Op amp design help needed

Thread Starter

mehmeter3

Joined Dec 19, 2013
13
Hello

I worked on the following question but couldn't solve. Could you maybe help me about it?

Sketch an amplifier circuit for measuring currents up to 1 μA from DC to 1 kHz. Input impedance shall be lower than 100 Ω. Output voltage at 1 μA shall be 1 V.

Thank you/Red.
 

Thread Starter

mehmeter3

Joined Dec 19, 2013
13
Yup. To amplify the signal voltage at low fequency I used an operational amplifier. If the source impedance is 100 Ohms, 1 uA through it generates 100 uV. To get one volt output I need a 10 000 gain using non inverting op amp. But then I don't know how to operate it from DC to 1 kHz.
 

shteii01

Joined Feb 19, 2010
4,644
Yup. To amplify the signal voltage at low fequency I used an operational amplifier. If the source impedance is 100 Ohms, 1 uA through it generates 100 uV. To get one volt output I need a 10 000 gain using non inverting op amp. But then I don't know how to operate it from DC to 1 kHz.
One of the features of opamp is that it can take dc voltage and it can take alternating voltage. So you have 1 kHz alternating voltage, no problem, input it to opamp. All you have to do is pick a model of opamp that is designed to accept 1 kHz. One of the features of opamp is that it can be used as a filter, to filter out high frequency inputs. An opamp datasheet will have an entry for Bandwidth. So. Take the bandwidth number and divide it by the gain you desire, the result is the frequency that will be the limit of your input. If 1 kHz is smaller than the frequency that we just calculated, then 1 kHz input will be fine. If 1 kHz is larger than the frequency we just calculated, then you need to use a different opamp.
 

Thread Starter

mehmeter3

Joined Dec 19, 2013
13
I took the bandwidth number as 100 MHZ and divided it by the gain I desired (10000 - I guess we don't represent it in dB). The result is 100 MHZ/10000=10000 where 1 kHz is smaller than the frequency I just calculated and it seems ok. Are there any other basic parameters I should calculate?
 
Last edited:

shteii01

Joined Feb 19, 2010
4,644
I took the bandwidth number as 100 MHZ and divided it by the gain I desired (10000 - I guess we don't represent it in dB). The result is 100 MHZ/10000=10000 where 1 kHz is smaller than the frequency I just calculated and it seems ok. Are there any other basic parameters I should calculate?
I think you got all the basics.

Do you have all the parts? I would try to build a test circuit.
 

GopherT

Joined Nov 23, 2012
8,009
I took the bandwidth number as 100 MHZ and divided it by the gain I desired (10000 - I guess we don't represent it in dB). The result is 100 MHZ/10000=10000 where 1 kHz is smaller than the frequency I just calculated and it seems ok. Are there any other basic parameters I should calculate?
When you say, "I took the bandwidth number as 100MHz...", does that mean you found an op amp model with 100M Hz of gain, or, did you just pick a random number?

100MHz is a very specialized op amp. A more normal number would be 3 to 5 MHz for classic general purpose op amps (TL072, NE5532, ...) and 15 to 40 for newer high end op amps (OPA1632).

All I'm saying is MikeML was on the right path up above if you want to stick to the more generic parts - and likely achieve the intent of your professor's challenge.
 

crutschow

Joined Mar 14, 2008
34,285
The critical parameter here, for determining the required op amp GBW is not the gain and frequency response, it's the required minimum input impedance and frequency response.

For a current (infinite impedance) source the voltage gain of the circuit is 1, thus the bandwidth of the circuit will be equal to the GBW product of the op amp.

A current input op amp circuit has no input resistor, just a feedback resistor. The 100Ω requirement is for the apparent or virtual input impedance of the circuit. For a 100Ω minimum (virtual) input impedance with a 1 MegΩ feedback resistor (to get 1V out with 1μA in) the amp gain must be 1MegΩ/100Ω = 10,000 min. This means the GBW of the op amp must be 1kHz * 10,000 = 10MHz min.
 
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