Voltage climb. - Page 1
Forum section: electronics / beginners

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 11:17:19 am
content: Looking for some help, I have built a voltage regulator PSU with a single darlington transistor, and a l7812 voltage regulator.The issue is when I use a base resistor of 25 ohms I get the expected 12 Volt output, but when I use a 100 ohm base resistor I get an output of 17 Volts.No components are changed in this circuit, just the base resistor. The transistor is a single MJ11015 TO3 power darlington transistor. Can anyone shed any light on this voltage increase with the 100 ohm resistor, yet normal output voltage with a base resistor of 25 ohms. Thank you for any help with this question.

author: Zero999
author_rank: Super Contributor
timestamp: March 17, 2016, 11:21:34 am
content: A schematic would be nice.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 11:36:13 am
content: This is the circuit, but I've used a single MJ11015 instead of the TIP2955, a 10 ohm resistor gives the expected output as well. I was advised to use a 100 ohm resistor, but that gives me an output of 17 Volts. A 25 ohm resistor works well, but I don't understand the voltage climb with 100 ohm resistor. Thanks for your help.

author: danadak
author_rank: Super Contributor
timestamp: March 17, 2016, 11:54:47 am
content: Looking at Fairchild datasheet they seem to do this differently -https://www.fairchildsemi.com/datasheets/LM/LM7805.pdfRegards, Dana.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 12:06:19 pm
content: I know this circuit works with a TIP2955 as I've built it, and it runs fine.But using the more powerful MJ11015 transistor, if I use a resistor between 10 - 25 ohms it works fine as well. What I am trying to find out is why I get 17 Volts output with 100 ohm base resistor that was suggested to me if using the MJ11015 instead of the TIP2955 package. I thought emitter followers amplifed current, but not voltage. As with 100 ohm base resistor I get an extra 5 Volts above the expected 12 Volts output. Hope that makes sense.

author: Delta
author_rank: Super Contributor
timestamp: March 17, 2016, 12:46:06 pm
content: What is the load?

author: SeanB
author_rank: Super Contributor
timestamp: March 17, 2016, 06:29:38 pm
content: Leaky transistor, the higher base resistor value means the leakage current from base to collector can turn it on more, as it is not being shunted as much by the 1R resistor and the base resistor. Change the transistor, and if there is no minimum load on the circuit add some, 10ma through a power on LED at the output should be enough to drop all the leakage current to ground so the voltage regulator can source current. Regulators do not often sink current, they all tend to source current so anything that adds more means it will rise to a high voltage or rise till some junction inside breaks down to limit voltage.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 06:54:26 pm
content: Ok, thanks for the reply, no load when I measured it, the voltage was 17 Volts, bravely Adding a 50 watt load the voltage dipped to 15.9 Volts be it with a very bright lamp. This was with a 100 ohm suggested base resistor. But as I said a resistor between 10 - 25 ohms gives a rock solid stable output.I just couldn't understand why it sent the voltage up that high, if I removed the collector from the output the voltage was normal with the 100 ohm resistor, but only having the voltage regulators 1 Amp ability. So if I wanted to parallel a couple, say two MJ11015 transistors for a bit more current, would 25 ohms do the job there ? I chose this transistor for its gain and higher power handling ability. Thanks for all the help. So is it this leakage thing with a base resistance of 100 ohms verses 10 - 25 ohms say ? Thanks again. This would make some sense because the voltage across the base resistor increases from 1.14 Volts to 1.9 Volts with a 24 ohm resistor.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 06:59:38 pm
content: Sorry that base voltage increases with a 50 watt load and that's the voltage measured across the base resistor of 24 ohms. Thanks again.

author: Audioguru
author_rank: Super Contributor
timestamp: March 17, 2016, 07:00:23 pm
content: The circuit needs a PNP transistor. Instead you are trying to use an NPN darlington that is completely different.The resistor in series with its base is not needed.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 07:37:11 pm
content: No the MJ11015 transistor is a PNP type. I would not have used an NPN transistor for this circuit.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 17, 2016, 10:53:53 pm
content: Quote from: SeanB on March 17, 2016, 06:29:38 pmThank you for your help, I'm going to research this more, I have tried to find the answer to this question in another forum, but made little progress there. You'd explanation offers real help, thank you !

author: Audioguru
author_rank: Super Contributor
timestamp: March 17, 2016, 11:59:37 pm
content: The datasheets are confusing about if it is NPN or PNP. It is actually a darlington and your 1 ohm base-emitter resistor must be re-calculated to about 2 ohms.If you use 100 ohms then the darlington will be turned on all the time as you saw.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 18, 2016, 12:49:46 am
content: I am using a 0.1 ohm 10 watt resistor, it's dissipating 2.1 Watts with a 50 watt load on the output. At the moment I am using a 24 ohm base resistor, it's measures 1.9 Volts across it with the 50 watt output load. No load the voltage across the 24 ohm base resistor is 1.14 Volts. By increasing the value of the ceramic power resistor it will draw far more current.Yes the data sheet shows both an NPN and a PNP transistor.

author: Audioguru
author_rank: Super Contributor
timestamp: March 18, 2016, 01:32:31 am
content: 50W at 12V is a current of 4.17A.With a 0.1 ohm resistor on series with the input of the 7812 and a load of 50W then the voltage across the resistor cannot be more than 0.417V, then a PNP transistor or a darlington will do nothing and the 7812 should limit the output current to its maximum of about 1.5A.With a 1 ohm resistor the transistor will begin to turn on when the current in the 7812 is about 700mA.With a 2 ohm resistor a darlington will begin to turn on when the current in the 7812 is about 650mA.You DO NOT want a resistor in series with the base because it messes up its base-emitter voltage drop that determines at what current the transistor turns on.With no load then the transistor or darlington should be turned off with no base current then the resistor in series with the base should have ZERO volts across it.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 18, 2016, 02:38:02 am
content: Those numbers you quoted are as close as my meter readings. Are you surggesting using no base resistor ? And just a power resistor of 1 or 2 ohms to the emitter junction ? So have I got that correct, no base resistor whilst using this darlington transistor. Thanks for your help.

author: Seekonk
author_rank: Super Contributor
timestamp: March 18, 2016, 06:59:41 am
content: Does anyone here really believe that if you connect a 100 ohm resistor base to emitter that this transistor should conduct heavily just on its own?

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 18, 2016, 10:17:58 am
content: Quote from: Seekonk on March 18, 2016, 06:59:41 amThanks for coming in there, what value transistor base resistor would you recommend, with a ceramic power resistor of 0.1 ohms 10 Watts to the emitter ? So what value for the base resistor to forward bias the MJ11015 transistor ? Thanks again.

author: mikerj
author_rank: Super Contributor
timestamp: March 18, 2016, 11:10:34 am
content: Are you certain you don't have the 0.1R and the 100R resistor swapped over in your circuit? This could explain the symptoms.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 18, 2016, 11:26:52 am
content: yes 100% sure on that. Later i am going to try a 1 ohm 10 watt ceramic power resistor with the 100 ohm base resistor, see if the higher value power resistor resolves the problem.Yet a 24R 0.5 watt base resistor, with a 0.1R 10 watt power resister works fine. I am only trying to find out why the 100R base resistor gives a 17 Volt output, yet a 24R base resistor works fine.With the 24R resistor and the 0.1R 10 watt power resistor i have had it powered up for a day with a 50 watt load, current measured 4.5 Amps, and no voltage drop, its rock solid on the output.Its just finding out why the 100R resistor makes the circuit behave in this odd way.

author: Zero999
author_rank: Super Contributor
timestamp: March 18, 2016, 11:49:28 am
content: Quote from: davelectronic on March 17, 2016, 12:06:19 pmThe additional transistor is not an emitter follower. It starts to turn on when the voltage across the base-emitter resistor is high enough to turn it on. Current is diverted around the LM7812, which should continue to regulate its output voltage as normal.Quote from: Seekonk on March 18, 2016, 06:59:41 amI certainly don't. Assuming the circuit is correctly assembled and none of the components are faulty, increasing the base resistor should not increase the output voltage. If anything it should increase the drop-out voltage and the output should fall, if the input voltage isn't high enough.The only thing which could explain this is the base-emitter resistor being too high, which would cause the bypass transistor to turn on, due to the LM7812's quiescent current, but it would need to be 150R or more for that to happen with a Darlington pair.

author: Seekonk
author_rank: Super Contributor
timestamp: March 18, 2016, 01:19:12 pm
content: My opinion is that this transistor is JUNK and not fooled with any further. I've heard there are Chinese clones of high power transistors like this.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 18, 2016, 07:05:30 pm
content: Thanks for the replys. I think the resistance and the idle current is the problem. I use a trusted uk supplier for componets, and in 10 years of purchasing from them have not had a bad component. I have another 5 of these transistors anyway. But that is not to say it might be a dodgy component, there is a first time for every thing.I plan to try 100R base resistor and a 1R 10 watt power resistor. All i know is with a 24R base resistor and 0.1R power resistor it works fine. But i am going to try this 1R power resistor with 100R as base resistor, and measure voltages across them.

author: Seekonk
author_rank: Super Contributor
timestamp: March 18, 2016, 07:29:33 pm
content: Of course, you may be oscillating like crazy with 100 ohm. Why don't you try it with transistor alone with 100 ohm B-E. Spec says less than 1ma leakage with 1000 ohm @60V. If you measure more than that it is a dud. Three terminal regulators do like to sing.

author: Zero999
author_rank: Super Contributor
timestamp: March 18, 2016, 08:45:48 pm
content: I agree, oscillation could be a problem. I'm surprised no one else has suggested it.Look at the output on an oscilloscope set to AC coupled.

author: Audioguru
author_rank: Super Contributor
timestamp: March 18, 2016, 09:47:28 pm
content: Semiconductor manufacturers agree on a simple way to boost the current of a voltage regulator IC. The circuit uses a 3 ohm Current-sensing resistor from emitter to base on a PNP power transistor.The regulator does not get hot because the transistor begins to turn on when the current in the regulator is 0.6V/3 ohms= 200mA.There is no problem with leakage current. There is no 0.1 ohm current sensing resistor. There is no 24 ohm or 100 ohm resistor in series with the base. It does not oscillate.The Fairchild datasheet was posted earlier but here is its circuit and the same circuit from National:

author: Seekonk
author_rank: Super Contributor
timestamp: March 18, 2016, 09:56:52 pm
content: The OP's input cap is a rather large 470uF and the darlington transistors gain is quite large. I wouldn't discount the possibility.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 18, 2016, 11:48:37 pm
content: With a 24R base resistor, and a 0.1R 10W power resistor to the emitter, its rock solid with a 50 watt halogeon lamp, drawing 4.5 Amps. I have run it solid for 24 hours, just over.I have tried the 100R base resistor, and 1R 10W power resistor to the emitter, it put up the voltage across the base resistor from 1.02 Volts to 1.20 Volts, but the output was still 17 Volts.With a 24R base resistor the voltage across it is 1.18 Volts, with a 50 watt load on the output, this goes up to 1.92 Volts.The voltage across the power resistor was 0.457 Volts with the same 50 watts on the output.Unfortunatley i don't have use of a scope.

author: danadak
author_rank: Super Contributor
timestamp: March 19, 2016, 12:24:47 am
content: The case where Rb = 0 implies the transistor turns on ~ Il x Rsense = Vbe,or when Il >= Vbe / Rsense. Il = Iregulator at turn on point.In the case Rb > 0, Il >= ( Vbe + Ib x Rb ) / Rsense, in other words Rb increasesthe effective turn on of the boost transistor. If it has very high beta then it has noeffect, but for devices with low beta the boost transistor will start to turn on at higherload current.As far as oscillation, a small amount of Rb helps kill the - Z that can appear in widebandtransistors. - Z is effectively a power generator/oscillator.Regards, Dana.

author: Audioguru
author_rank: Super Contributor
timestamp: March 19, 2016, 01:23:41 am
content: Why not get rid of the series base resistor and if you want a darlington transistor that has two transistor base-emitters in series then use a 6 ohm (or 5.6 or 6.2 ohms) resistor instead of 3 ohms for one transistor and a 0.33uF input capacitor on the voltage regulator IC instead of your 470uf capacitor? Of course the input to the circuit (the emitter of the PNP transistor) must be fairly smooth DC with a large filter capacitor.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 01:39:22 am
content: Quote from: davelectronic on March 18, 2016, 11:48:37 pmIt would be much easier to understand, if I/We could see an updated Schematic diagram of the latest circuit. Also if the parts were numbered (R1, R2, D1 etc), it would be easier to understand what changes you are making.I may be misunderstanding your changes, but ...I can understand why the 24 ohm resistor, would work ok. Because the Darlington needs MUCH less current, as it has a much higher current gain (Hfe), compared to the standard non-darlington PNP transistor.But raising that resistor to 100 ohms, could cause your 17 volts, because the quiescent current (which I have to GUESS, because I CAN'T see your latest schematic), could (conceivably) be creating enough voltage across the 100 Ohm resistor (even with NOTHING connected to the output/load. Because the regulator and other circuitry, use a small BUT significant current. Even with NO load), to turn the darlington transistor on TOO MUCH. Hence the 17 volts you are seeing.Hopefully the 0.1 Ohm resistors you are mentioning, are TOO SMALL (low valued), to make much difference.

author: danadak
author_rank: Super Contributor
timestamp: March 19, 2016, 01:51:59 am
content: Here is a spice simulation (I used LM317 because I did not have LM78xxx, shouldmake no significant difference in sim).I tried a base R of 40 ohms, circuit was completely out of regulation. The reg was incurrent limiting I think.10 ohms seems to be a good value.Regards, Dana.

author: Audioguru
author_rank: Super Contributor
timestamp: March 19, 2016, 02:52:48 am
content: Again, why do you have R5 in series with the base of the transistor? Get rid of it.Your latest schematic is missing the important small capacitors at the input and output of the regulator.

author: Delta
author_rank: Super Contributor
timestamp: March 19, 2016, 03:02:08 am
content: Just use the bloody industry standard circuit that Audioguru posted and be done with it!

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 03:33:25 am
content: Quote from: Delta on March 19, 2016, 03:02:08 amAlthough they (OP) do not seem to be mentioning it in THIS thread. Their longer term intention could be to use multiple output pass transistors, in the circuit. But I DON'T KNOW!. Best to wait until the OP mentions this.Hence the mentioning of having 6 of the power darlington transistors, and the messing about with the 0.1 Ohm (presumably) balancing resistors.From another thread (https://www.eevblog.com/forum/beginners/transistor-alternative/msg847748/#msg847748)...

author: SeanB
author_rank: Super Contributor
timestamp: March 19, 2016, 06:27:17 am
content: You can use multiple pass transistors ( all hopefully mounted on the same heatsink and with the regulator mounted there as well so you can abuse it's SOA and overtemperature protection) with emitter ballast resistors, and leave the base resistor out. The 3R resistor will provide protection, and the regulator itself will provide the current limiting into a short circuit as well, though the current will not fold back but will be pretty high, just not for long as the transistors will overheat and the regulator will shut down eventually from overtemperature ( though some regulators are very noisy electrically going into and out of thermal limiting, so best to add a 90C Klixon self resetting or manual reset thermal switch), and use the 330n ceramic ( not SMD, the voltage derating of capacittance will cause issues, use a larger capacitor or a polywhatever film unit) on input and output directly, and then the recommended protection diodes ( rate those across the pass devices so they can handle the full load current, though this will only be in fault conditions) and then a minimum load if the output can be disconnected.Years ago I designed a bench power supply to run off the 28VDC supply to the test bench ( capable of supplying 16kA, but the limit in the workshop was 60A due to the breakers used) and used a 723 regulator and a dozen paralleled 2N3773 transistors, with 10 as the final device, using 20cm lengths of 20SWG PTFE covered wire wrap wire ( lots of that around both new and used) and the other 2 as a darlington driver. Designed to handle 60A with no problem, drawback was the in wall wiring was not thick enough, the original contractor had skimped in pulling the cable. There was only a 2.5mm cable as supply, and at 30A the voltage drop was 15V. Still was able to weld using old battery carbon rods with it, though then it blanketed out all radio reception in the complex from the broadband noise it generated down the supply. Built all from salvaged and scrap parts, the heatsink was a massive aluminium extrusion originally intended for partitioning, and I just isolated it from the grounded bench with bakelite, as it made connecting all the collectors easy.

author: Zero999
author_rank: Super Contributor
timestamp: March 19, 2016, 08:46:09 am
content: Quote from: Delta on March 19, 2016, 03:02:08 amThe original poster's problem isn't getting it working (he's already done that) but trying to figure out why the circuit is behaving as it is.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 10:21:08 am
content: Thanks for all your replys. Yes that is the schematic i am using, i have built a two transistor version using the TIP2955's its powering my cb radio and amplifier, and has been for about a month now.I wanted to try and build a version using the MJ11015 transistors. Again the single transistor version of this i have built is a model to see if it works, before commiting to a 4 x transistor version. Nothing in the circuit has changed, except the transistor, and the number of them x 1 for testing.A value of R7 at 24R works, although there is an inital voltage climb by 0.7 Volts when first connecting a 50 watt load, but this settles down to 12.20 Volts with in a minute.The idea is to use the circuit above (which works) but with 4 x MJ11015 transistors. But i have built a single transistor version with the MJ11015 transistor to see if it will work, and how it behaves. And with a 24 R resistor at the transistors base it works, with a small voltage climb on inital load connection to 50 watts.With a base resistor of 100R i get 17 Volts output, i have even connected a 50 watt load to that, it dropped the voltage to 15.90 Volts, be it with a very bright lamp. So i am trying to find out why 100R does not work, as it was surggested to me to use it and it would work. Hope some of that explanation helps. Thanks for all your help.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:01:52 am
content: This is a picture of the single MJ11015 i am trouble shooting.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:24:47 am
content: I have tried a 10 ohm base resistor, it also worked fine, although there was a 0.2 Volts drop when loaded to 50 watts on the output. With a 24R base resistor there is no voltage sag when the output is loaded to 50 watts, just that intial 0.7 voltage climb that settles down in around 1 minute of connecting that 50 watt load.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:27:08 am
content: Last post wrong quote. It was ment for Dana simulation. And his mention of a 10 R base resistor.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 11:38:34 am
content: When R7 is 100 Ohms and (WITHOUT anything connected to the output, i.e. NO load) you get 17 Volts output. What voltage is there, measured across R7 ?

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:44:51 am
content: The voltage across R7 at 100 ohms is 1.02 Volts, with 24 ohms as the base resistor its 1.18 Volts, rising to 1.9 Volts with a 50 watt load.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 11:58:22 am
content: Quote from: davelectronic on March 19, 2016, 11:44:51 amThe datasheet seems to say, the voltage regulator will use about 8 mA's (with NO load).Other stuff, such as the electrolytic capacitor, could be using another couple of mA's, maybe.So the 1.02 Volts, could be caused by (Ohms law), about 10 mA's flowing through it.i.e. That seems perfectly reasonable.The 1.02 Volts, then going to the Base, maybe SHOULD have been JUST ABOUT low enough to NOT activate the darlington.But I'm NOT surprised, if it does/has.1.02 Volts is TOO HIGH, as it may turn on the pass transistor (darlington), and give you 17 Volts.i.e. I think the 100 Ohm value may be a DESIGN MISTAKE.R7 = 24 Ohms, under NO LOAD, should give you about a quarter of the 100 Ohm voltage reading (across R7), so 1.18 Volts, DOES NOT sound right. It should be more like 0.25 Volts (NO LOAD).

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 12:10:07 pm
content: Could the cooling fan and voltage drop series resistor be the reason for the 1.18 Volts with the 24R base resistor, i've a feeling it is that, as my measured voltages had that fan and series resistor running. Oops could that explain it ? My over sight if it does. But it still does not explain the 17 Volt output with 100 R base resistor, but your theory on the quiecent current.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 12:22:41 pm
content: Quote from: davelectronic on March 19, 2016, 12:10:07 pmDefinitely having the fan connected up to it, will use (I DON'T know specifically what fan you are using, but it does NOT matter too much) plenty of current. Such as hundreds of milliamps, so plenty of opportunity for the 24 Ohm resistor (R7), to have plenty of voltage across it.The original schematic above, using the TIP2955 and 100 Ohm resistors, with NO LOAD, seems to be a DESIGN ERROR, as shown.Because the 8mA (approx) for the voltage regulator, should give across R7, about 800 millivolts (0.8V), which is well above the 0.6/0.7V needed at the base to switch the TIP2955 ON.Darlingtons would/should be more tolerant of 800+ millivolts. But because of their HUGE (typical) current gain of around 10,000 (TYPICAL, speaking about darlingtons in general), and significant leakage current, and maybe other complications (as it has 2 transistors, 2 resistors and a diode, built in to it).

author: Delta
author_rank: Super Contributor
timestamp: March 19, 2016, 01:04:49 pm
content: So every time you've mentioned readings at "no load" you've actually had a sizeable fan as a load?

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 06:20:54 pm
content: Thank you for the help here. Yes i am going to have to re test with both the 100R and 24R base resisters with the fan and its series resistor dissconected. Ever felt stupid lol, compleate over looked on my part. I will post new voltages across both base resistors, less the fan.I've been reading up on transistor types and theory, although i do find some of the complex equations hard. Maths was never my strong point, i'm ok with ohms law and a bit more, but some of them complex equations i find quit tough.

author: Seekonk
author_rank: Super Contributor
timestamp: March 19, 2016, 07:41:23 pm
content: A Decision Analysis course stated that in spite of good intentions 2/3 of the information we are told is not true and we are too quick to jump to conclusions.In spite of knowledge that tells us the device should be going into current limiting, we accept the voltage is increasing. Since you don't have a scope it would be interesting to see that voltage measured through a 470-1000 ohm resistor feeding a 47-100uF capacitor your meter is in parallel with. If pure DC the voltage will be very close. If oscillating the voltage will be much lower. Just something to try.You seem genetically hard wired to put that resistor in series with the base like it is a FET. As the last schematic demonstrated, the transistor is a current device and even a darlington requires a current drive.There ought to be a sticky for three terminal regulators. These three terminal regulator threads show up every week and always run pages in length.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 10:08:03 pm
content: Well after some more testing i'm even more puzzled.No load on the output for the following.100 ohms base Volts across it, 0.55 Volts output will not come down below 17 Volts.24 ohms base Volts aross it, 144.2 mV . Its unstable at light loads, gaining at least 2 Volts with a 20 watt load, 0.7 Volt increase with a 50 watt load. But will drive the cooling fan and its series resistor with stable voltage.10 ohms base Volts across it 60.2 mV . When loaded to 50 watts its stable with 0.3 voltage drop. 20 watt load sends the output between 11.50 and 11.98 Volts seems unstable, drives the cooling fan and its series resistor stable output.The power resistor i used was 0.1 ohms 10 Watts. For 1 ohm resistor i would need it to be at least 20 Watts. I don't have one to hand at the moment. I tried 10 ohms base resister and a 1 ohm 10 watt power resistor. Volts across the 10 ohm base resistor was 5 Volts, i know way to much for half watt resistor. The voltage across the 1 ohm resistor was 4.1 Volts, again to much for that resistor. It was stable with all loads until the 10 ohm base resistor failed.I think from this a base resistor has to be higher than 10 ohms, maybe 15 ohms 3 watts, and the power resistor of 1 ohm 20watts. A 10 ohm resistor and 1 ohm power resistor does not limit the voltage below 5 Volts with a 50 Watt load. Hope some of this makes sense. Can anyone surggest a resistor combination for this single MJ11015 darlington power transistor and l7812 regulator circuit ? Thanks for everyones help so far.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 10:22:00 pm
content: As others have said, it could be badly oscillating and you don't have the equipment to develop stuff like that (no scope etc).As far as I know, there are MANY published (known to be working), PSU designs available, including for very high currents. Why not use or ask for, such a design ?It sounds to me, like you are just going round and round in circles, and not really getting anywhere.I think you can buy ready made PSUs (high current), somewhat cheaply. E.g. Around £50, depending on the exact spec.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 11:02:30 pm
content: Quote from: davelectronic on March 19, 2016, 10:08:03 pmThat reading makes little sense to me.17 Volts means that you will potentially destroy/burn-out/explode/overheat/damage whatever 12V expecting stuff you connect to it. Unless it has been designed to cope with a much higher voltage.If you disconnect JUST the base of the power transistor, does the voltage change ? (across R7 and/or output)If you remove the power transistor, does the voltage change ?But really I think you would be better off, taking one of the steps in my previous post.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:42:14 pm
content: I like building psu circuits, linear ones for now. I have no intention of connecting anything 12 Volts to a 17 Volt output circuit.No i don't go round in circles, rather exaust all posabilitys before giving up. I can see this particular transistor is not best suited for this application. But i gave it a try ! Thanks for the help people.

author: MK14
author_rank: Super Contributor
timestamp: March 19, 2016, 11:49:53 pm
content: Quote from: davelectronic on March 19, 2016, 11:42:14 pmYes, building PSU's can be real fun!The results you have been getting and the schematic circuit supplied, are NOT really making sense (agreeing).Hence my suggestion to try partially/fully disconnecting the power transistor, to try and find out why the strange/mysterious 17 Volts is occurring, apparently WITHOUT enough voltage to switch the power transistor on.tl;drOne or more things are wired up wrong (different to the schematic) and/or one or more components is bust/faulty. Or some other explanation. Such as it is oscillating e.g. Because 100 Ohms is too high a value for LM78XX series regulators (which is something I suspect may well be the case, BUT I'm NOT 100% sure).

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 12:29:35 am
content: Thanks for your help, i've built TIP2955 versions single and double transistors with very good results. I think maybe this transistor the MJ11015 is not best suited to this circuit application. I would like to build a unit with 4 x transistors for HF radio use, wanting to stay with a TO3 package, i'm looking at the PNP MJ2955 its not a darlington, the gain is not huge either, but if its as close to the TIP2955 then i would give the MJ2955 a try.

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 12:48:32 am
content: Quote from: davelectronic on March 20, 2016, 12:29:35 amYes, Darlingtons such as the MJ11015 can make things much more complicated. Their characteristics are harder to deal with (as you probably have just found out), because there can be such a huge variation in how much gain (Hfe), you are going to get, at a particular base voltage and temperature.I prefer to treat Darlingtons as on/off devices, as their HUGE gain (Hfe), can make them too sensitive/tricky to try and operate in linear mode and/or as a voltage regulator. It can be done though, some people do that. But somewhat often with op-amps, which close the loop so well, that they/it can cope.That basic circuit (from your schematic, but probably based on the datasheet application circuit, for external pass transistors), was not really designed to cope with darlington transistors. E.g. They need something like DOUBLE the base drive voltage, which probably worsens or even destabilizes, the circuit.The "standard" datasheet application circuits, all seem to use 3 Ohm resistors. So I would be VERY tempted to stick with 3 Ohms, and "normal" non-darlington transistors. There are probably a number of VERY good engineering reasons, and experimental experience findings, which make the 3 Ohms, a VERY good value to use.N.B. I'm getting increasing convinced that making R7 100 Ohms, is a VERY BAD idea. It may well cause terrible instability and NOT give the LM7812C enough current (etc), to operate well/nicely. Is there any evidence that the designer who made the value 100 Ohms, was sensible/experienced at doing circuits like that ?I would think (as you suggested) the PNP MJ2955, is well worth a try. It is possible that differences between that, and TIP2955's, might be significant, but I would hope NOT!The circuit should be able to cope with at least a small amount of difference (one would hope).Anyway, Good Luck!

author: Audioguru
author_rank: Super Contributor
timestamp: March 20, 2016, 12:51:48 am
content: I repeat, the circuit does not use a resistor in series with the base of the transistor or darlington. The circuit with the 100 ohm current sensing resistor has an error.You want to use a current-sensing resistor of 0.1 ohms so that it does not get as hot as a 1 ohm resistor? Then simply use Ohm's Law to see why it will not work. When the voltage across the resistor is 0.5V to 0.6V then the transistor will begin to turn on. Then the current in the regulator is 0.5V/0.1 ohms= 5A! (impossible with a 1A regulator). If you use a darlington then the regulator current must be 9A!

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 01:01:38 am
content: Quote from: Audioguru on March 20, 2016, 12:51:48 amWhat you describe, does not seem to fully match the supplied schematics.The 0.1 Ohm resistors, seem to be connected between the Emitters, and the raw incoming positive voltage supply, from the main smoothing capacitor.The 0.1 Ohm resistors, appear to be (load) balancing (current sharing) resistors, to allow slight/partial differences, between the multiple power transistors, base voltages. Without them, multiple power transistors, could be in danger of partial or worse thermal runaway.tl;drOtherwise one of the 4 or 6 power transistors, could have a base which activates some 0.15 volts, BELOW the others. So that transistor, would attempt to take ALL the current (and power dissipation), through itself. Potentially leading to that transistors early failure, due to over-heating.I.e. They are NOT current sense resistors, in this application.EDIT: In fairness to yourself, for most of this thread, there was only ONE (single) power transistor (mentioned). So no current/load balancing was needed. Adding to the confusion around 0.1 Ohm resistors. Sorry.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 01:25:25 am
content: This makes absolute sense 100% ! The theory for the 100R base resistor came from a member moderater from another forum ! limited as my knowledge is, i have learned a lot.Yes i tried no base resistor, all i got was the regulators rated voltage and current, as the transistor did nothing.Thanks again for all the help !

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 01:36:43 am
content: Quote from: davelectronic on March 20, 2016, 01:25:25 amBecause you disconnected the base, and left the rest of the transistor connected. We have apparently eliminated it being transistor leakage current.Also it shows that the 17 Volts IS being triggered by the Base of the transistor. Rather than some other cause.So either the base is getting a much bigger voltage than you measured, or there are oscillations when the base is connected, and the apparent lowish voltage (across R7), is because the meter is NOT showing the right voltage, because it is fluctuating (oscillating). Tricky to diagnose (Oscillations) WITHOUT access to a scope.Early indications are that this further shows the 100 Ohms is way too high a value, in this type of circuit. E.g. Terrible oscillations.IC Voltage regulators are somewhat notorious for oscillating at the best of times, anyway. Hence the (often) need for significant decoupling capacitors across their inputs and outputs.

author: Audioguru
author_rank: Super Contributor
timestamp: March 20, 2016, 02:20:14 am
content: PLEASE post the schematic you are using with its parts numbered.If the base of the PNP transistor is not connected to the input of the voltage regulator then the transistor does nothing.If you are using a darlington then please SHOW a darlington and the increased value of the current sensing resistor for it.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 02:53:18 am
content: You sure know what your talking about, and i'm no expert. For now i'm not going to continue to try and tame these monster transistors. It makes more sense to use a single transistor alternative PNP TO3 package.I did have bypass caps on the regulator itself, mounted with the regulator on a small three hole piece of veroboard.A tip was give to me to use a meter on an AC range to try and spot osallation, its one i will remember, but its time for me to move on, for now at least.Thanks again for all the help !

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 12:38:50 pm
content: Update:I seem to have found an older/same? version of your circuit, but complete with technical explanations on how it is suppose to work.http://www.zen22142.zen.co.uk/Circuits/Power/1230psu.htmhttp://www.eeweb.com/blog/circuit_projects/lm7812-high-current-power-supply-by-tip2955-pass-transistorsI can believe the original circuit can work out ok, and give 12 volts.But I think where things have gone wrong, is by your substitution of the standard power transistors, with the darlingtons.The original circuit (it confirms this in the explanation), was relying on the significant base currents, in order to provide the regulator with a decent load.But when you put in the darlington(s), they have a considerably higher gain (Hfe), of about 100 times higher. Maybe even more. Therefore that will potentially pass considerably LESS current, back into the regulator. Which could be worsening its tendency to misbehave/oscillate etc. I.e. Darlington(s) may make the R7/100 Ohms unsuitable/incorrect, in that situation.Without much more extensive analysis etc, I CAN'T be 100% sure, exactly how it will behave in practice. I've played about with some simulations, but that has only partially improved my understanding.tl;drProbably best to use the specified (or similar) power transistor type. The TO3 version, should still be ok as well, as it should give broadly similar base currents. But even that change is a small risk.Reportedly (as detailed on the web page(s), linked to above), it DOES WORK, when built EXACTLY as specified.Changing the transistors to darlingtons, is apparently a step too far.

author: Delta
author_rank: Super Contributor
timestamp: March 20, 2016, 06:52:45 pm
content: Mate, you have seriously fucked something up if your base resistOr is overheating. There should be next to no current flowing into the base of a transistor at DC, especially a Darlington pair.Stop what you are doing, and start again. An outboard boost transistor across a regulator should not have a base resistor anyway!

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 07:04:01 pm
content: Your spot on MK14 ! Yes that is a circuit i refferenced to build this circuit, and can confirm with the transistors in the schematic it works well. It was my idea to try this high powered darlington. Another friend has simulated it, and added an RC network gives a stable output with 100R base resistor.I did explain some wild voltage fluctuation to him on inital loading of the circuit with these darlington transistors. But for now with no scope i'm going to try the MJ2955 transistor instead, as with no scope i can not trouble shoot with the darlingtons in circuit. I might go back to this circuit with darlingtons once i have a scope.Along with the wild voltage fluctuations, i was also getting some odd buzzing from the cooling fan, but it was not constant, rather intermitent on connecting loads. Lighter loads seemed worse. Thank you for taking time to look into this, at least i have a better understanding of what was going on in the circuit with darlington transistors. Thanks again.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 07:29:29 pm
content: Quote from: Delta on March 20, 2016, 06:52:45 pmI am not continuing with the darlingtons for now, i'm aware i overloaded a resistor be it for a very short time, i replaced it once i relised what i had done. For now i'm continuing with more modest non darlington power PNP transistors. Thanks for the input.

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 07:32:34 pm
content: Quote from: davelectronic on March 20, 2016, 07:04:01 pmAssuming we are both right, that the problem is caused by the changeover to using darlingtons. You and/or your friend and/or someone else, can fix the new circuit, so that it DOES work with them, eventually, some time in the future. (As you said, a scope would help see what is going wrong, in better detail).Darlington transistors, can be (in theory) slower than standard transistors, which probably doesn't matter in a (somewhat slow) DC power supply. But it might (e.g. transients, stability etc).I guess after all this difficulty with them. It makes it an interesting challenge, to solve any issues with them, and make a darlington power transistor based PSU, work well.On the other hand, I DON'T think darlington transistors, offer any real advantages (in this application), to using standard power transistors. Their big advantage (for some applications), is their huge gain (Hfe), of e.g. 10,000 (typically). Hence needing minimal driving requirements. BUT that does NOT help with the circuit you are using. Because it already can drive, normal power transistors.But as you have just said, for the time being, going back to the MJ2955 (hopefully similar enough to the TIP2955's), is the way to go for now.I see getting a scope as a sort of investment (in Electronics), for the next 5 to 10+ years. You learn a lot more with a scope, and it does things that other test equipment can't really help you much with. By seeing EXACTLY how it is misbehaving, you can get a much better idea as to what needs to be done, to fix it.As you said previously, you can use AC mode on multimeters (to some extent), and other tricks. But it is a lot of messing about and hassle, just to get a tiny fraction of the information, that a decent scope would give you, in seconds.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 08:09:23 pm
content: I chose these MJ11015 darlingtons as i wanted a more higher maximum power devices, and less voltage sag or drop on the output, my oversight was not looking into darlington behavior in a psu type circuit first. I have looked up the MJ2955 and TIP2955, and compared them, there is little to no difference at all, other than higher power dissipation 90 watts for the TIP2955, and 115 watts for the MJ2955's All other specifications look the same, including hfe. So hopefully should be ok.The aim is to build a final 20 Amp linear psu for cb radio and linear amplifier use. I will need 15 Amps 12 Volts on about a 50% duty cycle, so figure 20 Amps for a final build should be plenty. When i did have the darlingtons running, for the same load on the output things run much cooler. But for now the instability is a problem with out that scope. But i should be able to utalise the MJ2955's hopefully.

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 08:18:16 pm
content: How about THIS ?STMICROELECTRONICS 2ST2121 Bipolar (BJT) Single Transistor, PNP, -250 V, 25 MHz, 250 W, -17 A, 80http://uk.farnell.com/stmicroelectronics/2st2121/trans-pnp-250v-17a-bipo-to-3/dp/1751964£2.36 + VAT each, from Farnell.I'm NOT 100% sure about compatibility, but a quick glance looks good.250 Watts!

author: Audioguru
author_rank: Super Contributor
timestamp: March 20, 2016, 09:47:53 pm
content: Yes, you can use the obsolete transistor. If the input is 17V, the output is 12V and you limit the output current to 17A then it will heat with (17V - 12V) x 17A= 85W and it will need a huge heatsink with a fan.You can use a 3 ohms current sensing resistor and NO series base resistor. Use the capacitors shown on the circuit I posted.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 10:40:36 pm
content: That looks like an interesting transistor. My neivity... Why does it have - before all the listed parameters, other PNP data sheets i've seen don't, or is that normal.Forgive my newbie questions, where in the circuit would the sence resistor go ? My seem obvious to some, but i've not built many transistor circuits, mostly followed known designs in previous builds i have done. In fact i've mostly used the lm338T and the lm338K and the 2 Amp 78S12cv. Also converted a few ATX psu's and server power supplys, although some are noisey electrically and don't lend themself well to powering HF radio gear.So i've done very little with transistos really, would have a link to an example schematic diagram ? Thank you.That transistor is about what i paid for the MJ11015 transistors, x 6 lol redundant for the moment.So this first two versions of the TIP2955 transistor circuits are the only ones i have built to date.First a single transistor model, second with two transistors, that one is powering my cb and linear amp, and has been for about a month now. Max current drawn from the two transistor version is about 7.2 Amps full load.But i would be interested in this transistor and circuit for it, but for where does the current sense resistor go in the circuit, also the transistors base emitter and collector, are they as per the original schematic that used the 6 x TIP2955's thanks again,

author: MK14
author_rank: Super Contributor
timestamp: March 20, 2016, 11:16:55 pm
content: The PNP transistors, are sort of complete opposites of NPN ones. So since NPN has got positive values on the datasheet. You can (if you want to be really strict), make them negative values (as they are opposites). But leaving them positive, makes them easier to understand.E.g. A cars top speed might be 125 MPH, so we could say the top speed (while in REVERSE), is -25 MPH. But most people would say the top speed in reverse is +25 MPH.Current sense resistor (On Diagram, Marked Rsc=Resistor Short Circuit) example follows, see page 20 on the data sheet, or image below:https://www.fairchildsemi.com/datasheets/LM/LM7812.pdfThe new (250W) power transistors, pin configuration, is shown (towards the top) on the datasheet.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 20, 2016, 11:49:58 pm
content: Ok, thank you for posting that, i understand the reason you explained about the data sheet now, thanks.Are both these transistors the 2ST2121 or is one a driver for the 2ST2121 ? And the RSC is that like 0R or a jumper link etc.Its certainly a circuit i could build. I will check out page 20 thanks.Ok after looking at page 20 there are sone interesting circuits there, i'm guessing as they both say TIP 42 that both transistors are the same. Does that configuration represent a darlington pair, i'm not sure. But it must be good as its in the data sheet.

author: MK14
author_rank: Super Contributor
timestamp: March 21, 2016, 12:08:09 am
content: The schematic was mainly meant to be an EXAMPLE to give you an idea, as to how to do it. Rather than an EXACT suggestion.Q1 represents the 2ST2121.Q2 (TIP42 may be ok) detects the overcurrent and limits the output current, accordingly. The resistor value depends on what output current you need, and maybe other factors.At VERY high currents, it would be something like 0.05 Ohms (suitably high wattage as necessary, SHUNT resistor).There is a formula on that schematic, but I appreciate that the formula is NOT easy to quickly understand.If you are NOT 100% sure how to design that part of the circuit, then I guess you can ask here, or somewear, or your friend.The maximum output current (limit) you want will need to be known. The design of the rest of the PSU, probably needs to be fixed as well, as that may change the exact design of the overcurrent circuitry. Then it probably needs to be tested.At REALLY high currents, it can get tricky. As a problematic current limit, could overheat the output transistors.

author: MK14
author_rank: Super Contributor
timestamp: March 21, 2016, 12:35:38 am
content: Designing the current limit circuitry, can be deceptively complicated.For such HUGE currents (which I think you intend), it makes foldback current limiting, or maybe better still, something which switches the unit off (disables the output), somewhat important.https://en.wikipedia.org/wiki/Foldback_(power_supply_design)The "foldback" concept is especially important. Otherwise the power transistors, could be badly overloaded (overheated), in certain current overload situations.A quick glance at the circuit/schematic I supplied above, seems to show that it is NOT a foldback type design.But I'm NOT 100% sure, because the regulator HAS got its own limiting capabilities built in, and it would take me longer, to determine the full circuit operation.tl;drProbably NOT a foldback current limit type, unless there is a complicated interaction with the regulator, which I have NOT spotted yet!

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 21, 2016, 12:36:09 am
content: Thank you, i will try and find out more about this type of circuit. My maths is not the greatest to be honest. I'm ok with ohms law and a bit beyond that. But sure i will research this as much as i can. Its all new, and i've got a lot to learn, although i understand most theory in say tutorials etc, i lack the real time practical skills as i've yet to do a lot of these circuits, and test and trouble shoot etc. Thanks again for all your help !

author: MK14
author_rank: Super Contributor
timestamp: March 21, 2016, 12:44:40 am
content: Quote from: davelectronic on March 21, 2016, 12:36:09 amOr an even simpler solution. A 30 Amp fuse (or whatever value you need).Bench/adjustable general use power supplies, benefit from the convenience of automatic built in current limits.But one-off, dedicated power supplies, are probably reasonably ok with just simple fuse(s).Or go for BOTH, and if your current limit circuitry fails, the fuse (at a slightly higher current), will pop, and save the day.Also, foldback current limiting (see previous post), can cause problems, with it inappropriately refusing to start up, if too much current is needed, early on.

author: davelectronic
author_rank: Frequent Contributor
timestamp: March 21, 2016, 02:41:06 am
content: I will have to look up foldback current limiting. I've used a zener diode and fuse up until know for over voltage protection, and a final current limit fuse. So fuse and zener on the regulators output, and final fuse after the high current output. But i have looked in to crowbar circuits, and will have a go at building one into a final purposed built circuit.