# Integrator Circuit Issue

Discussion in 'General Electronics Chat' started by Eoin_oc, Oct 24, 2015.

1. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
Hi all,

I need a single supply integrator circuit that will accumulate charge on a capacitor as long as any positive voltage is supplied at the input.
If the input goes below the current capacitor voltage, current should flow back out of the capacitor toward the input.
Basically if any voltage is present at the input the cap should continue charging (assuming there's enough voltage room in the supply)

Eoin

2. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
What will this be used for? You'll get more feedback if you show your circuit ideas.

3. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
It will be used a for a current meter, it will basically accumulate the voltage across a sense resistor (it is buffered)

4. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
Sorry sorry, i meant to say current should not flow back towards the input! It will be used for a current meter, the input voltage will be from a buffered sense resistor

5. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
I had a typo in my orginial post, the voltage on the cap should never decrease, only increase, a standard single resistor and cap integrator wont seem to do this for me.

6. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
Current is usually sensed with a resistor.

7. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
Now you seem to be describing a peak detector...

PeterCoxSmith likes this.
8. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
Thanks for that piece of information, irrespective of where this voltage is coming from, the circuit I need should continue charging a capacitor as long as an input voltage is present at the input, the voltage should never decrease on the cap.
Even if the current voltage on the cap is 5V and there is only a few mV present at the input, the cap should continue charging up.

9. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
Should the charging current for the cap be constant or depend on the input voltage?

Seriously, if you post circuit and your ideas, it won't seem so much like some tedious guessing game.

10. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
It should depend on the input voltage, I had a quick look and I think a peak detector is similar to what I need, but the voltage should keep increasing.
This is just a prototype for now I don't even have a schematic. Is there anyway around the diode drop of a normal peak detector?

11. ### #12 Expert

Nov 30, 2010
16,355
6,852
It's called a precision rectifier.
It's an op-amp circuit.

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12. ### dl324 Distinguished Member

Mar 30, 2015
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But a peak detector will only charge the capacitor if the input voltage is higher than the cap voltage.

13. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
Ok, any other circuits that will keep charging but have the non-discharge properties of the peak detector?

14. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
Or put the diode in the feedback loop.

#12 likes this.
15. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
See post #9.

Are you concerned with capacitor leakage?

16. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
Small leakage of the cap itself is fine, so no it will not be in an issue

17. ### dl324 Distinguished Member

Mar 30, 2015
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626
You still haven't stated how the charging current should, or shouldn't vary with input voltage.

18. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
It should depend on the voltage, I already stated this in post #10

19. ### dl324 Distinguished Member

Mar 30, 2015
3,250
626
For this situation, a few mV would not cause a capacitor already charged to a higher voltage to continue to charge. You need to define how capacitor charging current is to be determined.

20. ### Eoin_oc Thread Starter New Member

Oct 24, 2015
23
0
Ok I understand the problem, the charging current should be directly proportional to the input voltage.
Ic = Vin/ R.