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sureshot

Jul 7, 2012
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I think I've finally got there ! Making as close to 25 ohms as I could, two 12 ohm resistors in series, so tried 24 ohms for R7.

Voltage across R7 was 1.186 Volts unloaded, and 1.9 Volts with a 50 watt load on the final output.

The voltage across the 0.1 ohm 10 watt ceramic resistor was 0.469 with the same 50 Watts in the output. So I guess this value of AK's 25 ohms is ok. I'm not sure what was going on with the 100 ohm resistor value for R7 giving an output of 17 Volts. I am at a loss there !
 

sureshot

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Think I can move on from here, thank you for everyone's help, especially AK ! I've been over your post on page 2 a few times now, lol. This transistor the MJ11015 runs much cooler than the TIP2955 for the same load current. So whilst it's been a long thread, I got there in the end. Still learning... Thanks again all.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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0.98V across 10 ohms is 98mA

Are absolutely sure it's a 10 ohm resistor?

It's either a 100 ohm resistor or you have some load on the output.

Is it possible your 100 ohm resistor is a 1k resistor?
 

(*steve*)

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If you have the same number of transistors then the difference in temperature is due to heat sinking (perhaps the package radiates heat better, or perhaps you put it on a heat sink).
 

sureshot

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Yes Steve 100% sure, I went from a 100 ohm resistor down to a 10 ohm resistor. And the figures above where the result of that, that is with an input voltage of 24 Volts.

I always mount high powered packages on a heat sink, in this case a voltage regulator and TO3 transistor package on a modified CPU heatsink. And the figures for the 24 ohm resistor I tried are accurate also. As I said why 100 ohm resistor gave me an output of 17 Volts no load on the output, I have no idea. Nothing else In the circuit was changed.
 

(*steve*)

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the problem it that I can't believe you have 0.98V across a 10Ω R7 with no load assuming the circuit is constructed correctly and the 7812 is working correctly.
 

sureshot

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I have just put 100 ohms back in circuit and its 17 volts on the out put. then put 10 ohms back in and yes it reads the same 0.98 volts across it.

another thing is with a 10 ohm resistor for R7 , on the loaded output there is about a 0.2 voltage drop. but with a 24 ohm resistor there is no drop, and when first loaded on the output with 50 watts the voltage climbs to 12.92 volts, then slowly settles down to 12.20 volts. go figure i've no idea why its doing that. but zero voltage sag at 4.5 Amp load and even a slight voltage climb on connecting that 50 watt load. but i am sure its 0.98 volts across a 10 ohm resister for R7, but a 24 ohm resistor looks more favorable for an R7 value.
 

sureshot

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this is the circuit, best i can do with a picture. i had it running for over an hour drawing 4.5 Amps from a 50 watt lamp with 24 ohms for R7.
12.20 Volts and no voltage drop, just a slight voltage climb to 12.9 volts on i initial connection. but settles down with in the first minute or so.
 

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sureshot

Jul 7, 2012
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the problem it that I can't believe you have 0.98V across a 10Ω R7 with no load assuming the circuit is constructed correctly and the 7812 is working correctly.
Maybe this is the answer, as there is nothing i've left out or not mentioned. The only thing that is different is the l7812 is the 2 Amp version, its a 78S12cv, but i can't see that as a problem, as it worked fine in the TIP2955 transistor circuits i have built. The 2 x TIP2955 version i built has powered a cb radio and linear amplifier for at least a month now, it on 24/7.

Another thing using the MJ11015 transistor, is this small voltage climb on applying a load. I've read a bit about darlington transistors, does this climb in voltage represent a spike type transient, or is it some kind of compensation ?
I am not thinking its a transient because it takes at least a minute or just over to settle down. The voltage climb is from 12.20 Volts no load, to 12.90 Volts with a 50 watt load connected.

I should point out since trying this MJ11015 transistor, its as a single transistor version as a model to see how it would perform. And the heat for the same circuit with the same heatsink, with the TIP2955 verses the MJ11015 the latter runs much cooler for the same conditions and load. So i don't think the voltage climb is a transient, although not 100% sure. My equipment is not sensitive to this small climb in voltage, but is there a way to attenuate it for want of a better word. All the help is much appreciated.

PS. I was going to start another thread on the voltage climb issue, but thought it not a good idea to take up more forum space etc.
 

sureshot

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So i will make this a final post in this thread. Just one last question if anyone knows the answer. i'm just going to recap.

I built a psu circuit, a linear voltage regulator and a single pass emitter follower transistor. 78S12cv regulator and an MJ11015 darlington power transistor.

The circuit i used is on page 1 of this thread. My last question is again with the value of R7 and unexpected behavour from the circuit. With a value of 24 ohms resistor for R7 i get the expected voltage output, but if i use 100 ohm for R7 i gain an extra 5 Volts on the output giving me 17 Volts.

Can anyone offer an explanation for this strange behavour. I have searched extensively but can't find a reason. Many thanks all the help has been much appreciated. Any answer to this question would be a excellent, and fill in the blank !
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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yes, the reason can be found in a measurement I recommended you took but you decided not to.

quiescent current.

I can't really believe this reg has a quiescent current of 100mA, but apparently I does.
 

sureshot

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I've been told the high voltage with 100 ohms for R7 is leakage current. The only measurements that escaped me, or I forgot to take was the voltage across R7 at 100 ohms. It would have made no difference to the output voltage.

But I did measure them with the 10 ohm and 24 ohm resistors. This surley would have answered the increase in voltage with 100 ohm resistor.
 

sureshot

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As a newbie it can be tricky to take on all the terminology and understand exactly what's going on.

I'm an advanced diver, but would not expect a newbie to understand closed circuit gas mixtures.

The quesient current is 8mA according to the data sheet. I was having a hard time how a 100 ohm base resistor could send the output voltage 5 Volts above the expected 12 Volts output. I was just trying to make head and tail of what's going on.

And I've been told it's leakage current to do with base emitter. And I'm trying to understand this phenom. Thanks again !

Edit. Sorry leakage current from base to collector is what I've been told is sending the voltage high, as the regulator can't sourse current with a base resistor of 100 ohms. I'm kind of starting to understand this, I was told this leakage current needs shunting to ground to get the output voltage normal. And from what I can make out the lower resistance value does this, with some small load as well on the output.
 
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Arouse1973

Adam
Dec 18, 2013
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Here is something that might be of interest. I simulated approx. your circuit, see below. Everything works fine with 100R and with the circuit drawing 4.5 Amps. Now what's interesting is if I disconnect the common from the V Reg. Guess what voltage I get on the output. Yep 17 Volts. It only does this if I disconnect the common.

REG.PNG
 

Arouse1973

Adam
Dec 18, 2013
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Here is something that might be of interest. I simulated approx. your circuit, see below. Everything works fine with 100R and with the circuit drawing 4.5 Amps. Now what's interesting is if I disconnect the common from the V Reg. Guess what voltage I get on the output. Yep 17 Volts. It only does this if I disconnect the common.

View attachment 25674

Edit: After this I placed a 1K resistor in the common line and it did the same thing. This is possibly due to the internal 5.1V Zener lifting the nominal 12 Volts on the output up to 17 Volts.
 
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sureshot

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Tha
Edit: After this I placed a 1K resistor in the common line and it did the same thing. This is possibly due to the internal 5.1V Zener lifting the nominal 12 Volts on the output up to 17 Volts.
Thank you this is much appreciated ! I have ground coming in from the input filter caps, if I had no common surley there would be no output.

Why do I get 17 Volts with 100R resistor, yet 24R gives a rock solid 12 Volts. Once again if I had no common with the 24R in place I would get no output. I do appreciate your help here, but at no time have I removed - Volts to the regulator.

The zener theory is interesting here. If you had hold of and looked over my circuit you would see ground to the regulator centre pin.
 

sureshot

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I know as a newbie there is little more I can do, except read a research as much as I can. I have just gone over the entire underside of the board with a strong magnifier and used my continuity meter function, at yes it's all ok, as per the schematic. This really is my last post in this thread. Thank you to all you members that throwed out slices for me to chew on (-; as for the crumbs... Thank you again, but it helped little !

Thanks all (-;
 

(*steve*)

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ok, its now working, but be assured there is a fault
 

sureshot

Jul 7, 2012
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Absolutley final post ! i don't know enough to argue lol. its been run for just over a day with a meter connected to the output. And with that 24R R7 resistor it hasn't faultered. I've got a lot to learn.. but my source on this leakage current theory is a good one.

i'm confident it will be fine. After all 25 ohms was good on page two of this thread ! If it goes wrong i will learn from my mistakes. This is half the fun of learning electronics surely !
Thanks again to everyone !
 

Arouse1973

Adam
Dec 18, 2013
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Why do you keep saying final post, the whole point of this forum is to try and help people. I think we do a pretty damm good job bearing in mind we are in different parts of the world. With a strange fault like this it can take quite a while to understand and solve.

You will get an output with no common as I explained above. I can't tell you why it's happening with your circuit when you change the values of the resistor. If I had the unit here I could have a look, but I don't.

Thanks
Adam
 
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