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Dual slope ADC numerical (12 kHz clock)

Short answer

Using N2 = N1·( Vin/ Vref) gives 1000 counts. With clock period 83.33 µs, T1 = T2 = 83.33 ms, so conversion time = 166.67 ms.

A dual slope integrator with a clock frequency of 12KHZ, Vref = 100mv with 1000 clock pulses set for T1. Find out the digital counter output if analog input equals 100mv. Find the conversion time.

In a dual slope ADC the input is integrated for a fixed time T1 (fixed number of clock pulses N1), then the reference is integrated back to zero. The count N2 taken in the de-integration time T2 is the digital output.

Given

  • Clock frequency f = 12 kHz, so clock period T = 1/f = 1/12000 = 83.33 µs
  • Fixed count for T1: N1 = 1000 pulses
  • Vref = 100 mV, Vin = 100 mV

Basic relation

During T1 the integrator output reaches Vin·T1/RC. During T2 it falls back to zero at slope Vref/RC. Equating:

Vin·T1 / RC = Vref·T2 / RC
T2 = T1 · (Vin / Vref)
N2 = N1 · (Vin / Vref)

The RC value cancels, so it is not needed.

Digital counter output

N2 = 1000 × (100 mV / 100 mV)
   = 1000 counts

The counter shows 1000 (when Vin = Vref the de-integration takes exactly as many counts as the integration).

Conversion time

T1 = N1 × T = 1000 × 83.33 µs = 83.33 ms
T2 = N2 × T = 1000 × 83.33 µs = 83.33 ms
Tc = T1 + T2 = 83.33 + 83.33 = 166.67 ms

Answer: digital output = 1000 counts; T1 = T2 = 83.33 ms; conversion time ≈ 166.67 ms.

Note: this is also the maximum conversion time, because Vin = Vref is the full-scale input. A smaller input gives a shorter T2, while T1 stays fixed at 83.33 ms.

✓ Answer checked by Claude Opus 5.5 (high effort)

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