# AC-Coupled Current Input to Amplifier

1. Sep 29, 2015

### PhysicsGuy99

Hello all,

I am looking for a means of AC-coupling a current input to an amplifier http://www.thinksrs.com/products/SR570.htm . I found some methods to do this with a simple op amp set-up, but I am not sure how to accomplish this with this instrument. The reason I want to remove the DC current component is because I am trying to measure a very small (~6 orders of magnitude smaller) AC signal riding on the DC signal. The instrument does allow for a 0 to +_5 mA offset, but this is not a large enough range for my application. It seems that there is a potentially simple solution to this problem that I am overlooking.

Any insights would be much appreciated!

Thanks,
Alex

2. Sep 30, 2015

### Daz

Is there any reason why you cannot simply use a DC blocking capacitor? Frequency too low? Phase shift too great?

3. Sep 30, 2015

### Staff: Mentor

4. Sep 30, 2015

### PhysicsGuy99

I think I can use a dc-blocking capacitor, but from my readings I believe it is not just as simple as putting a capacitor in series with the signal. It seems that I need a sink for the dc component of the signal, but I am not sure how that all works, particularly because I am using a "black-box" amplifier. I know the input resistance for various gain settings, and there is a simplified circuit diagram in the user manual I provided above.

5. Sep 30, 2015

### Daz

You need a return path for the DC component - that makes sense. A large inductor across the input would provide that DC path while presenting a high(er) impedance to the AC signal. I guess it depends on the size of the DC component and the frequency of the (wanted) AC signal. A signal transformer might also work, as suggested above. You could also use a precision current source to subtract the DC current. One advantage of the current source is that there are no phase shifts to worry about.

6. Sep 30, 2015

### Baluncore

AC coupling implies a high-pass filter with a specified cut-off frequency. What is the cut-off frequency you require?

The SR570 has a “variable input offset current” adjustment. How much offset do you require?
If you null the input current before a series of measurements you will have eliminated the initial offset current.

7. Sep 30, 2015

### Staff: Mentor

Is the DC offset constant in time to 6 orders of magnitude?

Can you sample the signal with a 32 bit ADC, then remove the offset digitally?

8. Sep 30, 2015

### Baluncore

If the wanted signal is 6 orders of magnitude smaller than the bias current then any offset bias may need to be accurate to 6 orders of magnitude.

An alternative is to connect a big inductance across the current meter. That will act as a DC pass while the small changing AC signal is measured. The inductor will need to be screened to prevent it picking up stray noise. You may need a low-leakage flyback-diode across that inductor. You may also need to compensate for the series resistance of the inductor with the adjustable offset voltage of the meter.

The value of inductance required comes down to impedance and the bandwidth you need.

9. Oct 1, 2015

### PhysicsGuy99

The input current I require is order of 10-100mA. Well outside the range of the preamp. I am going to try just placing a capacitor in series with the signal and see if that does the trick.

10. Oct 1, 2015

### Baluncore

You would do better connecting a big inductor across the meter terminals.

11. Oct 2, 2015

### meBigGuy

The problem with just capacitively coupling is that the current source will saturate trying to produce the dc current.

Can you give a little insight with regard to the frequency range of the DC (you said it varied), and the frequency range of the recovered signal.

Have you tried the instrument's high pass filter function (That's a dc blocking filter, isn't it?)

Or, is that filter too far down the chain to not saturate from the DC component?