How to bypass regulators when testing?(was:Diagnosing a dying soldering station) - Page 1
Forum section: electronics / beginners

author: taste_tester
author_rank: Regular Contributor
timestamp: June 07, 2022, 08:58:38 pm
content: Let's say there is a clone JBC soldering station which has been in heavy use for ~3 years and has hundreds of working hours on it.It has a beeper on it and in the last month or so instead of powering on quickly and making a nice quick "BEEP" with a constant tone, it takes much longer to power on and then goes "....beeeEEEEEP" with a rising toneDoes it sound like bad capacitors? It uses no-name capacitors on the main board and I can't think of what else could be the problem. It has a big chunky internal transformer which then goes to the main board. (you can see here:https://youtu.be/omS5Tqs_yq8?t=1132)i will also be sure voltages are good and constant on the regulators before i try to replace any capsAs I understand it, it's not safe to monitor the voltages directly from the main board or directly off the transformer by scope without an isolation transformer?

author: ledtester
author_rank: Super Contributor
timestamp: June 08, 2022, 01:55:16 am
content: Quote from: taste_tester on June 07, 2022, 08:58:38 pmYou should be able to troubleshoot a transformer-based linear power supply with just a multimeter.I think the first step should be to find or create a schematic so you can know what points to probe in the circuit.Another approach is to see if you can power the station with a bench power supply. That would tell you if the MCU and analog control parts are working or not. You'll likely need a schematic to figure out how to replace the original power supply with a bench supply.

author: taste_tester
author_rank: Regular Contributor
timestamp: June 08, 2022, 02:04:17 am
content: Quote from: ledtester on June 08, 2022, 01:55:16 amThis would be great, but the main board also rectifies AC and then regulates it to DC. So I was not sure the best way to approach. Do you think i could simply connect the bench PSU to the output leg on the regulator and power it like that?

author: ledtester
author_rank: Super Contributor
timestamp: June 08, 2022, 02:16:11 am
content: It's really impossible to say until more is known about the circuit.

author: MikeK
author_rank: Super Contributor
timestamp: June 08, 2022, 02:33:34 am
content: Quote from: taste_tester on June 08, 2022, 02:04:17 amIf there's no protection diode that's a great way to destroy the regulator. Can you not *safely* check the output of the regulator?

author: taste_tester
author_rank: Regular Contributor
timestamp: June 08, 2022, 04:31:25 am
content: Quote from: MikeK on June 08, 2022, 02:33:34 amNo no , I know how to safely check theoutputof the regulator.what I was trying to figure out is how one safely applies DC power from a bench PSU to a PCB designed to take AC input and regulate it into DC? I don't know where one safely hooks up bench power to try to bypass the power supply section of the PCB in this case.

author: taste_tester
author_rank: Regular Contributor
timestamp: June 08, 2022, 05:46:49 pm
content: To ask my question in a different way, I've found a picture of a random 7805 circuit off google.I want to power the circuit after the 7805 from an external lab power supply, to see if behavior of the circuit changes with a different power supply.How would I safely attach an external DC power supply to the 5V circuit below? Would you attach the bench power supply output to the 0.1uf capacitor after the 7805?

author: ledtester
author_rank: Super Contributor
timestamp: June 08, 2022, 08:29:00 pm
content: Can you disconnect the low voltage side of the transformer from the circuit? If so you can apply 9V DC (perhaps a little more) to the leads marked "9V AC". It won't matter which one you connect to the positive terminal and which you connect to the negative terminal of your DC supply as the bridge will sort it out.But is this the entire circuit powered by the transformer? Are there any other regulators or parts of the circuit that the transformer powers? Where does the heating element of the soldering iron get its power from?

author: Ian.M
author_rank: Super Contributor
timestamp: June 08, 2022, 09:14:32 pm
content: For that specific circuit, after checking the bridge rectifier is good, simply inject DC from your bench supply, set to between 12V and 7V, positive to anywhere between pin 1 (+) of the bridge rectifier and the input of the regulator, negative to bridge rectifier pin 2 (-) or anywhere connected to it by a heavy track. If you wanted to inject a regulated 5.0V do exactly the same but short pins 3 to 1 of the regulator to bypass it.However that's specific to knowing the exact circuit topology and components used. Its only safe because the diodes in the bridge block reverse current through it from the DC side, and I know that a 7805 doesn't have a pull-down transistor in its output stage that will react badly if its internal feedback loop rails. With a different regulator you might need to disconnect it, and inject 5V at its output pad. It also gets a lot more complicated if the device has multiple power rails as the power sequencing can be critical (the order they come up in/shut off in) e.g. *DON'T* try to substitute the power supply on a $1000 FPGA board, unless you want to let out the really$EXPEN$IVE$sort of 'magic' smoke!Also note that soldering stations commonly use a TRIAC to switch low voltage AC to the heating element to power it and if so, it will definately misbehave if fed smooth DC without a line frequency reference pulse to the controller.

author: taste_tester
author_rank: Regular Contributor
timestamp: June 09, 2022, 12:31:31 am
content: Thank you both sincerely for the replies. Ok, so I'm not certain about how the heating element specifically is powered but I know it is powered by two back to back MOSFETs so AC is definitely possible. From there I will decide whether or not it's easier to remove parts and test them out of circuit to find the fault.