Changing PSU range with different pot - Page 1
Forum section: electronics / beginners

author: inNumbersPlz
author_rank: Contributor
timestamp: September 25, 2015, 09:29:41 am
content: Kia ora all,I've happened across an old DC variable voltage bench power supply and would like to reduce it's output voltage range. I've done a bit of digging around but couldn't find much on the topic (maybe it's just too basic). Surely just replacing the current pot with one of higher resistance will work right? It's an old unit with a transformer followed by two large electrolytic caps. Here are the specs on the unit:6-14.8V DC out~5 APot measured at 0.4 Ohm to 1.2 kOhmI would like to have it range from 1-10V and preferably less current, but I'm not sure what equations to be using. Tried Ohm's law with resistance of pot and voltage and it gave me silly values for current, so assuming it doesn't apply here. So if you know the equation, have some advice, or know this has been answered elsewhere it'd be of great help!Tena Koe

author: tautech
author_rank: Super Contributor
timestamp: September 25, 2015, 10:00:27 am
content: Welcome to the forum.There are many PSU designs and active elements used.It would be best if you find the schematic or reverse engineer the PSU.Photos might also help others recognise it.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 25, 2015, 10:56:02 am
content: tautech,thanks for the reply. Tried reverse engineering the circuit, but my inexperience and the odd PCB design led to disaster. I'll post a photo which will hopefully be of some help. The basic circuit seem to be:Positive in -> transformer -> cap1 -> cap2 and (resistor/transistor network -> pot -> resistor/transistor network) -> positive outTena koe

author: mikerj
author_rank: Super Contributor
timestamp: September 25, 2015, 12:15:52 pm
content: Quote from: inNumbersPlz on September 25, 2015, 09:29:41 amIt depends entirely on the design of the circuit. In most regulator designs, a fraction of the output voltage (the fraction being determined by the voltage control potentiometer) is compared against a fixed reference voltage, and the difference is amplified and used to drive the output stage. In basic designs that use a single voltage rail then the output can not go below the reference voltage e.g. if your reference voltage is 6v you may be stuck.

author: rs20
author_rank: Super Contributor
timestamp: September 25, 2015, 12:24:10 pm
content: Quote from: mikerj on September 25, 2015, 12:15:52 pmReplace the reference?

author: rdl
author_rank: Super Contributor
timestamp: September 25, 2015, 02:28:41 pm
content: It doesn't look like it was originally a variable supply, so you'd best reverse engineer it to figure out what's really going on.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 25, 2015, 08:59:24 pm
content: Ok, I didn't realise they worked like that. I will have another crack at reverse engineering it. Thanks all!

author: tautech
author_rank: Super Contributor
timestamp: September 25, 2015, 09:08:10 pm
content: Quote from: inNumbersPlz on September 25, 2015, 08:59:24 pmIdentify the active components, what is the TO3 device?It's very likely this PSU is some old common design and we can then point you to many of the online schematics.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 25, 2015, 10:52:31 pm
content: The TO-3 has written on it:2N 3055RCAA 7835I'm guessing it's made by RCA. The circuit seems to go through one cap, the resistors, then the TO-3 and then to the other cap and on to the output. Thanks for the input everyone.

author: tautech
author_rank: Super Contributor
timestamp: September 25, 2015, 11:03:40 pm
content: Quote from: inNumbersPlz on September 25, 2015, 10:52:31 pmJust as I suspected.Google for "linear PSU schematic" and see if you can match results with the various components on you PCB.It's likely you will find a perfect match, post the link and you're likely to get the advice you need.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 25, 2015, 11:42:47 pm
content: Found! It seems to be a kit set power supply, and the schematic does include a 1kOhm pot. I'll post the schematic with the pot circled. Anyone know if I can replace it with a higher resistance pot to get an output of 1-10V?Tena Koe

author: barry14
author_rank: Regular Contributor
timestamp: September 25, 2015, 11:53:45 pm
content: The schematic shows the reference voltage is derived from a 5V Zener. Including the base emitter drop of the transistor means that you will not be able to get less then about 6V from this supply. Also, since the supply was designed to supply a fixed 13.5V, the pass transistor (the 2N3055) may not have a large enough heatsink to dissipate the increased power when you reduce the output voltage much below this value. Since the supply is rated at 5 amps, you may get away with this if you don't try to draw anywhere near that value.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 26, 2015, 12:05:35 am
content: Darn, oh well. Thank you all for your input, I have so much to learn!

author: rdl
author_rank: Super Contributor
timestamp: September 26, 2015, 12:23:36 am
content: If 1.25 - 10 volts is good enough, just add a regular old LM317 circuit between the existing stuff and the binding posts. Output current would be limited, but you should be able to get several hundred milliamps or more, depending on how cool you can keep the LM317.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 26, 2015, 12:43:40 am
content: rdl, I was just thinking that. Would a pot between the existing circuit and the output posts work? Or do I need the LM317 for voltage regulation? The unit has a large heat sink I could mount this to.

author: rdl
author_rank: Super Contributor
timestamp: September 26, 2015, 04:09:23 am
content: Best to just add the simple LM317 circuit. If you use a pot only, wired as a voltage divider on the output, then all the current has to flow through the pot and most are not rated for very much power dissipation.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 26, 2015, 05:19:12 am
content: Great, thanks rdl. I'll wire up the lm317 on my breadboard and see how it goes. Thanks!

author: tautech
author_rank: Super Contributor
timestamp: September 26, 2015, 07:56:55 am
content: As it seems you have a good sized transformer, you could do somewhat better than a typical 1.5A LM317 supply.As with all stuff that is beyond ones experience, device datasheets are your friend and especially the typical applications often listed in them.http://www.ti.com/lit/ds/symlink/lm317.pdfI'd suggest you look at the LM338 for better current capabilities and it is just as easy to use as a LM317.It is commonly available in a TO3 package that will slot into where the 2N3055 was.http://www.ti.com/lit/ds/symlink/lm338.pdfIt is a stock item at Jaycar, altough probably dearer than most other sources.http://www.jaycar.co.nz/Active-Components/Regulators/Variable-Voltage/c/212BTo make this a quite reasonable basic PSU I'd advise you ditch the standard pot and replace it with a 5 or 10 turn unit that will give you much better resolution of voltage adjustment.If you wanted to go further you could ditch the moving coil meter and add a 3.5 digit LCD one, although they commonly need another isolated power source to run from.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 26, 2015, 09:24:30 am
content: tautech, thanks for the great suggestions. So the LM338 would be able to dissipate more power than the 2N3055 if I were to replace the pot as you suggest, with a higher resistance one? I'm a young enthusiast without much experience or budget, so really appreciate all the help!Tena Koe

author: tautech
author_rank: Super Contributor
timestamp: September 26, 2015, 09:42:57 am
content: Quote from: inNumbersPlz on September 26, 2015, 09:24:30 amIt's the current capability that is the advantage over a LM317, sure there will be more dissipation too.But honestly do you need 5A? 1.5 A with SC and thermal limiting is likely to be all you need for hobbyist use. And if it's not, look at upgrading the PSU later.Just follow the datasheet typical application, but I'd suggest you add an additional ~3A diode just before the output for protection. Connect your output voltmeter after the diode just before the outputs.Do some more Googling for basic LM317 circuits, they're quite easy and so simple you can just "rats nest" them between device terminals or tag strips.You might even be able to use part of the PCB, even just for launching connection leads from.You'll look back on this in time and wonder what all the fuss was about.

author: rdl
author_rank: Super Contributor
timestamp: September 26, 2015, 03:53:28 pm
content: Just to be clear, you wouldn't be changing or replacing any of the existing circuit. You will be adding an LM317 regulator circuit just before the output and using the existing power supply circuit as its input.Theoretically, all you need is an LM317, a resistor and a pot, but the better LM317 circuits will have a few more parts. Protection diodes and a bypass capacitor on the adjust pin should really be added. Still, it's not all that complicated.One thing every beginner should understand when building a power supply is the difference between current carrying capacity and power dissipation.

author: inNumbersPlz
author_rank: Contributor
timestamp: September 27, 2015, 03:11:42 am
content: I'm going for the LM317 approach added in just before the output. I will probably replace the current pot with a 500 Ohm resistor as this was in the pots range, and then use the hole for the old pot to install the new one in the LM317 circuit. The circuit I will use is one with 0.1 and 10 uF caps, a resistor and a pot, as well as the LM317.rdl I think I know what you mean, but correct me if I'm wrong. The LM317's current capacity is quite low so has to dissipate quite a bit of power as heat into the heat sink?

author: rdl
author_rank: Super Contributor
timestamp: September 27, 2015, 05:02:00 am
content: An LM317 can usually handle 1 amp just fine, as long as it doesn't get too hot.The power dissipation in a device is given by the current through it times the voltage drop across it. P = I x VSo basically, the difference between the voltage going into the LM317 and the voltage going out, multiplied by the current gives the power dissipation in watts. This is converted to heat which is mostly radiated away through the heat sink.(Vin - Vout) x I = WattsYour input voltage will be 13.5, so with 10 volts out at 1 amp, 3.5 watts of heat must be gotten rid of, but with only 5 volts out at the same 1 amp current, this rises to 8.5 watts. Big difference. A general rule of thumb is to keep things from getting too hot to touch.