# offset after coupling capacitor

Discussion in 'General Electronics Chat' started by iceman529, Jun 17, 2013.

1. ### iceman529 Thread Starter New Member

Jun 17, 2013
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Hello, i am working on a project and i have to read some values with an MCU from an the attached circuit i cannot understand why i have the voltage offset in the graph B, and if you could please tell me how to calculate that offset so i can integrated in my formula.

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2. ### kubeek AAC Fanatic!

Sep 20, 2005
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Perhaps because you are using a very lousy simulator, try micro-cap for instance where I celarly see 2V offset. What is the amplitude of the input sinewave?

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3. ### iceman529 Thread Starter New Member

Jun 17, 2013
16
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the same thing happens on the physical board, you can see the ac signal in graph A ±20.63V

4. ### kubeek AAC Fanatic!

Sep 20, 2005
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Ok now I see what those numbers mean. Imagine you delete the 1K resistor, since it doesn´t have any significant efect on the voltage at A, B and C. Now if the top voltage really is 2V, then at B you should have exactly what the sim tells you, i.e. the waveform is shifted up by two volts. Now at C the waveform should be attenuated to roughly +3.7V and 0.3V peaks.

This means you either have the circuit wired incorrectly, or there is something you are not telling like what is the actual supply voltage of the opamp.

5. ### LvW Active Member

Jun 13, 2013
674
100
Is there any dc current through the 82.5k resistor? No, there isn`t.
Thus, point B carries the same potential as the dc source (2 V).

6. ### iceman529 Thread Starter New Member

Jun 17, 2013
16
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the part right to the voltage divider doesn't influence the voltage in point B the voltages are very precise here is a link to the simulation click

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You seem to be missing an obvious point. Under steady state conditions, unless there is DC bias current flowing from the 2V source into the amplifier non-inverting input terminal, the DC potential at B & C will be the same and equal to 2V.

8. ### iceman529 Thread Starter New Member

Jun 17, 2013
16
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i do not know if i am making any sens, I'm not that good at electronics, i have this circuit connected to the ADC in a MCU, if i measure the circuit with the oscilloscope i get the same readings as in the simulation above, i know how to calculate the voltage without the 200nF capacitor. I do not know how to explain the change in signal with the capacitor there. From the simulation i can see that the signal in point B gets an offset of 2.065 and a gain of 0.98.

9. ### kubeek AAC Fanatic!

Sep 20, 2005
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The capactior blocks the DC level, but allows the AC changes to pass through it into the divider. Without any signal coming into it, the voltage at the opamp will be 2V. When you feed AC into the capacitor, it will get attenuated by the voltage divider made of the two resistors, resulting in 20.63*7k5/(82k5+7k5) volt sinewave being superimposed (added)on the 2V base level.

10. ### iceman529 Thread Starter New Member

Jun 17, 2013
16
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i did the math a couple of times, this is the formula that i got: $V_{OUT} = V_{REF}*(1-K)+V_{AC}*K$ where $V_{REF} = 2.048
K=\frac{R_{1}}{R_{1}+R_{2}}$

But this only applies if the capacitor isn't there.

11. ### kubeek AAC Fanatic!

Sep 20, 2005
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Now if the capacitor is there, the Vref*(1-k) changes to just Vref, since nothing is loading the divider at DC, so the DC component wil be the same as vref.

iceman529 likes this.
12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If you want your K factor to be correct for the AC component you will need to include the capacitive reactance which is of the order of 14.7k ohms.

13. ### iceman529 Thread Starter New Member

Jun 17, 2013
16
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So K will become:$K=\frac{R_{1}}{R_{1}+R_{2}+X_{C}}?$
I want to be as precise as i can with my calculations.

14. ### iceman529 Thread Starter New Member

Jun 17, 2013
16
0
if i use the formula that kubeek i will have a better precision but still its not good enough, the peak voltage will be at 3.767 not 3.75 as shown in the simulation, the resistors in the voltage divider are very precise 0.1%. I can see on the simulation that there is a slight negative gain after the low pass filter, the cut off frequency being at 1591hz i do not think it would affect it, how could i include that in the formula? i want to get a as precise as possible at this point, because the ac signal comes from transformer that has a ratio of 15.376 so the error will be multiplied at the end; i can see that my former formula, with the Xc, is wrong after implementing it in the code.