Potentiometer burning - Page 1
Forum section: electronics / beginners

author: AngraMelo
author_rank: Regular Contributor
timestamp: August 27, 2019, 06:44:51 pm
content: Hello guys,Im making this project, it is an unregulated power supply to use to test amplifiers.The issue is that my 100k pot keeps burning. I have no idea why. I used the 2k2 resistor to limit the current to the pot but after using it for a couple of minutes de pot burns and sometimes the MJE340 goes short as well.I have no idea what is going on. could you help me?

author: GerryR
author_rank: Frequent Contributor
timestamp: August 27, 2019, 11:32:22 pm
content: I believe you need a different base drive circuit. When the pot gets anywhere above about 8 volts (worst case Vbe for all the transistors set up in the Darlington configuration), you have no base current limit and you load your 47 volt source down. When you get near the top of the pot (actually anywhere above about 8V, the 47 V is bypassed because it is shunted by the base-emitter junctions. It's like putting an 8 V zener in parallel with your 47 volt zener. Just a quick glance.

author: AngraMelo
author_rank: Regular Contributor
timestamp: August 28, 2019, 01:53:40 am
content: Quote from: GerryR on August 27, 2019, 11:32:22 pmThank you Gerry! Could you please explain why is it 8V?Also, I tried putting a 10k resitor between the wiper and the base of the mje340 and now the pot is not burning anymore but the 340 or even a Tip142 are going short. Should I limit the current on the collectors of the 340 and tip41 with resistors?I dont understand why even using a 10k resistor on the base of the 340 can make the all the following transistor going short. They are all mounted on big heatsinks and even with a load around 500mA they are still going bad.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 28, 2019, 02:16:14 am
content: You have the emitter of one transistor tied to the base of the following transistor in a darlington configuration. If you look at the worst case Vbe drop for each transistor, they add up to about 8 V. I don't know what you are trying to accomplish(??), a variable voltage output supply (??), and at what current?? If that is what you are trying to do, you need to change your approach.

author: Andy Watson
author_rank: Super Contributor
timestamp: August 28, 2019, 02:21:25 am
content: You need to allow for the possibility of leakage current between collector and base/emitters of the transistors. This is usually achieved by placing a resistor across the base-emitter junctions of the driving transistors - aim to sink a few milliamps or more such that it takes care of any leakage. The leakage will inccrease with increase in temperature. I believe that this is the mechanism by which the output voltage can rise above the "set" voltage.Whatever the mechanism is, you need to mitigate the effects of the output voltage being greater than the set voltage - the MJE340 can only stand 5Vreversevoltage between emitter and base. Attach a diode between the output and the base of the MJE340 to prevent its base becoming negative with respect to the output. And put a series resistor between the potentimeter and the base - to prevent excessive output voltage feeding back, through the diode and into the potentimeter.

author: AngraMelo
author_rank: Regular Contributor
timestamp: August 28, 2019, 02:37:04 am
content: Quote from: GerryR on August 28, 2019, 02:16:14 amThe idea is to have a variable voltage output PS that can go up to 7amps. When I made this I tried cascading the transistors so I could use a minimal current coming from the pot. given that the voltage is rather high (47V from the zener) I dont know of other suitable ways to vary the voltage without introducing noise in the circuit with regulation and creating a project way harder than this already is for me. So that is why i decided to go with the pot and cascading transistors.

author: AngraMelo
author_rank: Regular Contributor
timestamp: August 28, 2019, 02:44:39 am
content: Quote from: Andy Watson on August 28, 2019, 02:21:25 amThank you for your input, Andy.I put a 10k from the pot to the base of the MJE340 and the pot doesnt burn but now the 340 keeps going short.I should put the diode after the 10k resistor and then into the base of the 340? do I do that for the Tip41 and Tip142 as well?When you talk about a resistor between base and emitter, would it still be necessary with a diode from the 10k resistor going to the base of the 340?I dont understand how can the emitter be at a higher voltage than the base of the transistors in this circuit. Could please explain to me how can that happen?Im sorry Im such a beginner, Im trying to learn with you all. Thank you for your time!

author: Jwillis
author_rank: Super Contributor
timestamp: August 28, 2019, 05:24:21 am
content: The MJE340 has a broad range of gain (30- 240) with a 10k resistor at the base gives around 4.7mA at the base when 100k pot is at zero ohms. Consider that Ic=IbXHfe.This means that the current from the collector is equal to the current at the base multiplied by the gain (Hfe). At the lowest gain of 30 the Collector Current will be approximately 141mA but at the maximum gain of 240 the Collector Current will have approximately 1128mA which exceeds the maximum Collector Current of 500mA .Proper transistor biasing is crucial with BJTs to limit the current passing through them. Another issue is with paralleling transistors.You have emitter resistors to prevent "thermal run away" but base biasing is important o prevent "load hogging". Load hogging occurs because transistors of like type turn on at slightly different voltages.The lowest voltage one will turn on first and load more than the others.Perhaps even burning out then the next lowest then the next. A base resistors will limit the current at the bases thus balancing the out put current of each.Rb = (Vin-Vbe)/Ib Base resistor is equal to the input voltage minus 0.7 divided by the current you want at the base. The 0.7volts is a typical forward voltage from base to emitter.Ib= Ic/hfe Current at the base is equal to the current you want from the collector multiplied by the Gain (hfe) Collector current is approximately equal to Emitter current .You must also consider the Active−Region Safe Operating Area the Tip 142 at 47 volts should only pass a maximum of maybe 1.5 A eachThe Active−Region Safe Operating Area of the MJE340 states that at 47 volts only 300mA should pass across the collector and emitter.The Active−Region Safe Operating Area of the TIP41 is 700mA at 47 Voltshope this helps a littleSo its possible your exceeding the safe operation of the transistors .This is why they are burning out.

author: AngraMelo
author_rank: Regular Contributor
timestamp: August 28, 2019, 05:44:44 am
content: Quote from: Jwillis on August 28, 2019, 05:24:21 amThank you!! Im making the math and Ill try the modifications!

author: amyk
author_rank: Super Contributor
timestamp: August 28, 2019, 11:32:31 am
content: Note that potentiometers are not meant to pass any significant current.

author: Ian.M
author_rank: Super Contributor
timestamp: August 28, 2019, 12:42:04 pm
content: N.B. A potentiometer's power rating assumes a uniform current through the whole track. If you are only using 10% of the track length it only has 10% of the power rating. Divide the wattage by the total resistance and take the square root to get that max rated current. If you exceed that current through any terminal (including the wiper) you'll burn out the pot sooner rather than later. Assuming a 1W rated potentiometer, with a 100K pot, its good for max. 3.16mA.With 47V across it your pot only has 0.47mA goint through its track. That isn't much to supply base current for your triple Darlington emitter follower, especially as one generally doesn't want to draw more than 10% of the track current at the wiper. If you switch to a 10K pot, its going to have 4.7mA through it and is good for an abs. max current of 10mA. The down side is the greater load on the Zener shunt regulator circuit, as the 2K2 resistor only provides 8.6mA. Looking at the OnSemi 1N4756A datasheet, its tested at a nominal 5.5mA Iz, and 3.9mA remaining for the Zener will be good enough.When you rapidly turn the output voltage down, the previous voltage the output capacitor was charged to causes Vout to exceed Vwiper. If that's by more than 15V or so, all the B-E junctions will Zener, and pop goes your pot and probably the weakest transistor as well. The easiest fix is to put a diode in series with the pot wiper to block reverse current. As Andy points out above, you'll also need a B-E resistor for each transistor to handle its worst case C-B leakage current, and with the wiper diode in place, the B-E resistor also prevents the B-E junction zenering.You've still got the problem of protecting the transistors against the surge current during startup, or if the load draws too much current. However adding effective current limiting to the design becomes quite tricky, as it would need to be fold-back limiting to protect the transistors effectively, and it rapidly turns into a full-blown high current high voltage bench PSU design exercise which is far from being beginner friendly.As an alternative, if all you need is variable 'raw' DC for amplifier testing, consider scrapping everything to the right of the fuse and simply putting a variac on the AC input!

author: GerryR
author_rank: Frequent Contributor
timestamp: August 28, 2019, 12:55:04 pm
content: Quote from: Jwillis on August 28, 2019, 05:24:21 amYou just lost me here. When the 100k pot is at zero ohms, isn't the base, even through a 10k resistor at ground, therefore no base current?? Just asking.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 28, 2019, 05:49:24 pm
content: You can get a hi-voltage, hi-current op amp, OPA548, or similar, for ~$10 to $15, set up as a unity gain amp. Tie the input to the wiper of your pot and use the output to drive your transistor. You will have to make some changes to your circuit, but this will isolate the pot from the driver section. You should be able to get within a couple of volts of the rails on each end. Just a suggestion.

author: Ian.M
author_rank: Super Contributor
timestamp: August 28, 2019, 06:05:23 pm
content: An OPAMP isn't an easy option with a 66V supply rail. If you are going to the trouble of either getting a HV OPAMP, or providing a floating supply for a regular one, you might as well design a proper regulator using the OPAMP with feedback from the output.

author: jmelson
author_rank: Super Contributor
timestamp: August 28, 2019, 07:04:15 pm
content: Quote from: Ian.M on August 28, 2019, 06:05:23 pmThere are such items, like the LTC2057HV that can go about that high. Very nice op amp for high voltage supplies.Jon

author: GerryR
author_rank: Frequent Contributor
timestamp: August 28, 2019, 08:20:56 pm
content: Quote from: Ian.M on August 28, 2019, 06:05:23 pmApex makes some that are good to 400 V and 4 amps continuous, but they are quite expensive. That is why I suggested the OPA548, which will operate within a couple of volts of the rails and can support a 60 volt single supply. I do agree that a proper regulated supply would be optimal, but the OP is into this and just seems to want to get it working.

author: Jwillis
author_rank: Super Contributor
timestamp: August 28, 2019, 10:17:15 pm
content: Quote from: GerryR on August 28, 2019, 12:55:04 pmYour pulling current from the Zener rectifier which supplies 21.36mA at 47V.Ohms law. I=V/R .If you insert a 10K resistor between that and the base of the transistor the base will pull 4.7mA current across that 10K resistor at 47Volts .The 100k potentiometer will also pass 0.47mA current from the zener to ground .The potentiometer is acting as a voltage divder. As the resistance increases at the pot you have to consider pot resistance in series with the 10K resistor but the current across the pot to ground will always be 0.47mA. Since the pot is acting as a voltage divider the voltage at the base will be approximate 47 volts when the wiper is close to 10K resistor. When the wiper goes toward ground end of the pot then you'll have around 23.5 volts to the base resistor and the current will pass across 100k pot and 10K resistor limiting the base to around 0.2mA.But when you calculate the output of a transistor always use the maximum voltage and current you intend to pass through it based on your design.

author: GerryR
author_rank: Frequent Contributor
timestamp: August 28, 2019, 11:36:16 pm
content: Maybe I'm mis-interpreting where he placed the 10 resistor. I assumed he tied it from the wiper of the pot to the base of the MJE340. If that is what he did, then when the pot wiper hits ground, no current is going to flow into the transistor. Am I missing something here?? The zener / pot bias and currents are straight forward and the pot, if it was buffered, would supply 0 to 47 volts to the base of the transistor. Without a buffer (and some other changes), it is not going to work properly, as the OP has found out.

author: Jwillis
author_rank: Super Contributor
timestamp: August 28, 2019, 11:48:07 pm
content: Quote from: GerryR on August 28, 2019, 11:36:16 pmHey I could be wrong. I based my analysis on reply #6 assuming he placed the resistor between the wiper of the pot and base of transistor .Exact placement wasn't specified .If it was placed in a different configuration then that of course changes everything .Edit: The potentiometer will most likely have a small resistance between wiper and either end.But your right .My bad. Even if it was 10 Ohms that would only be 0.0047 so you are correct that it would be zero .Sorry about that. But either way I was considering the current specifically to the transistor at full voltage.

author: AngraMelo
author_rank: Regular Contributor
timestamp: August 29, 2019, 05:20:58 am
content: Quote from: GerryR on August 28, 2019, 11:36:16 pmYou are correct, I put the 10k resistor between the wiper and the base of the 340.Ill do the suggested modifications to the design (if you even can call it that lol) and Ill let you guys know.Thank you so much for you input!

author: GerryR
author_rank: Frequent Contributor
timestamp: August 29, 2019, 05:43:51 pm
content: Quote from: AngraMelo on August 29, 2019, 05:20:58 amJust remember that everybody starts at the beginning at one time or another; that's how you learn. (The only difference between a professional and an amateur is the professional knows how to cover up his mistakes!) I've been in electronics for about 48 years and I still am learning, and still having fun at it!