I have the below circuit with...

R1 100 Ohm
R2 10k Ohm potentiometer
R3 10k Ohm potentiometer
R4 100k Ohm
R5 4.7k Ohm
R6 47 Ohm
C1 470 uF
D1 1N4148
D2 1N4148
T1 BC337 NPN
T2 2N2222 NPN
IC1 555 timer
LED1 white, 25mA, 3.4V

Circuit schematic

...but no cyclic LED fading; on power up, it lights up and then fades out exactly once. However, when I wire the input of D2 to the 555's pin 3, I obtain cyclic "flashes" that I can adjust with the two potentiometers. What could be the problem here? Thanks!


2 Answers 2


I suspect that when power is applied C1 charges which fades out the LED but the voltage on C1 never reaches 2/3 of Vcc due to the potential divider formed by R1, R2, D1, D2, R4 and the base-emitter junction of T1. So the circuit just sits there with no further rise of C1's voltage.

You could try increasing the value of R4 to try and get C1's voltage to rise further but don't increase R4 so much that T1 has insufficient base current.

  • \$\begingroup\$ I added a 47k Ohm resistor in series to R4, and tried a 330k Ohm resistor instead of the original 100k Ohm resistor, but to no avail - a single lighting up and fade out. \$\endgroup\$ Mar 13, 2022 at 13:38
  • \$\begingroup\$ After arriving at 1M Ohm, it works as intended. \$\endgroup\$ Mar 13, 2022 at 14:29

T1 (thanks for the reference designators!) is wired as a switch rather than as a linear amplifier or buffer. As such, it is "converting" the capacitor's voltage ramp into a very rapid ramp that looks like a switch to the eye.

I don't see any reason for D2, and in fact I think it hurts what you are trying to achieve. I recommend:

  1. Eliminate D2. Connect the left side of R2 directly to pin 2.

  2. Disconnect the T1 emitter from GND.

  3. Disconnect the T2 base from the T1 collector.

  4. Connect the T2 base to the T1 emitter.

This turns the T1-T2 transistors into a Darlington pair, a current amplifier that maintains the input voltage wave shape. BUT -

The circuit probably will not run well on 5 V. The capacitor voltage ramps up and down between 1.67 V and 3.33 V. 3.33 V minus two Vbe forward voltage drops does not leave much to run the LED. You might have to adjust R6 to see anything happen. A simple thing to try is to run the circuit on 9 V or 12 V.

The problem is the high impedance and relatively small voltage change across the timing capacitor. A common way around this is to add a second R-C network at the 555 output (pin 3) and connect the darlington pair to that.

  • \$\begingroup\$ All right, thanks. As commented above, I just got it to work with a 1M Ohm resistor instead of 100k Ohm resistor for R4, but I may try your suggestion also. Still wondering why I was off with R4 by a factor of 10. \$\endgroup\$ Mar 13, 2022 at 14:33
  • \$\begingroup\$ With that large a base resistor, T1 is being starved out of saturation and into its linear operating region. T2 now is seeing an inverted ramp rather than a steep edge. Way back, this was called "dangle-biasing". Its only real value was to show physically that transistors do have a finite gain that can be measured. The problem with it is that a transistor's gain changes with temperature, barometric pressure, collector current, base current, Vce, and just about everything except the phase of the moon. \$\endgroup\$
    – AnalogKid
    Mar 13, 2022 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.