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Hello I'm trying to understand how adding a capacitor across R2 in below circuit increases the switching speed. I don't see how C2 supplies charge to C1.

With the additional capacitor C2, it seems the output voltage has to charge one additional capacitor. Hmm.. Or, does the speed increase because capacitors in series reduce the capacitance ? I'm very much clueless and my textbook doesn't explain it well.. Appreciate any help. Thanks!

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The positive feedback path provides the hysteresis that makes it a Schmitt trigger. However, it also forms an RC filter between the parasitic capacitance C1 and the resistor R1||R2. It's possible the hysteresis may not kick in quickly enough for some applications because of this delay.

C2 provides a momentary, relatively high current feedback path to make sure the positive input responds quickly enough to the output.

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  • \$\begingroup\$ Ahh so C2 basically bypasses R2 to charge/discharge the stray capacitance when the output changes ? Without C2, the stray capacitance has to charge/discharge through R2 which might take a while. I think I get this. Thank you :) \$\endgroup\$ – rsadhvika Jun 7 '18 at 16:49
  • \$\begingroup\$ @CristobolPolychronopolis Why do they specify C2 = C1 * (R1/R2)? \$\endgroup\$ – crj11 Jun 7 '18 at 17:22
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    \$\begingroup\$ we want R1*C1 = C2*R2---> C2 = R1/R2*C1 \$\endgroup\$ – G36 Jun 7 '18 at 17:23
  • \$\begingroup\$ That bit of math provides a FLAT frequency response, ignoring all other parasitics and charge injections (such as Miller Effect inside the comparator diff-pair). \$\endgroup\$ – analogsystemsrf Jun 9 '18 at 5:08

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