0-30V Stabilized Power Supply

[mendimano]

Well Tompa it is dificult to help you from here, maby you should check your output tranzistor/tranzistors,
i watched the video and your P.S worked then just fine, check for any wires soldering joints wich might broaken during the instalation of your P.S to enclosure.

[redwire]

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.

[denci]

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

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)
 

[audioguru]

.....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.

[denci]

I calculated the heatsink wit aprox. 0.5°C/W thermal resistance for one transistor, how much must be that for three 2n3055.
What fun you suggest?? what dimension and voltage?

[audioguru]

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.

[LEECH666]

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.

[audioguru]

With a 28VAC transformer the unregulated positive supply is +37.6VDC. When the output is shorted then the BD139 has a max current of 50mA and a voltage of about 36V so it dissipates 1.8W. It is in an old package so its ability to cool itself with a heatsink is poor.

[LEECH666]

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

[audioguru]

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.

[LEECH666]

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

[denci]

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??

[audioguru]

If three 2N3055 output transistors share the 5A of current then each has 1.7A and its minimum current gain is 40. Then the current in the BD139 is a max of 125mA. Its max voltage is 36.2V so its max power dissipation is 125mA x 36.2V= 4.5W.

The max allowed chip temperature of a BD139 is 150 degrees C but 130 degrees is safer. Your ambient might be 30 degrees. The thermal resistance from chip to case is 10 degrees/W and for thermal grease is about 0.3 degrees/W. Then with a perfect heatsink the case will be at 30 degrees C and the chip will be (4.5W x 10.3) + 30= 76.4 degrees C.

Your heatsink must have a thermal resistance of (100 degrees - 76.4 degrees)/4.5W= 5.2 degrees C/W or less.   

[denci]

I usually use this way to calculate the heatsink, it is wrong that way:
(130°C-30°C)/4.5W - (10C/W+0.3C/W) = 11,9°C/W