I found a schematic with a 55 timer driving two transistors, each of which switches an LED (see below).

When the 555 output is low, LED1 is OFF and LED2 is ON.

When the 555 output is high, LED1 is ON and LED2 is OFF.

I have the query why it uses the Box 1 circuit, which looks like a common collector circuit, instead of the Box 2 wiring, which is more common.

Does the Box 1 circuit have some advantages over the Box 2 circuit?

Also, I think capacitor C2 is to stop oscillations but I can't completely understand what oscillations it is for. And would so small a capacitance value have any impact?

enter image description here

  • \$\begingroup\$ Given that C2 has a value of 0.01 pF, clearly the circuit cannot be trusted so, it's of no interest unless it has some vital provenance or some relevance to some quest in your life. A circuit on its own (with no provenance) is neither wrong nor right; it just is what it is and we can't read the mind of the creator. \$\endgroup\$
    – Andy aka
    Commented Apr 13, 2021 at 15:54
  • \$\begingroup\$ The circuit, you suggesting, inverting signal. The rest is matter of personal test of designer. \$\endgroup\$
    – user263983
    Commented Apr 13, 2021 at 15:58

2 Answers 2


If C2 were a larger value such as 10nF, it being in parallel with R2 to a transistor base, would be called a speed-up capacitor.

Generally this is done to "speed-up" the base current transition. Remember, capacitors resist a change in voltage. Without the capacitor, R2 completely determines how much current goes into the base. And transistors have parasitic capacitance to the base, so R2 is also charging and discharging this capacitance. That is what limits transition speed, and why a speed-up capacitor is sometimes used.

The bigger the C, the harder it will "push" the base at the moment of voltage change, overcoming it's parasitic capacitances, resulting in faster switching.

As for figure 1, that must be an error. Q1 should be a PNP type (NPN will not work correctly.)

  • \$\begingroup\$ I could see setting C2 to that value in a simulation program so that it would have no real effect, but I could switch it to something larger, later, without lots of tedious editing. But the capacitance across nearly any real-world resistor is going to be way more than 0.01pF, so -- it's meaningless. \$\endgroup\$
    – TimWescott
    Commented Apr 13, 2021 at 19:48
  • \$\begingroup\$ I think this was a classic case of "create a schematic as quickly as possible and post it", i.e. insufficient error-checking. \$\endgroup\$
    – rdtsc
    Commented Apr 13, 2021 at 21:17

A couple errors, one already noted by Sir Andy.

LM555 has a low output push-pull impedance which can easily drive a LED with Vout hi so the entire circuit of Q1 is redundant and non-inverting (hi=On).

Whereas your cct. for Q1 is fine if you want inverting.

Also when driving say a 2V LED to 5V from 5V logic levels, the Rb/Rjc ratios should account for reduced hFE when saturated. Fully saturated at rated current Ic/Ib=10 but here 20 will be fine then allow for Vbe and Vf of LED.


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