How to understand the circuit in MS3110

[Chengjun Li]

I will use MS3110 to measure tiny capacitance difference in a MEMS device. For better use it, I try to understand the theory of operation. Could anyone tell me how to understand the MS3110 IC's circuit?

Transfer function

I have some idea after studying and discussing with others.
1. CS1IN//CS1=CS1IN + CS2IN=CS1T CS2IN/CS2=CS2IN+CS2=CS2T
so the capacitor bridge on the left side can be simplified as two capacitors,CS1T and CS2T, connected in series.

2. A clearer diagram of the two sources is shown below,

The two switches are 180 degree out of phase. When the upper switch connects to 2.25V, the lower switch connects to 0V, vice versa.
I thought this may be considered as switch circuit, and a capacitor in a switch circuit can be consider as a resistor with resistance R=1/Cf, f is the switching frequency.
So the circuit can be further simplified as 2 resistors connected in series.

3. The noninverting input of the amplifier V+ is equal to the inverting input V-, ie; V+=V-=2.25V.
So the center of the capacitor bridge has the same potential with V-, which is 2.25V.

The final version of simplified circuit is like below.
when the upper source is 2.25V, the lower source is 0V, we have a simplified circuit like the following:

Current flow from high potential to low potential(p3 to p2), i=2.25/R2;
when the upper source has 0V and lower source has 2.25V,

Current flow from p3 to p1, i = 2.25/R1;
Suppose the simplest situation, R1=R2, then we have a constant current flowing through feedback capacitor Cf , we know that i = Cf*d(Vout-2.25)/dt, from this equation, Vout will become bigger and bigger as time goes by, that definitely is not the real case.The amplifier should output a constant voltage, like shown in the transfer function. I think I must done something wrong. Hope anyone could point out my mistake.

 

[crutschow]

You can't convert the input capacitors to fixed resistors to calculate the output in that manner. The circuit generates a pulse of charge through the capacitors when the voltage is switched by the two switches, with the amount of charge at the op amp input proportional to the voltage step and the capacitance difference. This charge is integrated by the integrator to give a step voltage. The continuous switching of the switches gives a square-wave step at the integrator output which is filtered to give the IC output.

 

[Chengjun Li]

You can't convert the input capacitors to fixed resistors to calculate the output in that manner. The circuit generates a pulse of charge through the capacitors when the voltage is switched by the two switches, with the amount of charge at the op amp input proportional to the voltage step and the capacitance difference. This charge is integrated by the integrator to give a step voltage. The continuous switching of the switches gives a square-wave step at the integrator output which is filtered to give the IC output.

Hi,
I make some analysis according to what you said.

Here is the simplified circuit for MS3110.

When S1 is connected to 2.25V, S2 connected to ground, C1 has 0 potential difference, according to Q=CV, there should be no charge on the C1 parallel plate. C2 has potential difference of 2.25V, so positive charge with amount of C2*2.25 accumulate on the upper plate of C2 while same amount negative charge accumulate on the lower plate. Positive charge on the upper plate is the result of same amount electrons leave the upper plate, these electrons can only move to the left plate of Cf since C1 has no charge accumulated, so the potential difference Vout - 2.25 = C2*2.25/Cf.


When S1 is connected to ground,S2 is connected to 2.25V, then C2 has no 2.25-2.25 = 0V, so no charge accumulation on the parallel plates of C2. C1 accumulate charge with amount of 2.25*C1, Cf has correspondingly same amount charge which led to Vout = 2.25*C1/Cf + 2.25.

If my analysis is reasonable, just like you said, the output voltage should be a square wave, but in the MS3110 datasheet, the output is a DC voltage

Why output voltage is proportional to the capacitance difference? And where does the 1.14 term come from? Does it come from the low pass filter?

Thanks.

 

[crutschow]

The low-pass filter averages the square-wave value to give a DC output.
Not sure where the 1.14 factor comes from.

 

[Chengjun Li]

The low-pass filter averages the square-wave value to give a DC output.
Not sure where the 1.14 factor comes from.

The signal going into the low pass filter is a square wave with a high level of 2.25*C1/Cf +2.25 and low level of 2.25*C2/Cf +2.25(suppose C1 is bigger than C2). This DC information contained in this square wave is the average of high level and low level which is 2.25*0.5*(C1+C2)/Cf +2.25, this is different from what's shown in the transfer function even if we ignore 1.14 factor. () . In the transfer function the output voltage is proportional to (C1-C2) . What do you think about this？