Induction motor controller. where does this circuit get DC power from? - Page 1
Forum section: electronics / repair

author: tolland
author_rank: Newbie
timestamp: June 28, 2015, 07:16:31 am
content: We recently received a donation of a laboratory incubator shaker, Stuart model SI50. The Incubator warms up ok, but the shaker is not working. The shaker platform is driven by a Shaded Pole Single Phase Induction motor, which is controlled by feedback from a optical encoder to read the RPM, and a potentiometer to set the desired speed level in RPM.I took the shaker apart, and I can see that a resistor is burned out on the Induction motor controller board. I can also see that the speed control is being done by some sort of phase control using a Z0107 Triac. The feedback is being controlled by a PIC16C711However there appears to be no low voltage DC power supplied to the board, so I can't see how the PIC is being powered. Is this using some sort of parasitic power from the MAINS voltages into CON1, or am I missing something?There was nothing connected to the JP4 and JP5 connectors, so the power must come from the mains...

author: Psi
author_rank: Super Contributor
timestamp: June 28, 2015, 07:56:27 am
content: Hard to say without seeing the tracks on the under side of the PCBBut i would speculate that the big yellow capacitor and those power diodes forms a mains voltage capacitive DC powersupply.The yellow cap looks physically about the right size (like around 300nF) to take 230VAC and provide ~20mA at whatever DC voltage is needed.

author: SeanB
author_rank: Super Contributor
timestamp: June 28, 2015, 08:13:23 am
content: Big yellow capacitor to the left, D1,D2, R6-R10 are likely that supply. Probably R9 fried because it is being overloaded, but check the big capacitor value as well, it probably will be low. Replace with a proper class X capacitor only.

author: amyk
author_rank: Super Contributor
timestamp: June 28, 2015, 12:31:28 pm
content: Looks similar tohttp://www.eevblog.com/2013/03/06/eevblog-435-3d-rocker-teardown/

author: LukeW
author_rank: Frequent Contributor
timestamp: June 28, 2015, 01:06:16 pm
content: Looks like a classic transformerless power supply.http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdfI'd expect D1 and D2 to be a standard rectifier and a zener.But there doesn't seem to be a high-power resistor in series with the cap as you might commonly expect.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 28, 2015, 01:48:15 pm
content: Quote from: LukeW on June 28, 2015, 01:06:16 pmR10 looks like it might be the deal, though it's not high power.Tim

author: tolland
author_rank: Newbie
timestamp: June 28, 2015, 10:42:24 pm
content: Quote from: Psi on June 28, 2015, 07:56:27 amI've attached an image of the underside, and my attempt at a composite of both sides done in GIMP

author: tolland
author_rank: Newbie
timestamp: June 28, 2015, 11:31:45 pm
content: It's kinda hard to tell what that resistor is gonna be, because the bands are burned in the centre;

author: tolland
author_rank: Newbie
timestamp: June 28, 2015, 11:47:32 pm
content: Quote from: Psi on June 28, 2015, 07:56:27 amthe big yellow cap on the side is this one;

author: tolland
author_rank: Newbie
timestamp: June 29, 2015, 02:15:37 am
content: Quote from: amyk on June 28, 2015, 12:31:28 pmWOW! that is clearly based on a very similar design and configuration. I'd best watch the rest of that podcast and see if Dave shares any brilliant insights on how to fix it.Thanks!

author: tolland
author_rank: Newbie
timestamp: June 30, 2015, 09:41:03 am
content: This seems to be the transformer-less power supply arrangement

author: Gyro
author_rank: Super Contributor
timestamp: June 30, 2015, 04:38:52 pm
content: It looks as if D2 is in the wrong place on your schematic (won't work where it is) so you probably want to re-trace that bit. I think you've written "(filter)" next to R8? That resistor is there to ensure that C4 gets discharged (shock hazard) when the unit is unplugged.As for the reason for R9 failing, I think the most likely reason is C4 shorting on a spike and not self-healing quick enough. If so, the zener D3 might or might not have been man enough to clamp the current (check if it's shorted). C4 will normally act as a AC 'constant current source' such that a short on the PIC supply wouldn't make any significant difference to the current draw (5V difference relative to full mains voltage). Check if C4 is shorted (it might still have self-healed though).R9 may (should) be a fusible resistor, these are sometimes rated a bit close and can slowly degrade over many switching cycles until they fail - surge current through C4 if turned on at a mains peak is higher than in continuous operation.After checking the zener and other diodes you may well find that replacing R9 with a suitably rated fusible replacement (same body size) and also replacing C4 with a suitable X2 rated part as a precaution restores normal operation.

author: tolland
author_rank: Newbie
timestamp: July 02, 2015, 06:28:14 am
content: (I guess I've kind of worked out how to replace this circuit now, but I am using this circuit as an opportunity to reverse engineer and improve my understanding of it)Quote from: Gyro on June 30, 2015, 04:38:52 pmI've retraced and double checked with the DMM continuity, and it looks accurate. Here is a my overlay of the tracks;

author: Gyro
author_rank: Super Contributor
timestamp: July 02, 2015, 09:11:11 am
content: QuoteAh, looks like you haven't traced the circuit quite far enough - if you go just a bit further you'll see that D2 and C6, in combination with Zener D4 are generating a negative supply rail. The schematic as drawn shows D2 charging C6 but nothing discharging it.At a guess U2 is an Op-amp using both positive and negative rails.