duncan webster | 06/10/2018 14:10:53 |
5307 forum posts 83 photos | Some time ago I bought a bench top power supply, which turned out to be not one of my better investments as the volt/amp dispaly on the front died. Yes I should have sent it back, but I decided to fit another display. The voltage display works fine, but the ammeter reads 0.55A when it should be reading 1.75. It's wired up as shown except that the red wire (Vcc) is fed from a separate 16v supply as instructed to make it work on <3V (the transformer had 2 secondaries) Anyone got any ideas, or have I bought a duff display as well? I've just noticed that the bridge recifier on the power board is running very hot as well, this really wasn't one of my better buys! I'm not giving up that easy. Next step a new power board. |
Grizzly bear | 06/10/2018 14:55:40 |
337 forum posts 8 photos | Hi Duncan, I'll jump in and risk losing some body parts. Ammeter should be in series with the load. Also I can't see the unit being happy with the red & yellow wires joined. Regards, Bear.. |
Speedy Builder5 | 06/10/2018 14:55:53 |
2878 forum posts 248 photos | Hi Duncan, what happens if you connect the thin red wire to the supply DC 4 - 30volt, instead of a separate power supply. (OK, I read the reason why you would want a separate supply, but just to test it out). |
Neil Wyatt | 06/10/2018 15:04:20 |
![]() 19226 forum posts 749 photos 86 articles | The red and yellow leads are so you can measure a different voltage id desired, no hazard using it to measure the supply voltage. Shouldn't GND be connected as well? I suspect the answer is that you should use a ceramic screwdriver to turn the little potentiometer marked Iadj until it reads correctly... Neil |
duncan webster | 06/10/2018 15:26:31 |
5307 forum posts 83 photos | Grizzly, the ammeter is in line with the load, the I+ and I- wires carry the current and there is a shunt on the board behind the connector. The yellow wire is the voltage you are measuring and the red wire is the voltage supply to the display electronics. You have to pwer this from a separate supply if you want to measure <3v. Neil, the ground wire is connected. I've tried twiddling Iadj and can only get it to chnge over the range 0.3 to 0.4 A when it is actually passing 1A. I'm going to take the display out from this unit and test it on another known to be OK psu, as I just noticed that the big series transistor which is controlling the voltage is also getting hot at 1A and the psu is rated at 2A
Edited By duncan webster on 06/10/2018 15:27:23 |
duncan webster | 06/10/2018 15:48:22 |
5307 forum posts 83 photos | So I've now connected the ammeter to another psu and load, but powered the display electronics from the suspect psu and it seems a lot nearer, within Iadj territory. Next step is to junk the power control board and make a new one from a 317 and a power transistor. There won't be much left of the original at this rate. At least your comments above helped me to know I hadn't committed some obvious goof |
Peter Simpson 1 | 06/10/2018 18:05:18 |
![]() 206 forum posts 9 photos | Cut the Red wire..........Anybody fancy a pint. I'll get me coat. |
Grizzly bear | 06/10/2018 19:49:07 |
337 forum posts 8 photos | Hi Duncan, Thanks for the update, I suspected I was pushing my luck. I trust the case/enclosure is OK. Good luck, Bear.. |
Phil Grant | 09/10/2018 22:41:52 |
107 forum posts 21 photos | A few thoughts: - 1. Has the shunt in the display replaced(displaced) the old shunt in the power supply or is it in series with it? 2. Are the grounds of both supplies connected together? 3. Have you put a separate meter in series with the display to check the current, if it's different then it's probably duff. |
mechman48 | 10/10/2018 16:52:49 |
![]() 2947 forum posts 468 photos | Don't know why but this thought immediately popped up in my ancient grey matter... |
duncan webster | 10/10/2018 23:39:57 |
5307 forum posts 83 photos | Things have moved on a bit. It appears that dumping what I thought was the display board seriously disrupted the 'power' board causing it to get very hot. I also found several more issues:
I've finished up dumping everything apart from the case, a fuseholder and one of the potentiometers and fitted a new (old) transformer I had in stock, a new rectifier with a huge capacitor I had in stock, a 5V linear regulator for supply to Arduinos, and a buck converter to give a separate 1.5V to 18V supply. It's all working fine except that the new display shows amps about 20% higher than my Fluke. I've also compared it with a moving coil ammeter. Yes I've tried twiddling the Iadj pot, not enough change. Next step is to experiment with a second shunt in parallel with the built in one to reduce the volts drop and presumably the displayed current, but I might just live with it for a bit, bored with this job! The new transformer was 14/9/0/9/14, so I've left it as centre tapped and used 2 separate rectifier/smoothing cap set ups for the linear and buck circuits. Why? Because I could! At least it is usable now. When I've made some supports for the cap I'll post some photos, not fit for human consumption at present with the cap supported on cable ties. |
SillyOldDuffer | 11/10/2018 12:15:07 |
10668 forum posts 2415 photos | Can't offer much other than sympathy but:
Dave |
Neil Wyatt | 11/10/2018 16:56:45 |
![]() 19226 forum posts 749 photos 86 articles | Posted by SillyOldDuffer on 11/10/2018 12:15:07:
Can't offer much other than sympathy but:
Peak value of rectified sine wave is about 1.4 x RMS. This is independent of transformer regulation, which is typically about 10% for a decent sized unit. 1.4 x 20V = 28V Knock of two diode drops for a full wave rectifier 0.6 x 2 = 1.2V That leaves you 26.8V off-load with no other losses. 20V with a 2A max current is a minimum load resistance of 20/2 = 10 ohms. Time constant for 10 ohms and 3,300 uf is 0.0033 x 10 = 0.033 seconds. The ripple frequency of rectified 50Hz AC is 100Hz, or of 0.01 seconds peak to peak. 0.033 is > twice the peak to peak time, which is the normal citeria for smoothing capacitors for non-critical applications. Sorry Duncan, it seems to have been reasonably specced from here!
Neil |
duncan webster | 11/10/2018 19:24:07 |
5307 forum posts 83 photos | 2 amps from a 3300uF cap gives a rate of change of voltage of 606 V/sec, which is 6v every cycle. Seems like a lot to me, 23%, but I bow to your wisdom. It did have a voltage control following so I don't suppose a bit of ripple matters. |
Neil Wyatt | 11/10/2018 21:44:40 |
![]() 19226 forum posts 749 photos 86 articles | Hi Duncan, I've managed to get it into an online SPICE simulator. I've assumed that as you measured 26V off-load (or thereabouts) that the off-load peak voltage is 28V to allow for the diode drops (some values in the above circuit have been changed). You can see the output voltage is fluctuating between 26V and 21V. bearing in mind that this represents an output current of ~23.5V/10 or 2/35A, then there should be sufficient headroom for a linear regulator at a 2A load.
|
duncan webster | 12/10/2018 00:45:31 |
5307 forum posts 83 photos | I'm most impressed with your simulation, it gives almost the same ripple as my sum. However, the 26V was AC RMS, so with a cap I'd expect 35V DC which would have been challenging for my buck converter. When loaded with 2 off 21W 12V bulbs in series the voltage dropped to 19V, a 27% drop, well above the 10% suggested by SOD. I should have given up then, but I wanted to understand why the DC volts drop was so much higher then the AC. I suspect it might be to do with the fact that the transformer is now effectively giving out short pulses into a very low impedance (capacitor) and so when it is conducting the current is high and so the voltage drop through the windings is high, but if anyone has a better explanation I'm all ears. I forgot to measure the AC volts whilst the DC was connected, but did attach my scope I really don't understand the vertical drop. Again if anyone has an explanation I'd be grateful. At least part fo the joy of making things is learning something new. I did try another set of diodes, didn't help, if anything it got worse. There is very little ripple on the DC, not surprising with my enormous cap. I've now given up, one transformer too high, one too low. I'll buy a complete 24V switch mode supply and connect the existing buck converter to that. I have a linear regulator for the 5V. Edited By duncan webster on 12/10/2018 00:46:09 Edited By duncan webster on 12/10/2018 00:47:40 |
duncan webster | 12/10/2018 00:49:15 |
5307 forum posts 83 photos | I forgot to mention that te ammeter behaves a lot better on this set up, still reading about 20% high, but I've had a play with shunting the shunt and it can be sorted. |
Neil Wyatt | 12/10/2018 10:25:35 |
![]() 19226 forum posts 749 photos 86 articles | This is all quite relevant to me at the moment, as I find myself needing a hefty 'power pack' for my astronomy, with the need for various supplies with different capabilities. I could just package up a PC power supply in an outer case, then hang various things off it, but it would be better if I could get something giving about 18V - 20V at 10amps and then hang an assortment of buck and PWM controllers off it. |
duncan webster | 12/10/2018 11:31:57 |
5307 forum posts 83 photos | I've got some humungous toroidal transformers which I think have 2 off 35V secondaries. You're welcome to one if it's any use. At a pinch I could take one to MIdlands show. If these sound vaguely interesting I'll dig them out from their hidey hole and check the output. The advantage of toroidal is that if you're careful you can unwind some of the secondary turns to lower the output volts. I've even been known to add another winding when I wanted another very low power secondary. |
Neil Wyatt | 12/10/2018 13:21:24 |
![]() 19226 forum posts 749 photos 86 articles | Thanks Duncan, but as it will get a lot of lugging around I think I will look for a switch-mode PSU to suit - I may still have something lurking under the bench suitable despite the recent clearout... Neil |
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