents...
 
yes, probably is for that reason...so what's the solution?
i have to put another op amp model(with a better range for input bias current) or can i insert something else in this ua741 circuit to obtain a better result?
 
I found the problem with your AC analysis. At the time you are doing the frequency sweep, the transistor has no bias (is effectively turned off), so when you do the AC analysis, you are seeing the open loop gain of the 741.

I re-ran my simulation of your circuit, but I specified that during the time I do the AC analysis, I first put 10mA into the current-to-voltage converter, and then let the AC analysis vary the current +-9.9mA centered on 10mA. This turns on the PNP transistor, and give a true picture of the frequency response. Note that in my attached sim, the d.c. bias and the a.c. amplitude are both specified independently (10m and AC 9.9m)

I am plotting the magnitude and phase of the trans-impedance (gain) =V(out)/I(I1). Note that at frequencies below ~100kHz, the gain is correct 66 @ -180 degrees (-66 comes directly from R1), and then some funny stuff happens between 100kHz and 10Mhz. Note that I am using a much better opamp than a 741. To get rid of the peaking, I would put a 10nF capacitor in parallel with R1.

D29b.jpg
 
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What kind of frequency response do you want? It's advisable to put a capacitor across R1 (Feedback resistor).

Pick an OP amp with a low Vos.
 
I found the problem with your AC analysis. At the time you are doing the frequency sweep, the transistor has no bias (is effectively turned off), so when you do the AC analysis, you are seeing the open loop gain of the 741.

I re-ran my simulation of your circuit, but I specified that during the time I do the AC analysis, I first put 10mA into the current-to-voltage converter, and then let the AC analysis vary the current +-9.9mA centered on 10mA. This turns on the PNP transistor, and give a true picture of the frequency response. Note that in my attached sim, the d.c. bias and the a.c. amplitude are both specified independently (10m and AC 9.9m)

I am plotting the magnitude and phase of the trans-impedance (gain) =V(out)/I(I1). Note that at frequencies below ~100kHz, the gain is correct 66 @ -180 degrees (-66 comes directly from R1), and then some funny stuff happens between 100kHz and 10Mhz. Note that I am using a much better opamp than a 741. To get rid of the peaking, I would put a 10nF capacitor in parallel with R1.

View attachment 83797
Thank you MikeMl!!!!
There was two problems with my configuration:
-the BJT was turned off as you said
-the current AC source in my configuration was in the wrong direction! It was in the opposite way than in the circuit that i made for the dc sweep!!

With these two solutions also ua741 works good. Than i try the LT1001 and the result is better than before!
Thank you!
 
Put the second figure here: https://www.linear.com/docs/1683 of the LT1010 etc within the feedback loop of your I-V converter at the output and it will work like a charm. Basically the input is from the output of your OP amp (Pin 6).
The feedback resistor gets moved to the output of the LT1010.

Then put a capacitor across your feedback resistor to restrict the bandwidth of your I-V converter to whatever by the relationship F=1/(2*PI*Rf*Cf)

For really precise measurements, you could add zero check and zero correct and use the offset pots., but 15 uV of Vos will make a BIG DIFFERENCE. e.g. V = -I*Rf; You have 15uV/Rf which is probably around 100 ohms or so which the OP amp can't drive without a booster. The 15 uV translates to about 15 uA of error. The closer you can get that to zero, the better the converter. The bigger issue is Vos is temperature dependent.

I have used this configuration with another OP amp and it worked really well. You also get current limiting for free.
 
Put the second figure here: https://www.linear.com/docs/1683 of the LT1010 etc within the feedback loop of your I-V converter at the output and it will work like a charm. Basically the input is from the output of your OP amp (Pin 6).
The feedback resistor gets moved to the output of the LT1010.

Then put a capacitor across your feedback resistor to restrict the bandwidth of your I-V converter to whatever by the relationship F=1/(2*PI*Rf*Cf)

For really precise measurements, you could add zero check and zero correct and use the offset pots., but 15 uV of Vos will make a BIG DIFFERENCE. e.g. V = -I*Rf; You have 15uV/Rf which is probably around 100 ohms or so which the OP amp can't drive without a booster. The 15 uV translates to about 15 uA of error. The closer you can get that to zero, the better the converter. The bigger issue is Vos is temperature dependent.

I have used this configuration with another OP amp and it worked really well. You also get current limiting for free.
Thanks for the help, can you show me just the picture of the final circuit?
I link components as you said but i'm not sure.
Thanks!
 
I'm somewhat computer crippled right now. Tough to draw/scan etc. Running Linux off a CD since my WIN 7 machine crashed.

The non-inverting buffer that isolates capacitance is the circuit you want to add. It has an input and output which you can identify. (other stuff like bypass caps and power are needed too).

Place the input of the buffer directly to the output of the I-V converter, pin 6. Move the feedback resistor from your pin 6 to the output of the buffer.

Tailoring the bandwidth requires a parallel cap across Rf.
 
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