Home made Power supply giving me fits

Discussion in 'The Projects Forum' started by El3ctroded, Mar 26, 2008.

  1. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    I built this power supply. The input is from a 24V 12.5A switching power supply. I then send that to the circuit attached. It works quite well, except when I draw high currents with lower voltages one or more 2n3055s blow real quick, or the 2n2222 driving them does. They are mounted to an aluminum heatsink that is about 5" long, 1.5" tall and with 1" long fins all across.

    I don't know what kind of thermal ratings the heatsink has, it came out of my junk pile. I have noticed the mounting screws get too hot to hold, so I'm pretty sure it's a thermal issue on the 3055s. Not sure why the 2n2222 blows.

    Opinions?

    Here's also images of the PSU built. The image on the right you can see the heatsinks where the 4 TO220 2n3055s are.

    On the front panel you see 2 sets of terminal posts. The upper set is for a LM317 regulator. You see it's heatsink on the left in the last image. Each regulator has 2 pots. One is coarse adjustment, the other is fine.

    The switch above each set of terminal posts is the output on/off switch. It doesn't turn off the regulator, just the + side of the terminal post, and the voltmeter is hooked up so you can still see the voltage with the output turned off.

    The multimeters are cheap $4 meters from Harbor freight, but I've found them to be pretty good; especially for this sort of thing. The meter on the left is for voltage. The one on the right is for amps but is actually reading voltage across a 0.01ohm shunt for the LM723 regulator and a 0.1ohm shunt for the LM317 regulator, thus the instructions for reading amperage on the top of the box.

    The two switches between the multimeters are on-off-on switches. The top one is for the ammeter and the bottom is for the voltmeter. The switches, when down, read volts and amps on the bottom set of terminals, and when up read the top set of terminals. When in the center, the multimeters read nothing.

    I'd really like to go to a switching regulator instead of the linear one because I would like to be capable of about 24 Amps at 12Volts and 30A at 10V. With the linear one, I'm limited to about 12Amps :( and keep frying the 2n3055s and 2n2222, but that's for another discussion, unless we resolve the transistors blowing problem.

    Thanks,
    El3ctroded
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Go to ON Semiconductor's site, and download their datasheet for the 2N3055 & MJ2955.
    http://onsemi.com
    Direct link to datasheet:
    http://www.onsemi.com/pub/Collateral/2N3055-D.PDF
    Look on Page 3, Figure 3. DC Current Gain, 2N3055 (NPN). Let's for the moment just look at what happens if you had only one 2N3055 in the output.

    You'll notice that the gain (hFE) of the 2N3055 varies significantly with temperature and the amount of current being passed, and starts dropping quite dramatically (below 50) around 3 Amperes. By 5 Amps, it's below 30. Your 2N2222 is being "ordered" (biased on) by the Vreg to supply Iout/hFE, so at the point of a single 2N3055 supplying 5A out, your 2N2222 is passing 166.6mA. Since it's Vcc is 24V, that's about 4 Watts! Even if you kept the case of the 2N2222 at 25°C, that's still more than twice it's rating. If the 2N2222 is not heatsinked, you're at about 8x it's power rating. (This is all assuming a low Vout. As Vout increases, the power dissipation across the 2N3055 and the 2N2222 decreases.)

    Now let's look at having four 2N3055's in parallel.
    You have 0.1 Ohm resistors on the emitters to attempt to keep the current balanced, which is good. However, you may not have matched the characteristics of the individual 2N3055's, and have no method of limiting/adjusting the base current.

    One of these transistors is going to conduct more than the other three. As it passes more current, it's going to get warmer, which increases it's gain, which causes it to conduct even MORE of the load. This is known as "thermal runaway".
    Looking back on the datasheet at the DC current gain chart, you'll notice that the gain of the transistor is dependent on not only the current being passed, but the temperature.

    I suggest a few things here:
    1) Exchange your aluminum heat sink for a copper one, or at least one that has a copper base. Copper has roughly twice the thermal conductivity of aluminum.
    2) Use heat sink compound, if you aren't already. You first need a good physical connection between the transistor and the heat sink. The heat sink compound "fills in the holes" and considerably improves the heat transfer.
    3) Devise a method to even out the base current between your four output transistors. Obviously, a single 2N2222 without a current limiter is not going to work very well.
    4) If you manage to get your 2N3055's perfectly balanced where each is carrying 1/4 of the load, at 20A output each will have a gain (hFE) of roughly 28, which means you will need roughly 715mA total through the bases. With a Vcc of 24V, that's roughly 17 Watts. If your output is 10V, your base driver will be dissipating roughly 14v*715mA=~10W.

    With linear supplies, as you have discovered, the lower the output voltage for a given current, the more power the output transistors must dissipate. If you are outputting 10V @ 10A, your output transistors are dissipating 14v x 10A = 140 Watts. That's a lot of power. You need a very large heat sink.
     
  3. beenthere

    Retired Moderator

    Apr 20, 2004
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    There are a couple of other things. There is no resistor in the 3055's base circuit to limit short circuit current. If the 2222 pops, then there can be several amps of current into each 3055 base. Using something more like a 2N3054 for the driver is going to be more successful than a single 2N2222.
     
  4. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    Ok guys, thanks a lot. Wookie that walk-through was just what I needed, and beenthere, good catch. I had noticed that when the 2n2222 pops full voltage and current hits the outputs, so you are correct.

    I didn't design the circuit, just built it as per a schematic I found; I assumed it would be an Ok design, and the designer said it was good for 10A with only 2 2n3055s. Guess he didn't use it extensively...

    I don't have a source for 2n3054... So lets say I use some other transistor, 1 for each 2n3055, and add a pot and hook it up as shown in the new schematic, would that work better? I could then adjust the amount of current the 2n3055s get...

    The only other heatsinks I have on hand are old PIII (slot style) black anodized heatsinks and they aren't big enough either, seeing as how those processors generated only 20-40 watts... And none have copper on them.

    Side thought: The datasheet for the LM723 lists a switching regulator circuit, but it's fixed and I need it variable. Also they used a transistor for the output and I would want to use N-ch FETs. Finally I and need it to be capable of 300W.
     
  5. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    Wookie, I just re-read the part about the 2n2222 dissipating 4W. Are you sure that's right? I know P=V*I, but I think you're using the wrong V...??? The 2n2222 isn't dissipating all 24V in heat. It's only dissipating what voltage it is dropping, which is about 1.4V (2 diode drops). So P=1.4V*0.1667A = 0.23338 Watts.
     
  6. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Hi,

    Assuming that the transistor is saturated, you can use the Vcesat specification from the datasheet. This does vary with current, so you need to know roughly what current you expect. For ~150mA, this is 400mV. So, you'll be dissipating 60mW.

    The problem with your circuit must lie within the lack of current limiting resistors into every base of every transistor.

    Steve
     
  7. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
    63
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    Thanks Steve. So do you agree that the circuit from my post 2-up from yours is a much better way to go? Wait a minute... Is that Pot wired wrong on the driver transistor?
     
  8. Søren

    Senior Member

    Sep 2, 2006
    472
    28
    Hi,

    With a large heat sink, 2N3055's should max. dissipate around 40W each (the 115W or thereabout the datasheets usually mentions takes cooling with liquid nitro to reach).

    An NPN shouldn't be used for high side regulation.

    If you wanna use linear regulation with high currents, you could consider a tapped transformer and switch terminals (automatically or by hand) to keep the voltage drop down to manageable amounts.

    Another way to keep losses down is to use a chopper front-end between the transformer and your (way too small) filter cap bank (controlled by the amount of voltage drop you want).

    For large currents, make a supply with the needed voltage and then another for variable voltage (usually different applications), or go switch mode.

    Personally, I do both - with a 1.5kW+ switcher for automotive electronics, a linear 0..30V/0..7A lab supply and a 19" rack box capable of 0..60V and 0..20A (not counting the purpose made variable batteri simulators and what not) - it is very hard to get too much power :D


    Your supply is rather badly designed (and quite obsolete) - I'll repeat myself: build a single voltage supply for whatever high current you need, build a couple if you need more than one voltage, that's the best way to keep losses down.

    If you feel like wasting power, you could use a tunnel heat sink with a fan each end of the tunnel, to keep your transistors alive, or at least go with a fan on whatever heat sink you are using - it does wonders for the apparent K/W figure for the setup.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    The graphs of current gain on the datasheets of transistors is their typical current gain. you can't buy typical transistors, you get whatever gain that they have. The minimum current gain at 4A for a 2N3055 transistor is only 20 so it needs a base current of 200mA. Four of them need a total base current of 800mA.

    The 2N2222 has a voltage drop across it that is 1V less than the 2n3055 transistors.
    If the input is 24V, the output is 5V, the output current is 16A and the current gain of the 2N3055 transistors is 20 then the 2N2222 dissipates about 14.4W. Its max allowed dissipation is only 0.5W. It should be a power transistor like a TIP31 that has a pretty big heatsink. A little 2N2222 can drive it.
     
  10. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    Why not?

    I didn't want to deal with any transformers. I don't have easy access to any and didn't want this thing to be like moving a cinder-block. Also, I know just enough about transformers to know that I don't know enough to size one or select the core type for the current I need, or how to wind one to do what I want. And buying one? Forget it. All I ever find are low wattage transformers for high dollar amounts. So I used a 24V 12.5A switching regulator for the input.

    I know. I was running out of space and out of caps. What I'm using it for it isn't critical.

    No can do. I need the high current to have a variable voltage to do what I need to do. I need 20A at 7V-15V, but it needs to be variable to above 15V (I can deal w/ decreasing currents at higher voltages). What I would really like is a 0-30V variable 500W switching psu. But to buy, I haven't ever seen one. Even if I found one it would be out of my price range right now.

    Agreed!

    Ok. I knew it was old, didn't know it was bad design. Like I said, I didn't design it, I only tried to make it work (I did redraw the schematic in my software). The problem is: I don't need different voltages so much as I need continuously variable voltages from 7-20V.

    Not particularly :) The reason I did this was that it was quick, simple, and inexpensive to do. I spent a total of about $25 on parts plus the case, (the 24V 12.5A switching psu, multimeters and some other parts I already had), and a weekend. So if you know of something else fairly inexpensive and quick to implement that meets my needs, please do tell :D

    Here's why I need this:
    To simulate conditions in a vehicle. Yes, I know, I still can't simulate alternator dumping or the noisy environment, but I can simulate the voltage levels of a battery charged, drained, car running and car being cranked. So I need to be able to move from 12v to 7v to 14.4v fairly quickly and without interruption of the circuit at high current levels. Can't do that with multiple supplies. Plus I need to know what happens to something I'm working on when it has sustained high voltage levels and sustained low voltage levels. The "something I'm working on" can pull as many as 20A from the vehicle... so far.

    Audioguru: If I understand you correctly, in your scenerio, the 2n2222 and the 2n3055 is dropping 24-5=19V. And the 2n2222 is dropping 1V less than the 2n3055, so that means the 2n2222 drops 9V and the 2n3055 drops 10V. Right? If so, then that all makes sense to me.

    Thanks everyone!
     
  11. Pich

    Active Member

    Mar 11, 2008
    119
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    I don't think you should use the second circuit you show, it makes no sense. Replace the 2N2222 with a TIP120 or 121 and mount it to the heat sink if you have room for one more 2N3055 install it. Check your wiring, the first circuit shows current limiting using the .1 ohm resistor, that means the the regulator will start limitimg current at ~.7 volts, I=E/R = 7a x 4=28 amps overall limit, if the transistors are balanced properly it should be OK. Also the .001 resistor at the output does nothing for you from what I can see. It is not necessary to have such high capacitance at the output one 3300uf should do, use the rest at the input.
    I'm impressed on the use of the meters, the 4 watts is correct
     
  12. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    Wow, I'm getting very differing statements from everyone :confused:

    The 2nd circuit I showed was done with the idea of having a drive transistor for each 2n3055 and the pots on each drive transistor would be trimmer pots to adjust the drive transistor's current so that each 2n3055 could be tweaked to output the same amount of current. Did I do it wrong? Or is it a bad idea?

    The 0.001 ohm resistor is a shunt. It's so I can measure voltage across it to calculate the total output current. I did it this way because the multimeter has a 10amp limit as an ammeter. This method lets me measure way more than 10A.

    Side thought: The shunt was supposed to be a 0.001 ohm but they sent me a 0.01 ohm, so I used it. It makes reading the meter harder than it should have been, 2 digits off instead of 3, but gives me better resolution on low current levels.

    Ok

    :eek: I just figured out why that is. I was thinking voltage=sum of voltage around the loop, but I was doing the loop wrong! I feel like such an idiot sometimes :oops:

    How do I know if the transistors are balanced? Can I measure that? Edit: Are we talking about the hfe? If so, the multimeter does that, np.

    Do I need any limiting resistors to the bases of all the 2n3055s like someone else mentioned?

    Thanks!!
     
  13. Pich

    Active Member

    Mar 11, 2008
    119
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    The .1 ohm resistors are for balancing the current flow in parallel transistors. it is common practice, you can install a small resistor less than 10 ohm at the base of the 2N3055s but that would cause a greater volteg dorp. If you have no wiring problems your biggest problem as was mentioned here was the 2N2222 and the output drawing too much current after the 2N2222 blew. There are a couple of things you can do to make the voltalge a little more stable, like the voltage sensing point and not bringing the adjust pot to ground. watch out for the 120w of heat dessipation maybe use a small fan. But I think what you have is sound.
     
  14. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    Thanks Pich. I'll swap out the 2n2222 and add a heatsink pending this question:

    Can I just get rid of the 2n2222 and replace the 2n3055s with an n-ch mosfet??? If so, would I need to make any other changes?
     
  15. Pich

    Active Member

    Mar 11, 2008
    119
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    Yes you can, add a 100 ohm resistor in series with each gate.
    Good luck
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    Well, power MOSFETS don't "like" to be in partial conduction. They'll get hot mighty quick, just as the transistors are. All the way on, or all the way off, as quickly as possible.
     
  17. beenthere

    Retired Moderator

    Apr 20, 2004
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    There is also the basic difference between FET's and BJT's. A FET is a voltage controlled device, while a BJT's conduction depends on the current in the base-emitter junction. A scheme to control current through a BJT will not necessarily produce an equivalent result with the substitution of a FET.
     
  18. El3ctroded

    Thread Starter Active Member

    Feb 4, 2008
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    I know the FETs will get hot in linear operation... I think I'll try out the fets next time the 2n3055s blow, as I'm low on the 2n3055s but have plenty of n-ch fets handy, and it will cut out one generator of heat: the driver trans.

    Thanks for everyone's help!
     
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