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0-30 Vdc Stabilized Power Supply

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Ok, my PSU is nearly finished... and what I've tested works fine.
I've used a "Lochrasterplatine" (don't know the English word for the boards with many holes).
Tomorrow I'll use some car lamps to test the maximum current.

I think I will change the R1 resistor to a 4W version. The 1W version has at 3.7V and 0.28A a temperature about 40 C.

I have still one question...
I have a 100uA analog amperemeter. I want to connect it parallel to the circuit. Which resistor I have to use in series to I-meter to get max. deflection at 3A overall current?
I=3A
I(M)=100uA=0.0001A
I(L)=2.9999A
U(M)=U(L)=U
R(M)=?

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Capricious,
R1 is across the unregulated supply voltage. Maybe you mean R2 or R3 gets hot with 3.7V across it.

A current meter is connected in series with the load to measure the current that is flowing in the load (and flowing through the current meter). You simply connect a very low value resistor in series with the load and use your meter to measure its voltage drop.
Instead of adding another resistor to this project, use your meter to measure the voltage across its R7, which is already in series with the load.
With 2.9999A flowing through the 0.47 ohms R7, its voltage drop is 1.41V. For your 100uA meter to have full-scale deflection with 1.41V, its total resistance must be 14,100 ohms. Add a resistor in series with your 100uA meter that is 14,100 minus the meter's resistance. Then connect the meter and its new series resistor across R7.

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Capricious,
R1 is across the unregulated supply voltage. Maybe you mean R2 or R3 gets hot with 3.7V across it.

A current meter is connected in series with the load to measure the current that is flowing in the load (and flowing through the current meter). You simply connect a very low value resistor in series with the load and use your meter to measure its voltage drop.


No, it's R1 and it is the output voltage. But I understand that the problem is not the output power. So the temperature will be constant at about 40 degrees C. I think it is too "warm".

Ok, I know that a current meter is connected in series. I will connect it in series but in parallel to the "shunt". And the "shunt" is a piece silver wire on the board from circuit. What I want to do is to split the current in two parallel flows. One with max. 2.9999A and the other with max. 0.0001A which is together the max. output current 3A. Then I can put in the A-meter in series to the low current part.
Now I only need the right resistor value to split it up like that.
Is this not possible?
I have the problem that I do not really know the resistance of the wire. It is too low for measurement and I think the "spezifische Widerstand" (specific resistance?) for silver isn't accurate enough to calculate the resistance of the wire in practice.
But maybe this way is too extreme so there must be a resistor with known value?

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Capricious,
Are we talking about the same circuit?
C7 is with the output voltage of this project across it, not R1. R1 is 2.2K ohms, and has the rectified and filtered supply voltage across it. If you are using a 24VAC transformer, R1 will have about 35VDC across it when the project has no load, and will dissipate only 0.56W. A 1W resistor will be quite warm, but not too hot. Use a 2W resistor if you are worried about its temperature.

Why make a "silver shunt" when R7 makes a perfect shunt that has a known resistance? The current will be split because R7 will pass up to 2.9999A, and your 100uA meter and its series resistor wil pass the remaining 0.0001A.



post-1706-14279141650806_thumb.gif

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Capricious,
Are we talking about the same circuit?
C7 is with the output voltage of this project across it, not R1.


We are talking about the same circuit ... and you are right but misunderstand me. I've never been talking about voltages across resistors. I've simply measured the output voltage and current of the PSU to know the load.

Thanks for the hint with R7.

My thingy is finished ... some technical data:

It has a max. output power of 73W.
30V => 2.5A
24V => 3.0A
14V => 4.3A => 60W
The max. current is limited to about 4.3A.
The max. output voltage is about 34V.

I try to find someone with an oszi to check the output quality.

post-2612-14279141651195_thumb.jpg

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Hi all
Im new to this forum

Im attempting to build the 30 V supply

2 questions.

1 -> can i put a LED & R right after the diode rectifier bridge in parallel with the rest of the circuit or will the say 20mA reduction(parallel current flowing away) caused by the LED have an effect on the rest of the circuit?? If so any suggestions.

2 -> Has anyone started on the negative supply that can be connected in junction with this circuit with same specifications and which has a common ground? If so where can i find schematics. If not any suggestions on how this can be accomplished.

Thanks

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Capricious,
That's good that you finished it and its wiring looks excellent.
A few questions:
1) Did you use a 24VAC, 3A transformer?
2) What is the total supply voltage of U2 and U3 without a load?
3) How hot does Q2 get with 5V or less output at 3A?
4) How hot do the rectifier diodes get at full current output?
Thanks.

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Kobus,
Welcome to our forum.

1) You can put an LED "power-on" indicator in series with a 1.5K, 1.5W resistor in parallel with R1/C1. Since it draws 20mA, the output current of the project may be restricted to only 2.98A, instead of 3.00A. The difference will not be noticeable.

2) You could build 2 of this project, each one using its own transformer, and connect them in series for 60V output. If you use their junction point as ground, then its output becomes + and -.

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Kobus,
In regards to the LED, you could also put it on the primary side of the transformer. Hotwaterwizard has posted a diagram here on the site showing how to do so. Let me know if you cannot find it and I will search.

MP

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Capricious,
That's good that you finished it and its wiring looks excellent.
A few questions:
1) Did you use a 24VAC, 3A transformer?


I use the 24V secondary of a 75VA trafo.


2) What is the total supply voltage of U2 and U3 without a load?


Both 41.5V.
Specified max. of TL081 is 36V.
Is there a simple pull out and put in replacement in DIL8 with higher max. ratings?


3) How hot does Q2 get with 5V or less output at 3A?


1.75V 3A => 130 deg. C
4V 4.3A => I switched it off at 175 deg. C
Should buy an heat sink ...


4) How hot do the rectifier diodes get at full current output?


14V 4.3A => 120 deg. C

The 2N3055 gets really hot, too.
But it is below the spec. max. of 200 deg. C.
Nevertheless I should not touch the heat sink at the back of my case. 120 degrees at heat sink are hot enough.

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Why the 2n3055 produces so much heat in this PSU?
I've another 13.8V 10A PSU with two 2n3055 and a similar heat sink. I remember that I can pull out 8A without burning my fingers at the heat sink.

How hot is the junction when the case of the part has about 130 degrees?

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Capricious,

The 13.8V psu have less voltage difference between the rectifier and the output than this variable one. To calculate the power burning off to heat (rectifier Voltage

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Capricious,

Well, Ohm:s law apply on every component in a circuit. The voltage drop over the 2N3055 * current (Ampere) is the power loss. If you like you could say that you burn off the difference. The lower the output voltage is the higher the loss (at the same current). If you have say 34Volts to the 2N3055 and 5Volts output at 2.5A the heat produced is 72.5W (That

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Capricious,
1) A pull out and put in replacement for the TL081 (change all 3) is either the MC34071AP or the TLE2141CP. Both are rated for 44V max.
2) Your project has heat problems:
a) The 2N3055 can dissipate 115W if you can somehow keep its case at no more than 25 degrees C. Its junction temp will be 200 degrees C. But it cannot conduct all its heat to your heatsink, because the surfaces are not perfectly flat (heatsink compound grease helps) and you have an insulator between them. The heatsink itself gets hot because it does not conduct nor radiate heat perfectly. A graph on the 2N3055's data sheet shows how much power must be derated from 115W in order to keep the junction temp at 200 degrees C, with increasing case temp.
Since your heatsink is 120 degrees C, the 2N3055's case temp must be much more. Assuming that the case temp is 140 degrees C, the transistor should be derated to dissipate only about 35W. If you set the output voltage to 4.0V and have a 4.3A load, then the 2N3055 has about 24.6V across it and 4.3A through it, which is a dissipation of 106W. Since it is dissipating 71W too much, its junction temp is probably about 308 degrees C. So you need a larger heatsink and/or forced air cooling with a fan.
b) The 2N2219 can dissipate 3W if you can somehow keep its case at no more than 25 degrees C. A heatsink won't help much because of its shape. If you set the output voltage to 4.0V and have a 4.3A load, then the 2N2219 has about 23.6V across it. If the 2N3055 has a gain of only 20, then the 2N2219 conducts 215mA. Therefore the power dissipation of the 2N2219 is 5.1W. Its junction temp is probably 323 degrees C.
I recommend replacing the 2N2219 with a TIP31A that is bolted to your project's metal case with an insulator and heatsink compound grease. It will not get hot. Also change R15 to 100 ohms for a reduction of its voltage loss.
c) The rectifier diodes are also much too hot. Replace them with a 6 or 10A bridge rectifier module that is bolted to your project's metal case.

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Capricious,

2) Your project has heat problems:
a) [...] So you need a larger heatsink and/or forced air cooling with a fan.


I think it is not possible to cool this thingy with passive heat sinks.
Please answer my question
http://www.electronics-lab.com/forum/index.php?board=15;action=display;threadid=1022;start=0;boardseen=1
about heat sink specification, too ...

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Capricious,
The author of this project used a pretty big heatsink. He even used a clip thingy on Q2. He didn't show his huge fan. Is his fan used for cooling, or for blowing away the smoke? (Or for blowing away your hair that you tear-out when the transistors melt)
See my reply in your other post for a discussion about heatsinks. Maybe you need to add an additional output transistor or two.

Many transformers have a center-tapped secondary. Have a switch that selects its full voltage, or half voltage at the tap. An extra pole on the switch will change the range of the output voltage adjustment control. The negative supply will need revision or elimination.
Then the heating will be halfed (but not for the rectifiers).

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Hi everybody
I am currently building the supply. I my opinion 3A on a single 2N3055 is to much. It may have to dissapate more than 100W.
Anyway the trafo I'm using is 1A and I believe that is about right for the ciruit..

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Hi Gosi,
Welcome to our forum.
I agree with you that a single 2N3055 has a serious thermal problem in this project.
Your 24VAC, 1A transformer is rated for a maximum power output of only 24VA. Since it must supply a peak voltage of 34V into the rectifiers at a high momentary current, then it can supply only 707mA DC current maximum, at its rated power. That's why I recommend using a transformer that can supply the DC current times 1.414 (peak to RMS ratio).
Also, your low current transformer probably has poor voltage regulation, so its voltage with no load may be high enough to damage the circuitry. I recommend using opamps rated at 44V for a 24VAC high current transformer. Without a load, the positive rectified output voltage of your low current transformer plus the 5.6V negative supply voltage, is probably much more than 44V.

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Hi audioguru and thanks for your comments.
Actually I just found out that the trafo is only 20va, it is a wall-wart type from a decoration. I'm still in an early stage of the construction, having made the pcb and mounted most of the R's. I had not given thougt to potential problem of exceeding the V specs on the opamps. Thanks for mentioning this.
The purpose of this project is a small secondary bench psu, so even 0.5A is sufficient. (I have another 723 based capable of 2A)

I am also thinking about using this circuit(modified) in a 7A psu, I have a nice homemade enclosure ready with almost everything except the regulator, it even has provision for 6 TO3's
..but first I want to see how this version works..

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Capricious,
The author of this project used a pretty big heatsink. He even used a clip thingy on Q2. He didn't show his huge fan. Is his fan used for cooling, or for blowing away the smoke? (Or for blowing away your hair that you tear-out when the transistors melt)


Hehe :P
If the author's heatsink is on this picture:
I think my heatsink is bigger ;)


See my reply in your other post for a discussion about heatsinks. Maybe you need to add an additional output transistor or two.


Ok. I'll see. Maybe I'll use a fan. My trafo has multi secondary. So I could use the 12V secondary to power the fan. Is it for a fan enough to put in 2 diodes?

What is a real value for my heatsink with 2.4 K/W and the additional thermal resistances? Should I take 3 or maybe 4 K/W for calculations instead?


An extra pole on the switch will change the range of the output voltage adjustment control. The negative supply will need revision or elimination.


Do I understand right that I could use the circuit with 12 VAC for lower output voltages without any changes?

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