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0-30V Stabilized Power Supply


redwire

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  • 2 weeks later...

Can I add two potentiometers to the "fine tuning"? From what I know, just add the potentiometer in series with the potentiometer in the circuit that already exists, but do not know if this circuit is possible to contain the RV2 and RV3. If possible, which would then potentiometers?

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  • 2 weeks later...

Hi,
Couple of questions regarding the negative supply:
Why is it required
Why only used on U3
Isn't there a lot of ripple on it?

U1 has an output of +11.2V and its inputs are at +5.6V so it does not need a negative supply.
U2 has an output that goes as low as +1.0V and has inputs that work at 0V so it does not need a negative supply.
U3 is the current regulator. It works by reducing the output voltage so that the output current is what is set on the current-setting pot. In order to reduce the output current when the output is shorted then the output voltage of U3 must go to -0.7V because of the voltage drop across D9.
Then the output voltage can be reduced to 0V.

So U3 has a -1.3V negative supply.
I don't think the negative supply has much ripple because its load current is only a couple of mA. Besides, U3 is not affected by a little ripple.
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  • 4 weeks later...

I finished my version of this famous project :)
Here how it looks like: http://diyfan.blogspot.com/2012/02/adjustable-lab-power-supply.html
You may also download PDF files of the project on that page.


Hi,
i will use a power supply 24Vdc 3A instead of a Transformer. Can i eliminate the rectifier diodes, R2 e C2 ?
I didn't find MC34071, only MC34072. Can i change ??
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Hi,

I didn't find MC34071, only MC34072. Can i change ??


The MC24072 is a dual opamp configuration, it should work but its not a direct replacement part, as the pinout is different, but if you can work around this, then I'm sure you could.

PS, you will loose the offset too. but not a train smash though.

Thats just my opinion, perhaps wait for some  pro's to answer.
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Thenr,
This project uses a 28VAC or 30VAC transformer. The rectified and filtered unregulated DC is +37.6V to +40.4V when the load is 3A, not just 24VDC. The circuit might need changes to operate from only +24V and the max output voltage will be about +18V or +19V.

Go to www.farnell.com and click on the flag of your country. They might have a warehouse full of MC34071 or TLE2141 opamps in your country.

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Hello

I build modified version of this PSU... and i got 1 problem.... PSU worked fine untill today (when I put it inside the case/box)... something went wrong.. I cant regulate output voltage.. I got constant 36V.....

U2 is getting very hot after few minutes....


This is PSU when it worked... now is broken :/



sry for my bad english...
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Tompa,  How many and what type (T0-3 or T220 package)  of output transistors are you using.  The heat sink looks a bit small.    Without being able to see the amps clearly,  I am assuming you are pushing 4 amps  and the output transistors are  having to burn off over 80W.

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  • 2 weeks later...

So, this is my first post on this forum, first hi everybody!! ;D

I will make this project in near future therfore i am interested how to calculate total dissipation on transistor 2n3055.
I will use the 5A (REV3) version of this project, so i'll use three transistor and transformer 30VAC 210VA.
I need this calculation for calculate heatsink thermal resistance for three paralleled 2n3055.
Is this calculation is correct or not:
Pd = 5A * (40V - 0,55V - 1.35V) = 190,5W

i got 0,55 from total voltage drop of three 0,33ohm paralled resistor (0,11*5), 1.35 is voltage drop on shunt resistence R7 (0,27ohm*5A)

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.....calculate total dissipation on transistor 2n3055.
I will use the 5A (REV3) version of this project, so i'll use three transistor and transformer 30VAC 210VA.
I need this calculation for calculate heatsink thermal resistance for three paralleled 2n3055.
Is this calculation is correct or not:
Pd = 5A * (40V - 0,55V - 1.35V) = 190,5W

i got 0,55 from total voltage drop of three 0,33ohm paralled resistor (0,11*5), 1.35 is voltage drop on shunt resistence R7 (0,27ohm*5A)

Your calculations are correct. When the output is set to 5a and is shorted then each 2N3055 output transistor must dissipate (190.5W)/3= 63.5W.
A huge heatsink cannot provide enough cooling so a high velocity fan must be added.
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If you don't use insulators for the transistors when you can insulate the heatsnk from the chassis and prevent somebody from shorting it to ground or getting a 40V shock.

Thermal compound adds about 0.2 degrees C per Watt so the cooler is 0.7 degrees C per Watt.
Maybe the ambient temperature is 30 degrees C. Then 190.5W causes the chip of each transistor to become (190.5 x 0.7) + 30= 163.4 degrees C which is extremely hot but not too hot (200 degrees C is the  maximum for a 2N3055).

I buy heatsinks where the manufacturer lists dimentions and thermal resistance. If you make your own then you are just guessing.

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Hello once again. I still haven't managed to build this supply even though I designed my own PCB, which probably is still crap. Who cares.

The reason I post here today is because I finally want to finish the PSU and I am not sure what type of heat sink I should pick for Q2 (BD139). It only says "on a pretty big heat sink" in the parts list, which is a kind of vague assertion.

Some toughts on this topic (not necessarily correct or complete):
Q2 is used to drive the two power transistors Q4 and Q5. At the nominal 3A output current (IOUT=3A) the current gets shared between the two transistors and each transistor has to handle 1.5A of current.

I took a look at the OnSemi datasheet for the 2N3055 and the minimum DC current gain (B) seems to be 20 (worst case).

IB = IC / B
IB = 1.5A / 20 = 75 mA

Since we're driving two 2N3055s we have to supply two times the base current that I just calucated.

IBx = 2 x IB = 2 x 75mA = 150 mA

Since I sort of suck at electronics I have no idea how much voltage the transistor Q2 will see (worst case). However I need the voltage to calculate the total power dissipated by Q2, which then would allow me to calculate an appropriate heat sink.

Looking into the output transitor circuit more closely I think it kinda works like this:

Supply Voltage:
US = 30V

Q2 Emitter Resitor Voltage:
UR16 = UR24 + UBE_Q5 = (IOUT/2 * R24) + UBE_Q5 = (1.5A * 0.33R) + 0.7V = 0.495V + 0.7V = 1.195V

Q4 and Q5 Collector Emitter Voltage:
UCE_Q4 = UCE_Q5 = US - UR24 = 30V - 0.495V = 29.505V

Q2 Collector Emitter Voltage:
UCE_Q2 = US - UR16 = 30V - 1.195V = 28.805V

Power dissipated in Q2:
P_Q2 = UCE_Q2 * IBx = 28.805V * 150mA = 4.32W

Thermal resistance of (heat) sink to ambient:
RthSA = to be calculated

Thermal resistance of case to (heat) sink (thermal grease layer):
RthCS = 0.3K/W (don't know where this number comes from, http://sound.westhost.com/heatsinks.htm#s7 suggest 0.25 for beryllium oxide which is kind of grease like (?))

Thermal resistance of transistor junction to case:
RthJC = 10K/W (from ST datasheet)

Maximum ambient temperature:
Ta = 55°C (selected, ambient temp in summer in north germany hardly ever gets over 35°C, 20°C safety margin included)

Maximum operating junction temperature of transistor:
Tj = 150°C (from ST Datasheet)

Tj = P_Q2 (RthJC + RthCS + RthSA) + Ta
(equation from http://irf.custhelp.com/app/answers/detail/a_id/91/~/heat-sink-selection-and-thermal-calculation.)

Solve for RthSA:
(Tj-Ta/P_Q2) - (RthJC + RthCS) = RthSA

RthSA = ([150°C - 55°C] / 4.32W) - (10K/W + 0.3K/W) = 11.69K/W

RthHS = RthSA * 0.80 = 11.69K/W * 0.80 = 9K/W (another 20% safety margin)

So the heat sink should have a thermal resistance of around 9K/W or lower.
Hope my calculations are ok, not entirely sure. I probably forgot to take the rectification factor into account and US is probably higher than 30V.

What type of heat sink did you guys use in your build of this PSU?

Florian

//Edit: Few typos corrected.

post-46370-1427914445012_thumb.jpg

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Ah thanks. I'll have to adjust my calculations then.

I've got a 160 VA transformer with 1x 230V primary and 2x 15V (5,33A) secondary windings.
Im going to put the secondary windings in series so the resulting RMS voltage should be 30V.

My rectifier bridge drops 1.1V for each diode inside of it. So I guess my total unregulated supply voltage should be:

Us = 30V * sqrt(2) - 2 * 1.1V = 42.43V - 2.2V = 40.23V

If the TO-126 package of the BD139 is so bad couldn't I just use some other general purpose NPN transistor in a TO-220 package? Or maybe drive some Darlingtons directly from the opamp instead of the 2N3055?

Florian

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The very old 2N2219 driver transistor in the original circuit had a good high frequency response so its transient response was good. I suggested replacing it with a TIP31 transistor that can be cooled well but another member showed that its slow response almost causes oscillations and rercommended the very fast BD139.
A darlington will be much too slow.

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I noticed that only the original Philips datasheet of the BD139 specifies a value for fT = 190 Mhz. So far I haven't found a transistor in the TO-220 package that's that fast.  I think the fastest one I've found so far is the 2SD1762 @90Mhz or so, but the dc current gain doesn't seem to be as high. MJF44H11 seems to have a similar minimal dc current gain and fT=50Mhz.

*sigh*

Not as easy as I thought. ;)

Guess I'll just stick to the BD139 for now and see how that transtor works out on a heat sink.

Thanks again for your help audioguru.

Florian

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So, what heatsink do you suggest, how much should be a thermal resistance, i have one heatsink with 5°C/W at home, which I intend to fix on housing of BD139.
I think that it is enought for condition in this stabilized power supply with max 5A current or not??

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