Cannot figure out the function of this small transformer - Page 1
Forum section: electronics / beginners

author: hgg
author_rank: Frequent Contributor
timestamp: November 10, 2016, 07:21:46 pm
content: Hi,I am trying to figure out the function/purpose of the small transformer L103 in the attached schematic.It measures 1.6uH on one side and 11.3mH on the other.This might be trivial but I cannot understand its purpose.Can someone shed some light ?Thank you.George.Schematic:

author: PA0PBZ
author_rank: Super Contributor
timestamp: November 10, 2016, 07:50:16 pm
content: I think it is a current transformer, also because I see nothing else that is measuring the current trough the mosfet.

author: xygor
author_rank: Regular Contributor
timestamp: November 10, 2016, 07:51:23 pm
content: It's used to sense current to feed back to the regulator IC, either to control the current or more likely to cause an abrupt temporary shutdown.

author: hgg
author_rank: Frequent Contributor
timestamp: November 10, 2016, 07:56:36 pm
content: Its connected in series with the positive DC output of the bridge rectifier D101.How can it sense anything since its on DC?Is it there to detect some kind of current change?

author: xygor
author_rank: Regular Contributor
timestamp: November 10, 2016, 07:59:11 pm
content: Yes. On each switching cycle, the current ramps up. This is where the DC gets changed to AC to get through the transformer where it is rectified back to DC on the secondary side.Edit: I meant the main transformer above, but the statement also applies to the transformer you are asking about. Also fyi, if you want to lookup more information, this is a forward converter (not a flyback).

author: jeroen79
author_rank: Frequent Contributor
timestamp: November 10, 2016, 08:11:22 pm
content: You mean L183?As it is behind rectifier D181, wouldn't it only sense the ripple current? (AC)The transformer goes to transistor Q102 which will conduct when current flows clock wise through the transformer's circuit.The will influence R108 and R107 which in turn have something to do with the FET control IC101.If I am not mistaking the transformer's polarity dots the windings are in counterphase and the clockwise flow in the secondary will occur when the AC component in the primary goes towards the rectifier and capacitor C105.

author: chris_leyson
author_rank: Super Contributor
timestamp: November 10, 2016, 08:12:35 pm
content: Hi, it's a current transformer. The 1.6uH winding is the primary or current sense winding and the 11.3mH winding is the seconary. From the measurments you've given the turns ratio would be sqrt(1.6E-6/11.3E-3) = 1:84 so the secondary current will be 1/84 th of the primary current.. Forward converter current, flowing from left to right in the schematic, will develop a negative voltage across the secondary load R111 C110. When the voltage across the secondary load R111 is greater than the drop across Zenner D105 + 0.8V(ish) then Q102 will start to turn on and that will act as a primary current limit and remove the drive to Q101. IC101 is probably a voltage mode controller because of the primary/secondary voltage feedback using a TL431 and opto coupler PC101. A part number for IC101 would be helpfull, does Q101 drive the voltage feedback loop in which case it would be an average current limit or does it drive a dedicated "over current" pin for fast protection ?.

author: xygor
author_rank: Regular Contributor
timestamp: November 10, 2016, 08:16:55 pm
content: Since it is a forward converter, the transformer is likely providing a current ramp to the control IC. The FET gets turned off when the current reaches a point set by the voltage feedback (from the secondary side).My abrupt shutdown statement would be more for the case of a flyback converter.

author: xygor
author_rank: Regular Contributor
timestamp: November 10, 2016, 08:20:20 pm
content: Or as chris_leyson said, it could be using voltage mode. Depends on IC101.

author: hgg
author_rank: Frequent Contributor
timestamp: November 10, 2016, 08:22:53 pm
content: IC101 is the following :

author: xygor
author_rank: Regular Contributor
timestamp: November 10, 2016, 08:38:09 pm
content: Do you need to know which type it is? You could scope it. (But not if you are beginner. This is a dangerous measurement to make.) Or short the output and observe how it behaves (again, not if you are a beginner.) Or, if possible, find the data sheet for the IC.

author: hgg
author_rank: Frequent Contributor
timestamp: November 10, 2016, 08:43:48 pm
content: I can not scope it as it is.A capacitor (C106) exploded and was completely annihilated making a huge mess...I was trying to figure out how the circuit works in order to see what othercomponents could have been damaged.

author: hgg
author_rank: Frequent Contributor
timestamp: November 10, 2016, 08:55:31 pm
content: Thank you guys!chris_leysonthank you for the detailed explanation.

author: xygor
author_rank: Regular Contributor
timestamp: November 10, 2016, 09:03:10 pm
content: Which is C106? (I can't make it out on the schematic.)Change the bad parts and feed maybe 15V to the line input. Monitor this current and limit it initially to maybe 100 mA. Short the necessary resistors to get 15V to IC101 pin 4. (pin 6 is GND) and see if you get gate drive.

author: hgg
author_rank: Frequent Contributor
timestamp: November 11, 2016, 06:11:34 am
content: C106 is connected to the MOSFET source Q101 and diode D104.I will try to replace all the damaged parts (a 4A diode, two wire-wound resistors and the capacitor)and repair the pcb solder connections. The MOSFET tests ok. The resistors were broken off theirsolder connections.Have a look at the damage done to the current transformer:The plastic cover and the potting compound broke off completely.Luckily no damage to the windings. That cap exploded with some force...By line input I assume that you mean 15VDC to the +/- of the bridge rectifier output, correct?

author: jonovid
author_rank: Super Contributor
timestamp: November 11, 2016, 06:46:18 am
content: see in your Schematic:current sensing transformer. idea for an ac power supply unit. I did think why not,add a little transformer back to front windings to drive a current sensing circuit.maybe Dave can do a fundamentals video on the best way to sense high current in a power supplywhere high voltage isolation is needed. ?

author: hgg
author_rank: Frequent Contributor
timestamp: November 11, 2016, 06:57:42 am
content: QuoteThat would indeed be a good idea.I don't see a lot of info on current sensing transformers.

author: xygor
author_rank: Regular Contributor
timestamp: November 11, 2016, 01:48:44 pm
content: Quote from: hgg on November 11, 2016, 06:11:34 amThat works too, but I meant the AC input.

author: hgg
author_rank: Frequent Contributor
timestamp: November 11, 2016, 03:07:39 pm
content: I think I have a 24VAC transformer to try the AC line input.Which resistors do I have to short for this test ?? R104/R105?They measure ok but I will replace those two.I don't have any wire-wound resistors though.Can I replace them with carbon ones?The thing is I don't have their power ratings.The small one measures 10.8 x 3mm and the large one 19 x 5mm.

author: xygor
author_rank: Regular Contributor
timestamp: November 11, 2016, 05:41:34 pm
content: No, short R117 and the other one in series with it that connects to the bridge rectifier. This is the bootstrapping supply that would normally power IC101 until the capacitor that blew up supplies the power. I don't understand why the capacitor blew so violently. The IC101 power supply (D104, C106) should not have that kind of energy available. Perhaps the main transformer or T103 primary to secondary arced or shorted.R104 is part of a snubber. R105 is the FET gate resistor. Yes, you can use carbon, but R104 especially needs to have sufficient power rating. They are not wirewound, they are film. What you see is a spiral cut.Edit: Correction: No need to short R117 for the test, just the other one. Don't forget to remove the short before going back to full voltage!

author: xygor
author_rank: Regular Contributor
timestamp: November 11, 2016, 05:44:35 pm
content: You can use the 24VAC to supply it, but you can also use DC. I suggested DC so that you can control the voltage and current from a bench supply. AC bench supplies with adjustable current limiting are not as common.

author: hgg
author_rank: Frequent Contributor
timestamp: November 11, 2016, 06:01:19 pm
content: QuoteYes, this is probably a better way to test it.Maybe a 50mA current limit would be enough. I wouldn't want to fry the IC as well...Thank you for the explanation.p.s. I thought they were wire-wound...

author: xygor
author_rank: Regular Contributor
timestamp: November 11, 2016, 06:07:04 pm
content: Quote from: hgg on November 11, 2016, 06:11:34 amIs D104 the 4A diode? That seems way too big for the amount of current needed there.

author: hgg
author_rank: Frequent Contributor
timestamp: November 11, 2016, 06:10:09 pm
content: No. D102 & D103 are 4A.