Power Supply question - Page 1
Forum section: electronics / beginners

author: pyroesp
author_rank: Regular Contributor
timestamp: August 20, 2013, 07:43:26 pm
content: So I have this EP-925 power supply, which doesn't work. Probably some of the ICs don't work anymore.(fromhttp://pa0fri.home.xs4all.nl/Diversen/EP925/ep925eng.htm)I don't know why they used five 0.1 Ohm, 5W resistors at the emitter of the power transistors (TR1~TR5). Why would you put those resistors there, what do they do ?I checked, my PS and those resistors are 0.1, not 0.5 like in the schematic.I'd like to try and make an upgraded version of this PS, controlled with a microcontroller, LCD/OLED display and stuff like that. I don't think I'll use anything but the power transistors and transformer. Shouldn't be to hard to drive those transistors.Anyways...- pyroesp

author: Zbig
author_rank: Frequent Contributor
timestamp: August 20, 2013, 08:02:19 pm
content: Quote from: pyroesp on August 20, 2013, 07:43:26 pmThese are current sense resistors (shunts). They (well, actually one of them) allow the ammeter to work by creating a voltage drop proportional to the current flowing through them. I don't want to sound like an a-hole but if you don't know this, I think it's probably not exactly the best idea to start with redesigning this power supply just yet.

author: pyroesp
author_rank: Regular Contributor
timestamp: August 20, 2013, 08:25:06 pm
content: Ok, but why put 5 of those, couldn't they just connect all the emitters together and use only one 0.1 Ohm resistor and use that as a shunt?I mean... it's kind of stupid to have five shunt resistors if you only need one.I can understand one shunt resistor but not 5 of those, that's why I asked.

author: Zbig
author_rank: Frequent Contributor
timestamp: August 20, 2013, 08:29:14 pm
content: Quote from: pyroesp on August 20, 2013, 08:25:06 pmThen all the power (R * I^2) would have to be dissipated by this one resistor. By doing it like in the schematic, they're sharing the load equally (just like the transistors) and that's why 5W ones are enough.

author: pyroesp
author_rank: Regular Contributor
timestamp: August 20, 2013, 08:33:51 pm
content: Hmmm.... The max output current is 3 amps, so with one shunt resistor it could have been enough. Those are 5W resistor and if you draw 3A, the power dissipated is still lower than 1W.

author: TerminalJack505
author_rank: Super Contributor
timestamp: August 20, 2013, 08:34:05 pm
content: It looks like they are using one for current sensing but the more important function of those resistors is as ballast resistors.BJTs have a negative temperature coefficient. As they get hotter their effective resistance decreases. Because of this, when you put two or more in parallel, you can get thermal runaway. Once one of the transistors get a little hotter than the others it will start drawing more current. More current makes it hotter, which lowers its resistance, which causes it to pass more current, etc. This is basically a positive feedback cycle and leads to eventual destruction.The resistors on the emitters will prevent this. More voltage will drop across the ones drawing more current. This provides a negative feedback to the transistor, thus preventing thermal runaway.

author: pyroesp
author_rank: Regular Contributor
timestamp: August 20, 2013, 08:41:51 pm
content: Quote from: TerminalJack505 on August 20, 2013, 08:34:05 pmHmmm, that makes sense. Very interesting too. Thanks

author: Zbig
author_rank: Frequent Contributor
timestamp: August 20, 2013, 08:48:41 pm
content: Indeed, thanks TerminalJack. I guess that's why I'm not designing my power supplies eitherBut pyroesp, I'm pretty sure this PS is capable of delivering whopping 25A through its banana sockets and the 3A limit only applies to the "speaker-type" terminals:http://www.conrad.com/ce/en/product/511492/VOLTCRAFT-EP-925-375W-Single-Output-Variable-DC-Power-Supply-Linear-Bench-3-15Vdc-25ASo while I'm sure TerminalJack's great explanation is valid, it's also true that only one shunt resistor wouldn't be able to pass all that current.EDIT:One 5W shunt resistor, that is.

author: pyroesp
author_rank: Regular Contributor
timestamp: August 20, 2013, 09:03:40 pm
content: I thought I saw it only could output 3A, but now that I took another look at the semi-broken case, it doesn't mention anything.Quote from: Zbig on August 20, 2013, 08:48:41 pmVery true. Now I know 2 good reasons why those resistors are there for.Thanks!

author: Paul Price
author_rank: Super Contributor
timestamp: August 21, 2013, 12:11:50 am
content: When you say it don't work, that's not telling much of the story, so I have to make more guesses than if I just had a few more facts about operation, i.e. whether you have any LEDs lit, any voltage output, etc.If this supply doesn't work it is due to:1) blown fuse or shorted D8(have you checked this? Very likely)2) blown LM723 IC (likely)3) shorted output on LM339 section IC2D(fairly likely)4) blown 22-ohm resistor (R6) going to base of TIP31 (TR6) (not too likely)5) open Base-Emitter TIP31 (TR6) (not too likely)6) open power transformer/bridge rectifier(very unlikely)7) bad AC power switch wiring or bad solder joint on PCB or from rectifiers to filter caps to control PCB(not too likely).Bad voltage adj. pot or connection wires thereof to PCB(unlikely).

author: w2aew
author_rank: Super Contributor
timestamp: August 21, 2013, 12:21:43 am
content: Quote from: TerminalJack505 on August 20, 2013, 08:34:05 pmThis is basically right, except it is the Vbe of the BJT transistors that has the negative temperature coefficient. The ballast resistors prevent the thermal runaway condition.

author: pyroesp
author_rank: Regular Contributor
timestamp: August 21, 2013, 10:14:44 am
content: @Paul Price: I'm not asking why it doesn't work, I don't want to repair it. I just wanted to know why there are 5 0.1 Ohm resistors at the emitter of those power transistors.