Dropping 52v to <30v to power an LED volt meter (17mA) on an AC-DC converter - Page 1
Forum section: electronics / beginners

author: TheBloke
author_rank: Contributor
timestamp: July 11, 2016, 05:26:16 pm
content: Hi allFirstly, I have a habit of writing at great length, so if the following is far too wordy for anyone, here's a quick summary: I'm trying to run a basic volt meter from a 52V DC supply; the volt meter draws 16-17mA and needs 4V-30V max. I've managed to power it using 6 resistors (enough that none exceeds 250mW), as shown in simulator screenshot/link at end of post, but would appreciate any advice and suggestions on whether this is the best way, and other possible solutions.I'm slowly learning about electronics, having started to get interested in the last couple of months. One of my first projects has been to get myself a decent bench power supply. Or rather, because I can't really afford a decent one, I'm putting one together out of cheaper components. Specifically, I have a Chinese programmable DC-DC buck converter, with constant current and voltage, LCD display, multiple memories, and serial RS232/TTL connectivity. It's quite a nice unit, able to provide 0-60V at 0-15A as long as I can feed it with a higher voltage. Therefore my second component isan AC-DC converter, also cheap Chinese, rated at 48V/10A/500W, and adjustable from 39V - 57V. I plan to use this at 52V, feeding the buck converter and the combination therefore providing a programmable CC/CV power supply of 0-50V and 0-15A, or 0-10A at higher voltages. (I'm aware it'll be rather noisy, but I think for my beginner purposes it should be pretty good, and usefully flexible.)The first step of my project is to do some upgrades on the48V AC-DC converter. It's one of those cheap Chinese "LED/CCTV driver" units that has the AC input and DC output both connected via a single screw terminal block, thus positioning the 240V AC input just millimeters away from much lower voltage DC output. So I've removed that, replacing it with an IEC socket for AC input on one side of the unit, and 4mm banana plugs for DC output on the opposite side. I've also added a power switch, beefed up the heatsinks on the transistors so I can put in a quieter fan, and am replacing the tiny adjustment potentiometer with a larger one with a proper knob. That's all going fine.For the final step of the upgrade I thought I'd throw in a cheap LED voltage display, so I can see what voltage it is putting out without always needing a multimeter. This leads me - belatedly! - to my question.I have a small, very cheap,three-wire LED voltmeter. Red, black and yellow wires. The voltage display can measure 0-100V, but needs between 4V and 30V to power it. So where the source is >30V, the yellow goes to the output to be measured and the red and black need to go to a lower voltage source on the same ground as the device under measurement. This is where I run into a problem, because my AC-DC converter will always be between 39V and 57V.So my question is: what method would you recommend to drop 52V DC to a maximum of 30V to power this volt meter, which I've measured to draw a current of 15-17mA (seemingly irrespective of the voltage it's running at)?My first thought had been that I might be able to simply avoid the problem, because the AC-DC unit has a 12V socket for the fan. There's no temperature measurement, the fan is just always on. So I thought I could take my red and black wire and run them from the 12V fan output, and then put the yellow voltage measuring wire to the positive DC output of the AC-DC supply. Sadly, while this powers the voltmeter fine, it doesn't give a useful reading. I believe this is because there's no shared ground between the 12V fan socket and the main DC output of the device, and so I always get a reading of around 1V.Next I looked into linear voltage regulators, but all the chips I could easily find - eg L7805, L7812, L7818 and L7824 - also have a maximum voltage that's too low. Either 35V or the highest max is 40V (on the L7824). So that just redefines my problem with a slightly higher max V, rather than actually solving it. I also briefly investigated zener diodes for clamping the voltage to eg 12V, but that seemed to require far too much power on the current limiting resistor (and didn't seem to have much point given I could just use series resistance, as described below.)Then I looked at voltage dividers, and this is where I found a working solution,first using a circuit simulatorand then on the real device. Although it didn't actually end up being a voltage divider precisely, because it seemed that just using series resistance worked better. My main issue with this was being able to drop the voltage without burning out my 1/4w resistors, so I worked out the following circuit that provided enough current and voltage drop, without any one resistor exceeding 250mW:In this simulation I used an LED set to Vf=5V to simulate my real load. And the resistor values are chosen so as to match resistors I already owned without exceeding any power ratings.Running this on a breadboard and measuring it, I get an RMS voltage of around 23V after the resistors, although my scope shows that it is wildly fluctuating between as low as 8V and as high as 37V. The frequency also varies, from around 500Hz - 1Khz.It seemed to work fine but I was a bit worried about those peaks of 37V, so I then stuck an L7812 after the resistors. I don't get 12V out (because of those 8V lows I guess), but I do get a reasonably stable output of 6V: with the L7812 in place, the scope shows a Vupper of 8.2V and a Vlower of 5.9V with RMS of 6.1V. The frequency is exactly 2.7Khz, which I guess is the internal switching speed of the L7812. The volt meter works fine and measures correctly, and although it seems a bit odd using a 12V regulator to get 6V, it works and I'm happier knowing I'm not exceeding its 30V limit at any point. (An L7809 and L7805 didn't output anything - again, I suppose, because the voltage dips as low as 8V.)The resistors get a little warm after a while running, but I think will be OK, and in the final circuit they'll be positioned reasonably close to a fan. I could always replace them with 0.5W resistors to be absolutely sure.So I do now have a working solution. But is this the best way? Could anyone tell me how they would do it? Since I first tested it I have bought a bunch more resistors, and now I have many more 0.25W values, as well as a selection of 0.5W and 2W resistors. I did try breadboarding it with a couple of 2W resistors but they were running at up to 100°C and I figured that, even though they could take that without dying, I'd rather use a few more resistors running at low temps (and save the 2W resistors for other projects.)Thanks in advance for any help (and apologies if my question is far too long/detailed!)

author: Ian.M
author_rank: Super Contributor
timestamp: July 11, 2016, 07:17:59 pm
content: Assuming the fan supply is a steady 12V (not PWMed or half wave rectified) an isolated 0.5W or 1W 12V DC-DC converter running off the fan supply to power the meter would probably be the simplest option.

author: TheBloke
author_rank: Contributor
timestamp: July 11, 2016, 07:25:38 pm
content: Quote from: Ian.M onTodayat 20:17:59Thanks for the reply.I believe the fan is a steady 12V DC - it certainly seemed stable on my DMM, although I haven't scoped it yet. But using that was the first thing I tried, and although it powered the volt meter fine, it didn't provide usable volt readings. The volt meter always showed around 0.9v I think it was, rather than the 52v actually coming from the main DC output.I assumed at the time that this was because there was no common ground between the fan output and the main DC output - certainly there wasn't any continuity when I ran a check between the fan ground and the main DC output ground. The info provided with the volt meter did mention that it requires that the ground used to power it and the ground of the device under measurement are common.Or is that why you mention a DC-DC converter on the fan output? Would that solve the bad reading problem? If so could you elaborate on that?

author: Curtis
author_rank: Contributor
timestamp: July 11, 2016, 07:41:52 pm
content: If the meter can run at 30v @ 16ma then wont this work?(52v - 30v = 22v) / .016ma ) = 1375 ohms which = (.016ma x .016ma) x 1375 ohm) = .352 milliwatts. 1/2 watt resistor should be fine. A 1 watt would be better.

author: Ian.M
author_rank: Super Contributor
timestamp: July 11, 2016, 07:51:34 pm
content: The trouble with a resistive dropper is the voltmeter's supply voltage will fluctuate according to its current consumption. A 12V Zener across its supply terminals to clip the peaks and a large enough reservoir cap to smooth out the drops to 8V would be one option but you'd still have the resistors running hot.The DC to DC converter, powering the voltmeter on its Red and Black wires isolates it so you can connect its Yellow and Black wires across the output you want to measure's +V out and 0V (ground) terminals, without any problems due to differing grounds. It effectively turns the meter into a four wire meter with separate isolated power and measurement connections.

author: TheBloke
author_rank: Contributor
timestamp: July 11, 2016, 08:12:04 pm
content: Quote from: Curtis on July 11, 2016, 07:41:52 pmYeah this does work, and I ran a couple of similar tests: one with 1k + 480r in 2W, and another with 1.5k in 5W. It runs fine, and actually allowed me to use a L7809 instead of the 7812 because the output voltage was a little higher.The only issue - and perhaps it's a non-issue - was that I measured the temperature on the resistor(s) and they were up to 100°C after only a few minutes. I was certainly within the power limits of the resistor (very far within in the case of the 5W) so I guess it's no problem. I just felt uncomfortable about having such a hot component, and as I already had 6 x 0.25w resistors working and barely getting warm, that seemed preferable to me. Plus I thought I my as well keep the better, higher power resistors for time when they were essential.Quote from: Ian.M on July 11, 2016, 07:51:34 pmYeah, the volt meter's current seems to fluctuate by around 0.5mA and I see this causing the large fluctuations in voltage that I see.But with six resistors sharing the load it doesn't seem to get that hot. That's good info on the Zener and cap, thanks. Actually I guess that's basically what I'm doing already, indirectly, by including the L7812 or L7809? I think I read that those linear voltage regulators are ICs based around a Zener diode clamp + transistor(s) + cap(s)?To be sure, the device seems to work fine even with the large voltage swings. I just wanted to be safe and stop it exceeding 30V and the voltage regulator achieved that. But yeah I'm going to try the Zener + cap as well, just so I can see that I understand that too.Quote from: Ian.M on July 11, 2016, 07:51:34 pmAhh, that's very interesting, thanks. I ordered some mini LM2596s DC-DC buck converter PCBs from China a couple of weeks ago - each one only 4cm x 2cm - so I guess that would do the job perfectly. And at £0.60 each, they're basically not much more than the cost of a linear voltage regulator or a couple of (good) resistors.I'll definitely try that solution as well once I receive the PCBs.Out of interest, how complex would it be to do the same job without a DC-DC converter? I mean, do it myself with discrete components - what components would I need to put between the fan output and the volt meter? I've seen schematics for buck and boost converters and they're fairly complex, but in this case I don't actually need to change DC voltage, only achieve the ground isolation - so maybe that's simpler? Not that I need it for this case, given how cheap the little buck PCBs are. But for educational purposes it'd be cool to understand how it would be achieved. Is there a standard circuit schematic I could search for? Or does it have to be a full DC-DC converter to achieve the ground isolation?Thanks again guys, all very helpful.

author: edavid
author_rank: Super Contributor
timestamp: July 11, 2016, 08:26:43 pm
content: Quote from: TheBloke on July 11, 2016, 07:25:38 pmIf the fan supply is isolated, you can just connect the grounds together.QuoteIt would be needed if the fan supply were not isolated.Quote from: Ian.M on July 11, 2016, 07:51:34 pmNot a problem, since the voltmeter has a regulator built in (it can be used as a 2 terminal device).

author: edavid
author_rank: Super Contributor
timestamp: July 11, 2016, 08:28:14 pm
content: Quote from: TheBloke on July 11, 2016, 08:12:04 pmNo, they won't help at all, since the input and output are not isolated.

author: Ian.M
author_rank: Super Contributor
timestamp: July 11, 2016, 08:39:56 pm
content: If the fan supply was fully isolated, the voltmeter would read 0 with its yellow wire connected to the output terminal and its red and black wires to the fan supply. However it read 1V so I recommended an isolated DC-DC converter module.

author: Seekonk
author_rank: Super Contributor
timestamp: July 11, 2016, 08:44:14 pm
content: Any old 5V cell phone wall wart converter will easily run on 52V DC. I've been running all my control electronics and meters on my 36V solar system on wall warts for years.

author: edavid
author_rank: Super Contributor
timestamp: July 11, 2016, 08:55:02 pm
content: Quote from: Ian.M on July 11, 2016, 08:39:56 pmSince the voltmeter has high impedance input, a small amount of capacitive coupling could be causing the nonzero reading. Perhaps OP could try connecting the grounds through a resistor.

author: TheBloke
author_rank: Contributor
timestamp: July 12, 2016, 10:21:16 pm
content: Hi all. Thanks so much for the continued help and info.Here's two updates regarding the fan output method:Firstly, earlier I said that running the volt meter from the fan output gave readings of 0.9-1.0v. Re-testing tonight, it was showing 1.5v. I don't know if this is important or not!Secondly, I took edavid's advice and tried joining the grounds via a resistor. I joined them via a 5k - I have no idea what sort of resistor he had in mind, but that seemed suitably high value so as to provide protection, if protection was needed.The good news is that it worked! The volt meter powered, and showed a realistic reading. The minor bad news is that the reading was slightly inaccurate.For example: when the main DC output was reading 52.060 on my multimeter, the volt meter would read 52.1 when connected directly to main DC output via series resistance. But when connected via the fan output with the shared ground, it would read 51.7. It always seemed to be around three or four counts low when connected that way, regardless of the exact value of the DC output (tested from 40v - 53v.)I wondered if this was related to the resistance between ground. So I replaced the 5k with a 3.3k, and I think this did indeed change things - it then read 51.8 instead of 51.7, ie 2-3 counts out instead of 3-4.I don't quite follow why a resistance between the grounds is affecting the measurement?Now maybe I'd be completely fine using a much lower resistance, like maybe 100r or so which would presumably give a fully accurate reading. But I'm slightly hesitant to try that without further validation, in case there is still some risk of joining the grounds without protection. I don't mind if it kills the volt meter, but I do want to protect the AC-DC converter itself. But maybe the fact that I've got this far indicates there is isolation and it's safe?Anyway even if I ultimately go back to the basic method of series resistance on the main output, I'm learning lots of new things which was the main object of my question and much appreciated

author: edavid
author_rank: Super Contributor
timestamp: July 12, 2016, 10:36:39 pm
content: Quote from: TheBloke on July 12, 2016, 10:21:16 pmIf you don't measure any significant voltage across the resistor, that means the circuits really are isolated, and you can replace it with a low value resistor or a short circuit.

author: TheBloke
author_rank: Contributor
timestamp: July 12, 2016, 10:45:28 pm
content: Quote from: edavid onTodayat 23:36:39OK thanks. Measuring across the 3.3k resistor currently in place shows 0.3822V. That certainly doesn't seem significant to me, though it's not zero? Maybe a low value resistor, 100r or so?

author: Ian.M
author_rank: Super Contributor
timestamp: July 12, 2016, 11:05:36 pm
content: That's about 0.11 mA DC, which is good evidence the fan circuit is *NOT* fully isolated. Its possible the 12V supply is also used to power control circuits in the SMPSU (48V LED PSU). I would be very reluctant to short fan ground to output ground. It *may* work, but why risk it electing a Pope?Use a suitableISOLATED12V DC-DC converter as I suggested and you'll have an accurate readout + peace of mind.

author: edavid
author_rank: Super Contributor
timestamp: July 12, 2016, 11:18:17 pm
content: Quote from: Ian.M on July 12, 2016, 11:05:36 pmAgreed.QuoteThey are a little expensive though ($3). It would be cheaper to use a wall wart as previously suggested.