Diy bench power supply trouble

[andrejsstrumfs]

Hi all!

I'm trying to build the power supply based on this circuit.

It kind of works, but the problem is when the power supply switches between voltage limit mode and current limit mode, the opamp output can swing to negative rail and the power transistors get reverse biased and burn out. I actually do not see any protection against reverse biasing the transistors in this circuit. Hovewer my understanding could be quite limited Anyone got any ideas how to improve this?

[Yansi]

The most negative they can go against the emitter is -5V. Realistically speaking, about -4V at best. You are not going to damage a NPN with 4V reverse on  BE junction. I think the issue you are describing is non existent.

There are other issues instead, that are not caught in the design: Like when you force external voltage on the output terminals, especially with the mains powered off.

[andrejsstrumfs]

I've got a bunch of burnt out transistors already The voltage regulation works and the current regulation works but as soon as the modes change - the transistors burns out. i just can't image what else could be wrong. About the external voltage on output - TIP142 have got internal diode between emitter and collector. Shouldn't that protect the transistors?

[Kleinstein]

For this circuit there is a chance the it can start to oscillate. When the circuit was designed and build first they likely also observed this and added those 4.7 pF and 100 pF caps - but just at poor positions (C21 is at a good position but to small). So they can help with external RF coupling in but for oscillation of the regulator it makes things even worse.

The current regulator can take quite some time to activate as the OP has to come from it's positive side down to about -1 V. So a short might already damage the output transistors from too much current before the regulators sets in. With oscillations or when turning on the circuit the same might happen. To reduce the turn on problem it might help to change R3: instead of going toward GND an additional 5.6 V Zener in series and than connect to the -5 V would delay the turn on until the -5 V supply has at least about -3 V.

Normally one should have a second fast current limit, e.g. with an extra NPN transistor driven directly from the shunts to limit the current to a fixed about 2-3 A per transistor.

[Yansi]

Or probably be VERY aware of the origin of those TIP142.  I would be suspicious if they burn out just that quickly. It takes some effort to knock out a decent power transistor, let alone by the microsecond or two when the CV/CC changeover happens.

H.

[Jwillis]

You have no base resistors on your TIP142 to prevent Load hogging.Your transformer voltage is 30 volts but after the filter capacitor it will be around 40 Volts.You need to calculate The current output of the Parallel TIP 142 at that voltage .Based on the data sheet the maximum current at the emitters at 40 volts will be around 2.5 amps each. And that's pushing them Hard .  You have ballast at your emitters resistors to prevent thermal runaway .But they don't prevent load hogging .You stop load hogging by putting a resistor of equal value at the base of each TIP142 .The current across the collector and emitter is determined by the Beta(Hfe) multiplied by the Base current.This means That is if the BC558 is delivering its maximum current of 100mA then the current out of the TIP 142 will be 0.1A  X  its Hfe of 500 or 50 amps This far exceeds the TIP 142 Safe Operating Area.Since the Ic (collector current)is around the same as the Ie(emitter current) you only need the formula below.

Ic=Hfe X Ib   Ic = 500 X 100mA   Ic = 50A   To much!

So lets put some reasonable over head on the TIP 142s and try to keep them at an 1A to start.You can go as high as 2A if you like but not suggested.To give  1 amp at the emitter The base only needs 0.002 amps.Based on the above formula.

Now lets calculate the resistor required to limit the base to 0.002 amps. The Vbe is generally the same for most BJTs of around 0.7 volts but can be higher for darlingtons.For the TIP142 the Vbe is 3 volts
Base resistor is equal to the maximum volts of 40 minus the Vbe of 3 volts (darlington) divided by the base current or

Rb = (Vin-Vbe)/Ib       Rb =  (40-3)/0.002)     Rb=18500 Ohms at each TIP142
 
These resistors will keep your power transistors stable.
This is of the assumption that T1 is is indeed providing a maximum of 100mA but not less than 2mA and you are getting 40volts at the collectors.
You can adjust the base resistor according to the values you have or need but keeping within the safe operating are of the TIP142's

Always calculate a base resistor for BJTs even if you don't think one is required or even if a simulator tells you you don't need one.
NO transistor of the same model is exactly alike. Data sheets are based on a few samples and averaged and simulators will only tell you so much based on those datasheets.
Trust in the math based on your measurements.

[EntropyWizard]

Pictures and a description of your heat sink would also be helpful to rule out problems there.

[Yansi]

You have no base resistors on your TIP142 to prevent Load hogging.Your transformer voltage is 30 volts but after the filter capacitor it will be around 40 Volts.You need to calculate The current output of the Parallel TIP 142 at that voltage .Based on the data sheet the maximum current at the emitters at 40 volts will be around 2.5 amps each. And that's pushing them Hard .  You have ballast at your emitters resistors to prevent thermal runaway .But they don't prevent load hogging .You stop load hogging by putting a resistor of equal value at the base of each TIP142 .The current across the collector and emitter is determined by the Beta(Hfe) multiplied by the Base current.This means That is if the BC558 is delivering its maximum current of 100mA then the current out of the TIP 142 will be 0.1A  X  its Hfe of 500 or 50 amps This far exceeds the TIP 142 Safe Operating Area.Since the Ic (collector current)is around the same as the Ie(emitter current) you only need the formula below.

Ic=Hfe X Ib   Ic = 500 X 100mA   Ic = 50A   To much!

So lets put some reasonable over head on the TIP 142s and try to keep them at an 1A to start.You can go as high as 2A if you like but not suggested.To give  1 amp at the emitter The base only needs 0.002 amps.Based on the above formula.

Now lets calculate the resistor required to limit the base to 0.002 amps. The Vbe is generally the same for most BJTs of around 0.7 volts but can be higher for darlingtons.For the TIP142 the Vbe is 3 volts
Base resistor is equal to the maximum volts of 40 minus the Vbe of 3 volts (darlington) divided by the base current or

Rb = (Vin-Vbe)/Ib       Rb =  (40-3)/0.002)     Rb=18500 Ohms at each TIP142
 
These resistors will keep your power transistors stable.
This is of the assumption that T1 is is indeed providing a maximum of 100mA but not less than 2mA and you are getting 40volts at the collectors.
You can adjust the base resistor according to the values you have or need but keeping within the safe operating are of the TIP142's

Your circuit analysis is nice, however completely out of reality.

The BC558 is about 8 mA current source. It will never supply more than that.

Your hFE calculation is also good, but has nothing to do with the current limiting. In fact, it is good you have excess of current gain. Trying to design for a maximum current limit based on under-sizing the base drive is very wrong approach. At minimum due to the very very large spread of hFE of each transistor. I do agree that some base resistors should be used when the transistors are not paired for Vbe, but their value two orders lower than you're suggesting, see below.

Also, you do not seem to understand much how the circuit works, as the base drive circuitry (in fact the BC558) can only supply up to some few volts (compliance range of only about 2V*) with regards to the emitter of the TIP142. Putting a 18k resistor in the base will render the circuit completely non-functional, as there would not be a way to supply enough current from the few V limited base drive supply.

Please look closely again at the circuit diagram: The TIP142's collector node is connected nowhere else in the circuit. The base is driven from the aux supply of +-5V, referenced to the positive output terminal. This is a standard widely used topology well established and proven to work in many PSUs.

Current limiting should never be done by limiting the base current. It is a wrong approach.  In this case, there is a current regulating loop, that if correctly set, will protect the transistors reliably. In fact, will make the output of the PSU behave as constant current source - which is the wanted property here, obviously.

You might add additional high speed current protection, utilizing another NPN to sense the voltage across the current shunt and to shunt the base current off the TIP142s with it. Personally I consider this unnecessary fort supply such as this, however it is only a few parts added to the BOM and would not discourage anyone against doing so.

Much more realistic values of power dissipation is like 30W per TO247 device maximum for practical reasons.  So at 40V Uce, it is
like below 1A. A pair of TIP142s will be barely enough for a 2A 30V output power supply. It will likely not survive a long term output short -the built-in temperature protection is indeed good idea.

That is why most linear PSUs have some transformer tap switching implemented, to lower the power dissipation of the power transistors. Having a relay to switch here between 15 or 30V secondary voltage would be a nice feature - the remaining OPamp could be used to control the relay action.

Please also note, that a 30V transformer secondary voltage is not enough for a reliable 30V DC output. 32-36V range would be more like it. You need to account for mains voltage variation (up to +-10%) and account for the voltage ripple on the filter cap. Making the cap deliberately large won't help much either, will only add excessive load on the rectifiers and transformer (lower power factor, higher VA and effective current pulled from the winding).

Always calculate a base resistor for BJTs even if you don't think one is required or even if a simulator tells you you don't need one.
NO transistor of the same model is exactly alike. Data sheets are based on a few samples and averaged and simulators will only tell you so much based on those datasheets.
Trust in the math based on your measurements.

That is a wrong assumption. Datasheets are completely fine when followed correctly (understood correctly).  If a base resistor is not required, do not put one. If you are not sure, use math to be sure. Adding random resistors to circuits "just because" is not the recommended practice.

*I would modify the bias node of the BC558 to increase its compliance range. 2V seems not enough to me, if the Vbe of the darlington plus the emitter shunt voltage is considered. (For example R1=68R, R2=1k and R3=3k3)

//EDIT: Fixed many typos.

[andrejsstrumfs]

Thanks!

Lots of information here. Those transistors are cheap ebay ones, so of course that can be the cause. Considering the heat sink - while testing i used not very big passive radiator, but it never got really warm.

[blackdog]

Hi andrejsstrumfs,

The first thing you have to do is place a 150 Ohm resistor direct on both basis of the TIP142!
And do not use spaghetti wiring Technic. 

One simpel 1N4148 kan help with the reverse voltage of the TIP142 Powertransistors.
The anode connected to the +output of the power supply en the cathode to the connection of D12 and D17.

There are a lot of stupid capacitors in this design, all for so called RFI protection, but almost all make the U and the I controle loop more onstable. 
I build this design a view years ago and first removed these capacitors to make the power supply stable.
C17, C18, C19, C20, C21 and C26, C27 "out the will go my young padawan" (Yoda) 

Do not use cheap components for a power supply, who are you kidding?
Blowing up expensive devices connected to this power wil learn you to do it better. 
Think about charging your mobile phone or charging a care battery and your power supply breaks down because you have used cheap, bad components...

Use a big heatsink and mount the TIP142 transistors thera proper way.


I hope this helps.

Kind regards,
Bram

[Jwillis]

As I explain in my original post that the calculations I made were base on the Maximum output of the BC558 and that the OP needs to take the appropriate measurements his/her self.I was simply addressing a possible problem and a simple solution to fix it based on the schematic presented.By presenting the formulas and how to plug in the variables based on measurements may not have been apparent .So its really a good idea to show the way the calculations work. You can't be afraid of the math or induce fear of the math because it's really not that hard or mysterious.
As for the use of a base resistor as a means to control the output of a BJT ,that is a common  method used because there is less heat produced there, because the amperage is lower, then if you attempted a control at the collector where the amperage is higher.If there is a solution that doesn't require a massive change to the design then by all means use it.I was presenting  the simplest possible method to use on the current design. Changing transformers for multiple outputs would be a solution but a bit extreme without massive changes to the design. But if the OP is considering it then that's fine.
As for  datasheets, I never implied that they were not an integral part to designing anything.I simply stated the component may not have the Exact characteristics stated in the datasheet an that adjustments may be needed to compensate for those variables.And that proper measurements and correct calculations are key to a successful out come.

[xavier60]

Hi all!

I'm trying to build the power supply based on this circuit.

It kind of works, but the problem is when the power supply switches between voltage limit mode and current limit mode, the opamp output can swing to negative rail and the power transistors get reverse biased and burn out. I actually do not see any protection against reverse biasing the transistors in this circuit. Hovewer my understanding could be quite limited Anyone got any ideas how to improve this?

"when the power supply switches between voltage limit mode and current limit mode,", could mean that a short circuit has been applied  across the output? This is stressful for a power supply that doesn't have fast current limiting.