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In this monostable circuit the C1 capacitor is reverse biased. When T1 is closed and C1 is discharged what will happen on the right side of C1?

Will it charge up to 0.6 volts so that T2 can be opened and then discharge again over the base-emitter line and than charge through R2?

That would mean that the polarized capacitor could be loaded reverse biased. Is that true?

  • \$\begingroup\$ What is a 'polarized resistor'? In a side note, the schematic looks as if it was NOT designed to work at 9V. T2 Base Emitter reverse voltage is being exceeded and may avalanche. I'd suggest this was originally designed to work at 5V or less. \$\endgroup\$ Apr 27, 2019 at 15:07
  • \$\begingroup\$ My question is not wether the capacitor can withstand. My question is will he charge reverse bias so that T2 can switch. \$\endgroup\$
    – silvan
    Apr 27, 2019 at 16:12
  • \$\begingroup\$ "question is not wether the capacitor can withstand" - And the answer is given in the duplicate question: electrolytic caps behave as normal unipolar caps up to about 1-1.5V. \$\endgroup\$ Apr 27, 2019 at 16:17

1 Answer 1


I entered this into LTSpice to help visualize what happens, which I highly recommend.

At DC with the switch S1 open, R3 provides base bias to T2 turning it on. LED2 stays lit. At this point T2 is saturated (collector ~= 0.2V). R5 holds the base of T1 at about 0.2V, turning it off.

The left side of C1 is around 8V and the right side is at approx 0.6V. So far so good.

When the user closes switch S1, T1 turns on grounding the left side of C1. This will cause the right side of C1 to transition DOWN by about 8V to somewhere around -7V. T2 will see negative Vbe of 7V which WILL be an issue for a BC548. The AbsMax VEBO rating is 5V and the circuit applies 7V. Not good.

Now the right side of C1 begins charging through R3. This is the monostable pulse portion. The right side of C1 slowly charges until reaching about 0.6V and T2 turns on. The collector of T2 goes back to 0.2V and R5 causes Q1 to turn off. This terminates the pulse.

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In the picture above:

Blue trace: Switch S1 (high = switch closed)

Red trace: Current through LED1 (the pulse)

Green trace: Voltage across the capacitor

It's true that C1 does become reverse biased by about 0.5V at the end of the charging cycle. The max reverse bias can be calculated as T2_Vbe - T1_Vce_sat. Use a non-polarized cap if you want to avoid trouble.

  • 2
    \$\begingroup\$ While your polarized capacitor is unlikely to have problems with a small (<1V) reverse bias, you can eliminate this reverse bias entirely by putting a diode (1N914 for example) in series with the collector of T1. This will keep the left side of C1 always positive with respect to the the right side of C1. This will alter the timing, but only slightly. \$\endgroup\$
    – Randy Nuss
    Apr 27, 2019 at 16:55

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