AN1812 datasheet(7/34 Pages) STMICROELECTRONICS

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AN1812

4 Circuit analysis

7/34

Circuit analysis

In this analysis, it is assumed that there is no noise present and the i/p offset voltage of the op-

amp is negligible.

Now,

I = (V1 – Vin) / R= C * dVc / dt (where Vc = Vout – V1)

Where, current ‘I’ is constant for a given input. Applying the Laplace transform:

(V1 – Vin) / s * R = C * (s Vc(s) – Vc (0+))

or,

(V1 – Vin) / s

2 = (R * C) * ( V

c (s) - Vc(0+) / s)

Applying the inverse Laplace transform, we get

(V1 – Vin) * T = (R * C) * ( Vc(t) - Vc(0+) )

------------------- (1)

Now as shown in the Timing Diagram -

At T = T2;

Vc(T2)= V2 – V1

And, at T =T3;

Vc (T3)= V3 – V1

So,

(V1 – Vin) * T2 = (R * C) * (V2 – V1 - Vc(0+))

--------------- (2)

And,

(V1– Vin) * T3 = (R * C) * (V3 – V1 - Vc(0+))

--------------- (3)

Equation (2) and equation (3) can both be used as the characteristic equation for this converter,

but factors like Vc(0+) and other constant errors remain present. But if we use both

comparators, then we can remove these factors by subtracting equation (2) and equation (3).

Now after subtracting equation (2) from equation (1) and rearranging we get-

Vin = V1 - (R * C) *( V3 – V2 ) / (T3 –T2)

-------------- (4)

Let measured time T3 - T2 = Tm. So we get:

Vin = V1 - (R * C) * ( V3 – V2 ) / Tm

-------------- (5)

By using equation (5) we can measure the value of Vin depending on the value of T3 and T2.

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