# Inital charge in capactior for integrator

Discussion in 'General Electronics Chat' started by forbi, Nov 6, 2014.

1. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Hi,

I would like to know what can cause the initial charge inside my capacitor in an integrator circuit ?

Because for a simple integrator circuit with 10kohm input resistor and a 1uf capacitor on the feedback to the inverting pin. I also have a nulling circuit for the op amp to reduce the offset voltage to 0v before hand.

There is an DC offset voltage of appro 7.2V when I using the oscilloscope to measure.

All the ground are to a common ground from the power supply 0V.

2. ### crutschow Expert

Mar 14, 2008
16,569
4,475
When do you see this 7.2V?
What are the test conditions?

3. ### forbi Thread Starter Member

Sep 11, 2012
37
0

The circuit is similar this. with Rin as 10kohm and Capacitor at 1uf.

I did some testing, I saw this 7.2V DC value across the capacitor.
DC value across input resistor is 0.02V.

Supply voltage to op amp is +- 9V.

Input signal is a square wave of Vpp 15V.

AC value across input resistor is appro 7.5V similar to input signal voltage Vp of 15V .

How come there is this DC value across the capacitor ?

4. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
This is interesting but have you noticed that 7.5V is exactly half of 15V.
How can you be measuring a DC voltage across the resistor if you have a square wave of 15Vpp. Is that 15Vpp +-7.5V?
Also, why measure across the capacitor? Would you not measure across vout?

What is the frequency of the input square wave?

All the best!
Robin

5. ### forbi Thread Starter Member

Sep 11, 2012
37
0
The Capacitor is connected to a multiplexer 4053 at the output pin of Y and output of X. before connecting back to the Vout at the op amp .

FYI the capacitor is a ceramic capacitor without polarity.

6. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Yes. the square wave is 15vpp of +- 7.5v .

Result for Vout in AC condition are correct as per calculated but there is just DC offset which I have no idea where it came from.

7. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
Ok could you give me the precise model number for the 4053 (like CD4053G something).

If you want to use the voltage across the capacitor you may need to (and probably should), use buffers. The current capability of that capacitor is going to be weak and if a device draws too much then the voltage will change and as a result the whole biasing of the system will also change.

8. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,071
Could you post the schematic of how you have this hooked up.

Robin Mitchell likes this.
9. ### forbi Thread Starter Member

Sep 11, 2012
37
0
CD4053BE for multiplexer.

This is the diagram on which im connecting.

Supply voltage to op amp and multiplexer is +-9V
Pin 6 and pin 8 of multiplexer is tie to ground. Vdd is +9V and Vee is -9V.

10. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
Start by disconnecting the 4053 from the integrator. Then check the DC level across the capacitor. If it still present then let me know. If you are taking the voltage across the capacitor is that not the same as across Vout since V- is a virtual ground and should have the same voltage as the +pin which is also tied to ground. In which case, connect an output buffer to the op-amp integrator and then have that fed into the 4053.

All the best,
Robin

11. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
Might have found your problem and its to do with frequency!

What is the input frequency?

12. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
If you use a high frequency then the offset voltage arises:

If you use a lower frequency then you get this:

13. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
Initially your integrator starts at the highest output voltage. Because your resistor capacitor combination (RC), cant respond fast enough to the input wave (if that makes sense), the output cant reach all the way to -7.5V in time. (this is difficult to put in words).

So you need to choose an RC value that matches that of your input square wave. You cant stick any frequency in and expect to get a triangular wave out. If that is what you want then let me know and I can give you a two op amp triangular wave with no negative voltages!

14. ### forbi Thread Starter Member

Sep 11, 2012
37
0
im using 80 Hz, 10Vpp (+-5V) square wave.. and the capacitor will have this offset voltage.

I also notice if I put the frequency to be like 5Hz there isn't any offset voltage..

How come it is taking in such a low frequency ?

15. ### forbi Thread Starter Member

Sep 11, 2012
37
0
or it might be because the offset voltage is too small to be noticeable at lower frequency ?

16. ### forbi Thread Starter Member

Sep 11, 2012
37
0
My RC time constant value is 10ms for my configuration. So the frequency below 100Hz will me able to match up the my RC time ? But I'm getting it for at 80Hz, unless I set it to 5Hz I wouldn't see this DC offset voltage appearing.

17. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
The RC Constant is for 0.7 of the charge time. So that sounds about right. Try replacing the capacitor with one that is 0.5uF (or even 100n)

18. ### forbi Thread Starter Member

Sep 11, 2012
37
0
I'm confused. What it means by RC constant is 0.7 of the charge time ? isn't RC time constant = (Rin)(C) = (10k)(1uF) = 10ms ?

19. ### Robin Mitchell Well-Known Member

Oct 25, 2009
803
252
Use a 22k resistor and a 0.1uF capacitor

Oct 25, 2009
803
252