2
\$\begingroup\$

I took apart a power supply unit of some old kitchen lamp, and inside was this push-pull converter. It has an "optional" part, shown at the bottom of the schematic, which is not actually present in the PSU (no components soldered on the PCB, just silkscreen). I tried to reverse-engineer it, and from what I can gather, it is some sort of current limiting or short-circuit protection.

The resistors are of unknown values, as is the electrolytic capacitor. With the transistor I am not entirely sure which terminals are B/C/E, but I chose the configuration which seemed to make the most sense. What I do not quite understand is: how does discharging the starter capacitor help limit an overcurrent? Supposedly an overcurrent would happen once the converter is already self-oscillating. Or does this component serve a different purpose?

Schematic: (made in circuit.js because CircuitLab doesn't support creating transformers with custom windings) Circuit schematic

Edit the circuit. Note: The simulation doesn't seem to produce correct results.

PCB photos:

  • Front
  • Back (flipped, so it aligns with the front)
\$\endgroup\$
2
  • \$\begingroup\$ what type of a lamp? \$\endgroup\$ Commented Feb 21, 2024 at 3:09
  • \$\begingroup\$ @jsotola It's for a halogen lamp, although I suspect it may be a retrofitted flourescent lamp PSU. \$\endgroup\$ Commented Feb 21, 2024 at 3:42

2 Answers 2

Reset to default
2
\$\begingroup\$

Assuming your rendition is accurate (arrangement and type of components), my guess would be an alternative to the clamp diode (that's discharging the startup capacitor, preventing the DIAC firing during operation). There doesn't seem much purpose to base-circuit RC values other than to bias it into conduction after a time delay, whether as a leading-edge delay (100s ns time constant) or averaging over multiple cycles (100+ µs). In any case, it doesn't seem like it could provide current limiting; more likely it's activated by normal operating current flow.

Several components, seems like a poor alternative to a single diode, especially when high-speed diodes are already present in the design (E-C clamp diodes), so it may be that the component types are incorrect, and the function is different. I don't have any likely offhand alternatives to your interpretation of the schematic, though. It doesn't seem an adequate arrangement to replace the startup mechanism, in case the purpose were to substitute the DIAC for example.

\$\endgroup\$
2
  • \$\begingroup\$ I am not completely sure about the component types. The "transistor" could be any 3-terminal device, and the diode is any kind of diode (possibly zener). I added PCB photos to my question. \$\endgroup\$ Commented Feb 21, 2024 at 16:17
  • \$\begingroup\$ But if this component is activated by normal current flow, the voltage across the 1Ohm resistor from emitter to ground has to equal the sum of voltages across the diode, resistor and capacitor (equivalently, diode, both resistors and transistor). There would have to be around 300mA of current just for the diode alone (assuming it's a Schottky). \$\endgroup\$ Commented Feb 21, 2024 at 16:53
1
\$\begingroup\$

Is it something like this ?

Yes, it seems that it is a flourescent lamp drive PSU modified (a little) for halogen lamp ...

Just tried to make it work.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ This seems alright, my PSU is also rated around 30-40W. Where is the top terminal of X1 connected to? \$\endgroup\$ Commented Feb 21, 2024 at 16:06
  • \$\begingroup\$ Top X1 not connected. It is a spark tube for simulating a fluorescent lamp, but it does not work yet. \$\endgroup\$ Commented Feb 21, 2024 at 16:47

Your Answer

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

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.