Following is the analysis I have done for the circuit: enter image description here

This circuit is a part of the shift register and I am posting only a section which is required for the analysis (By the way, D flip-flop (74LVC1G80) is powered to 3.3 V ) and the output D is connected to another D-FF (74LVC1G80) using R(1 K Ohm) - C(20 pF) filter and so on.

  • Data signal, Din: Frequency = 400 kHz, voltage range = 0 – 3.3 V

  • Clock signal, CLK = 12 Mhz.

  • D-FF input capacitance = 3.5 pF

My intention is to analyze the RC filer and find the input voltage V at the pin D.

  1. Figured out that the purpose of the RC filter is to provide sufficient hold time. But I was trying to understand how the hold time will be improved with an RC circuit at the input and how I can I calculate the hold time mathematically. Guessed in this way, the data logic high or low will stored in the capacitance (20 pF) and this will be available to the D-FF, thus enough hold time. Basically I am not convinced with it and unfortunately there is no reference for this case from google as well, kindly share your views about the purpose of RC filter and how to calculate the hold time.

  2. What will be the maximum voltage (V) at pin D.

See the voltage division approach,

$$V_{max} = V_{Din} \frac{ X_{C\text{(20pF)}}}{ X_{C\text{(20pF)}} + R_{\text{(1k)}}}$$

If I take the voltage division rule, how it is going to be wrong, what is the criteria to select the VD rule?

Another approach:

$$ V_{max} = V_{Din} ( 1 – e ^ \frac{-t1}{R_{\text{1k}} C_{\text{20pf}}})$$

When can I use this formula over the VD rule? How can I select the time t1?

  • \$\begingroup\$ You are getting confused a bit. The RC filter does not have a hold time as it is continuously rising or falling. You need a "sample-and-hold" on the cap to hold its charge long enough to be read as a '0' or '1'. Note that reading a voltage at about 1/2 Vcc will cause an unknown reading, a.k.a. Meta-stable state. To get synchronous readings you need two flip-flops in a row. You have the basic's of more complex circuits. \$\endgroup\$
    – user105652
    Mar 2, 2018 at 6:30
  • \$\begingroup\$ @Sparky256, I viewed in this way, A series resistor and capacitor in parallel with the pin-capacitance would have contributed more to hold time \$\endgroup\$
    – vt673
    Mar 2, 2018 at 6:42
  • \$\begingroup\$ Well, you are oversampling by an extreme amount, but the output of the flip-flop will have jitter due to Nyquist noise. If you only want to know the 'D' input voltage, consider the timing of a 400 KHZ signal. \$\endgroup\$
    – user105652
    Mar 2, 2018 at 6:51
  • \$\begingroup\$ Anything twice the maximum signal will provide maximum spectrum spacing in frequency domain and thus there wont be any aliasing issue, But seems like there should be an upper limit for sampling frequency as well to limit the Nyquist noise. \$\endgroup\$
    – vt673
    Mar 2, 2018 at 6:58
  • \$\begingroup\$ What is the waveform of the 400 kHz signal? Is it synchronous with the 12 MHz clock? \$\endgroup\$
    – Dave Tweed
    Mar 2, 2018 at 12:00

2 Answers 2


This approach is not going to work. You have no relationship in your circuit between the 12 MHz clock, and the incoming signal (which I assume is a nominal square wave, but this doesn't change much).

This means that for any particular input edge, you have no idea where the clock edge will fall, and if you will see a setup/hold violation.

The correct approach here is to cascade your flop with another flop. Although the first flop may not get a clean input, it will resolve to one of the two legal states. The second flop then gets a dramatically improved setup/hold at it's input. The cost here is increased latency - it takes 2 clock cycles on average to sample the input change (could also be 1 or 3).

Metastability affects the flops flowing this one more significantly than the capture flop.

  • \$\begingroup\$ Basically this is the shareable section of the circuit I am working on. And the intention is to understand the use of RC filter and input pin voltage maximum value as well in detail (Kindly see question - 2). \$\endgroup\$
    – vt673
    Mar 2, 2018 at 10:59
  • \$\begingroup\$ Your approach is never going to work. Based on your comment, I need to vote-to-close as 'unclear what you're asking'. \$\endgroup\$ Mar 2, 2018 at 11:01
  • \$\begingroup\$ I believe that shared information is enough to answer the question I have asked. But I could edit the question if there is any modification required. \$\endgroup\$
    – vt673
    Mar 2, 2018 at 11:09

If the 400 KHZ clock is synchronous with the 12 MHZ clock, the issue becomes the time delay set up by your RC filter. Though it is not a sine wave, the math says it creates a delay of 125 nS and a -3 dB roll-off at 7.957 MHZ. A 400 KHZ square wave has edges 1.25 uS apart, for a period of 2.5 uS.

A typical 74AC74 flip-flop has a set-up time of 5.5 nS, and a hold time of 0.5 nS, during which the 'D' input must be stable during the rising edge of clock.

With a 12 MHZ sampling rate you have a rising clock edge every 8.33 nS. With the RC values you have you should get an output that is toggling between '1' and '0' almost constantly, with a 30 HZ jitter rate. I added a drawing to include a jitter filter, also synced to the 12 MHZ clock, but with a 8.33 nS delay it will NOT pass along any noisy output from the first flip-flop.

NOTE: With a synced square wave and small values of R and C, it is possible NOT to have any jitter from the first flip-flop


simulate this circuit – Schematic created using CircuitLab


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