Period to Voltage Converter

ICA, R1, R2, R3, and Q1 form a current source. The current that charges C1 is given by:


I= VDN x R1
(R1+R2) x R3

= 15 x 3 kW
(3kW +12kW) x 470kW

= 6.4mA


The input signal drives ICD. Because ICD's positive input (V+) is slightly offset to +0.1V, its steady state output will be around +13V. This voltage is send to ICC thorough D2, setting ICC's output to +13V. Therefore, point D is cut off by D1, and C1 is charged by the current source. Assuming the initial voltage on C1 is zero, the maximum voltage (VCmax) is given by:


VCMAX= I x T
C1

= 6.4 x t
0.0047

= 1326t


If t=1 ms, then VCmax=1.362V

When the input goes from low to high, a narrow positive pule is generated at point A. This pulse becomes - 13V at point B, which cuts off D2. ICC's V+ voltage becomes zero. The charge on C1 will be absorbed by ICC on in a short time. The time constant of C2 and R5 determines the discharge period -- about 10us. ICB is a buffer whose gain is equal to (R8+R9)/R9=1.545. ICD's average voltage will be (1362t x 1.545)/2=1052t. R10 and C3 smooth the sawtooth waveform to a dc output.

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