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Greetings,

This is a scheme of a comparator with dynamic hysteresis.

Green line is the output Red line is the input Blue line is the positive feedback.

I want to ask for the explanation why C1 does not start charging up at 5.3V (like the green line) but has its peak at 4.3V and goes down from there?

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With lots of practice, you may be able to do dynamic KVL in your head.

let's state my assumptions without any of your graphs.

  • Output swing = 4.6+0.7 = 5.3V with Rout=2700 Ohms and Iout= (15-5.3)/2k7=3.6mA
  • feedback loop resistance = (R5+R3+R1 )//R2 (neglect R2 for now)
  • the peak is defined when the voltage output is at peak and input across R1 is near 0 due to R1/R2 ratio
  • initial voltage drop across C is thus determine by KVL ratio of R5 to the rest of the loop.
  • lowering both R3 R5 to 0 would restore full amplitude across C1 with a 10.6v transient from -5.3 to +5.3

Also note relevant to your previous question on same circuit, R5C1 reduces the slew rate, not the Op Amp but only after the zero crossing.

Conclusion: To make V(c1) = 50% peak of total output swing of 10.6V choose R5=R1+R3=3.5k (not 2700)

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  • \$\begingroup\$ I have chosen R5 by the Vo of the op-amp and Iz so my zeners would work with a constant output with any input and frequency. I just wanted to know why the cap doesn't start charging up with the Vout. \$\endgroup\$ – atkristin Jan 8 '18 at 21:01
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    \$\begingroup\$ I thought I explained what you wanted to know. its the ratio of the peak to peak voltage. The cap is not in parallel with the output it is in series with the chain of R's from output which swings 10.6V not 5.3 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '18 at 21:08
  • \$\begingroup\$ So basically ratio 2.7k/3.5k approximately 77% of 0 to + 5.3V due to KVL if I understood correctly? \$\endgroup\$ – atkristin Jan 8 '18 at 21:14
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    \$\begingroup\$ 2.7k/(3.5+2.7) or 43% of 10.6V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '18 at 21:29
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    \$\begingroup\$ Bonus simulation .. feel free to change any value tinyurl.com/yczdbau9 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '18 at 22:13
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It helps to look at this in two independent ways, the instantaneous AC response, and the long term DC response.

In the short term AC case, consider the voltage across the cap as not changing. Any step output will show up at the positive input attenuated by the resistor divider of (R2//R3) and R1. This attenuates by 2.33. A 30 V output step would therefore cause a 12.9 V step on the positive input.

In the long term DC case, just consider the capacitor open. The feedback is thru R2, and attenuated by R1. The positive input, your blue line, will always be close to ground since the R1,R2 divider attenuates by 670 times. 15 V on the output, for example, becomes only 22 mV at the positive input.

In between these two cases, the instantaneous AC effect decays to the long term DC effect roughly according to the time constant (R1 + R3)C1, which is about 300 µs.

The labels on your graph are unreadable, so the numbers can't be verified. However, the general shape of the signals is as expected.

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