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KCC's Quizzes AQQ289 about a pipeline Analog-to-Digital Converter

1. First a funny quote to start the week-end: "Coworkers are (sometimes) like Christmas lights. They hang together, half of them don't work and the other half aren't so bright" anonymous

2. New quiz AQQ289 about a Pipeline ADC:

Sorry for our non-electronic colleagues, but this one is a tough one...

Here is a block diagram of a 3-stages pipe line ADC.

With Vref = 5 volt and Vin = 2 volt, fill-in the various voltages observed at various nodes.           

In particular, indicate the status of the bits B2, B1 and B0 at the end of the third active clock.

It is assumed here the components used (S/H, subtractor, amplifier, mux, comparator are all ideal (i.e. zero delay, no offset, no loss)

Good luck, and try to be among the first ones!

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  • CLK Vx2 Vy2 Vw2 Vx1 Vy1 Vw1 Vx0 Vy0 Vw0 B2,B1,B0
    1 2,0V 2,0V 4,0V B2=0
    2 4,0V 1,5V 3,0V B1=1
    3 3,0V 0,5V 1,0V B0=1

    The status of bits B2, B1ubd B0 at the end of the 3rd active clock is 011

  • To fill in the voltages and bit values for a 3-stage pipelined ADC with ideal components, we need to understand the basic operation of each stage. Each stage typically consists of:

    1. Sample and Hold (S/H)
    2. Subtractor
    3. Amplifier (usually gain of 2)
    4. Comparator (1-bit ADC)
    5. Multiplexer (MUX)

    Given:

    • Vref = 5 V
    • Vin = 2 V
    • Each stage resolves 1 bit, so B2 is MSB, B0 is LSB.
    • Ideal components: no delay, no offset, no loss.

    Stage 0 (First Clock Cycle)

    • Vin = 2 V
    • Comparator compares Vin to Vref/2 = 2.5 V
      • Since 2 V < 2.5 V → B2 = 0
    • Subtractor: 2 V - 0 × 2.5 V = 2 V
    • Amplifier: 2 V × 2 = 4 V
    • This becomes input to next stage.

    Stage 1 (Second Clock Cycle)

    • Vin = 4 V
    • Comparator compares 4 V to 2.5 V
      • 4 V > 2.5 V → B1 = 1
    • Subtractor: 4 V - 1 × 2.5 V = 1.5 V
    • Amplifier: 1.5 V × 2 = 3 V
    • This becomes input to next stage.

    Stage 2 (Third Clock Cycle)

    • Vin = 3 V
    • Comparator compares 3 V to 2.5 V
      • 3 V > 2.5 V → B0 = 1
    • Subtractor: 3 V - 2.5 V = 0.5 V
    • Amplifier: 0.5 V × 2 = 1 V (not used further)

    Final Table

    CLK Vx2 Vy2 Vw2 Vx1 Vy1 Vw1 Vx0 Vy0 Vw0 B2 B1 B0
    1 2.0 2.0 4.0
    2 4.0 4.0 3.0 0
    3 3.0 3.0 1.0 0 1 1
    • Vx: Input to comparator
    • Vy: Input to subtractor
    • Vw: Output of amplifier
  • CLK    1    2      3

    Vx2    2V  2V    2V

    Vy2    2V  2V    2V

    Vw2   4V  4V    4V

    Vx1    0V  4V    4V

    Vy1    0V  1.5V 1.5V

    Vw1   0V  3V    3V

    Vx0    0V  0V    3V

    Vy0    0V  0V    0.5V

    Vw0   0V  0V    1V

    B2      0    0      0

    B1      0    1      1

    B0      0    0      1

    The final conversion gives us an ADC output of 011, or a converted voltage of 1.875V.

  • clk vx2 vy2 vw2 vx1 vy1 vw1 vx0 vy0 vw0 b2b1b0
    1 2 2 4 0 0 0 0 0 0 0,0,0
    2 2 2 4 4 1.5 3 0 0 0 0,1,0
    3 2 2 4 4 1.5 3 3 0.5 1 0,1,1
  • 2, 2, 4,0,0,0,0,0,0; 0, 0, 0
    2, 2, 4, 4, 1.5, 3, 0, 0, 0; 0, 1, 0
    2, 2, 4, 4, 1.5, 3, 3, 0.5, 1; 0, 1, 1

  • The pipeline ADC uses successive compare, subtract, and multiply by 2 to get a complete conversion. Usually, the conversion is done with a recirculating remainder to achieve the result. The main contributors of error are the x2 multiplication, and any offsets in the S/H.

     

  • CLK Vx2 Vy2 Vw2 Vx1 Vy1 Vw1 Vx0 Vy0 Vw0 B2 B1 B0
    1 2 0 4 0 0 0 0 0 0 0 0 0
    2 2 0 4 4 1.5 3 0 0 0 0 1 0
    3 2 0 4 4 1.5 3 3 0.5 1 0 1 1
  • Unfortunately, a poorly described problem, as the initial state of the 3 sample and hold blocks
    is unspecified.

    Commonly, for this type of ADC, the clock is running continuously, and as the input is specified
    as 2 volts, the ADC reaches steady state after 3 clocks regardless of the initial state of the hold blocks.

    Alternatively, if the 3 hold circuits are at 0V initially, then the result is