Some huge problems paralleling MOSFETs (Linear Power Supply)

[Mathias (Guest)]

Hello to all here,

Since I restore and repair vintage transistor radios, radio recorders 
and vintage hifi items myself, the need for a laboratory power supply 
came up. Usual switch mode power supplies from China were unsuitable for 
me, since these as you surely know disturb the SW, AM and LW reception 
e.g. with a transistor radio massively. Therefore it had to be something 
"linear". So I had bought a up to 95% finished „old“ 2 channel power 
supply from an older guy who owned an electronics company here in 
Germany. He told me, it was built by one of his employees (according to 
him an electronics engineer) in the early 90s. It was built according to 
a circuit diagram from the University of Frankfurt and the engineer 
allegedly even built several devices of it. So he also handled me the 
ciruit diagrams attached and additional 4 other, new  IRF 540N he still 
did have. Much more he could not tell about the device.

The basic regulation of both boards (i.e. of both channels) works 
perfectly and over the whole range. Also the current limitation is ok 
and reacts well. But there are problems with the MOSFET. I have dealt 
with them intensively and also researched. And so I also found some 
videos on YouTube about paralleling MOSFETs. However, I am not an 
electronic technician, but my main focus is rather the restoration. Even 
if I made many self-construction devices in my young 20's, including PCB 
creation "home brewed".

The parallel connection of the MOSFETs seems to be not without problems: 
What is described in the internet in various (technical) places happens 
also here with this power supply: The currents are distributed very 
differently over the 4 MOSFETs. These are IRF 540N. During the (not very 
long "endurance test" with 24V and just 1.2 Ampere already as my first 
test one of these MOSFETs already broke down, the reason can be found 
further down in the text ...I think. I have exchanged (repaired) this 
one. Now it is so that e.g. with 2 running MOSFETs (connected as a test) 
and 9V / 300mA (bulb) at first some 150/150 mA measured through both 
MOSFETs. One of them quickly becomes much hotter and the current is 
quickly distributed much more unequally (200/100 mA) between the two as 
the temperature rises. With 2 bulbs connected, making i.e. 600 mA load, 
this is even so that 400/200 mA is distributed between both 2/3 : 1/3 
and the MOSFET with 200 mA even goes below 100 mA share of total load as 
the temperature rises. So almost everything over the one MOSFET then 
only runs. With all 4 MOSFETs connected, only 1 MOSFET gets hot at 600 
mA, because all load runs completely over only this one MOSFET. The 2A 
reading per channel can never be enough. This (from my opinion) also was 
the reason the one MOSFET was destroyed with the 1,2A – all current did 
flow over it

I made some tests: A resistor of 1K Ohm at the gate (like in the 
schematic, "original" was not available) did not change anything. Also 
connecting 2 or 4 trimmers (variable resistor, 4.7K set to 1K) in front 
of the gate of each of the 2 or 4 MOSFETs with various "try adjustment" 
did not change the current distribution.

Now you may have to select the MOSFETs a little bit, I had read this 
where. Or I try the BUZ11A as MOSFETs instead of these IRF540N (these 
are noted in the "hand-drawn" plan, but they are even more low power). 
There are also minimal differences between the two schematics. The PCB 
is the same as the hand drawn one, but with the IRF540N but without the 
1K gate resistors which are marked on the "professional" schematic.

So my question would be primarily, if there is a fundamental design 
error in the circuit? Which would mean, that it would be useless to try 
to get the unequal load distribution under control. Or is it only badly 
selected components (MOSFET) the problem. Or it makes sense to try to 
use BUZ 11A MOSFET or to change or add something to the circuit. Before 
I invest more time and money in the device, I wanted to make sure that 
there is no basic circuit design error. Because otherwise I rebuild the 
device with the good parts further used (housing, large toroidal 
transformer etc.) with another circuit without MOSFET.

May bee you have some idea for what I can do more. I'm glad for any 
tips.

Greeting Mathias

[Magnoval (Guest)]

Ad a small Value resistor in each source path before connecting them 
together..

[oszi40 (Guest)]

Magnoval wrote:
> Value resistor in each source path before connecting them
> together..

The temperature must goooood distributed of all MOSFETs T2...T5!

[MaWin (Guest)]

Mathias wrote:
> there is a fundamental design error in the circuit?

Yes.
It's a schematic of a bipolar power supply that simply replaced the NPN 
by MOSFETs.
This will never work.
Neither the direct paralleling of MOSFETs without current distribution 
resistors with about a voltage drop of UGSth at the full current a 
MOSFET will have to hande is ok.
Nor the Gate-drive that is not symmetrical to charge and discharge the 
gate.
And the whole signal response of the circuit is also not matched to the 
different behaviour of the MOSFETs. Replace them with 2 2N3055 in 
parallel with 0.22 ohms emitter resistor, and the power supply will 
probably work right out of the box.

Or read: http://www.dse-faq.elektronik-kompendium.de/dse-faq.htm#F.9.1 
and the chapter 
http://www.dse-faq.elektronik-kompendium.de/dse-faq.htm#F.22.2

[Mathias (Guest)]

Hello, thank you very much for this information. To be honest, I was 
afraid that something was simply designed together in an unprofessional 
way. I think it would be a very good approach if this could be replaced 
by normal power transistors. Actually I would prefer that. How should I 
connect it concretely? Would there be a diagram? That would help me a 
lot.

[Mathias (Guest)]

Would it be possible to use an alternative for 4 power transistors in 
the TO220 package? Because these holes would already be there. But 
otherwise I can install the 2 x 3055. There are 2 large heat sinks.

[hinz (Guest)]

Mathias wrote:
> Would it be possible to use an alternative for 4 power transistors in
> the TO220 package?

TIP31 od BD241 will do the job.

[matzetronics (Guest)]

Mathias wrote:
> Would there be a diagram? That would help me a
> lot.

There is probably no need for that, as the change is quite simple. The 
BD139 already provides enough current to drive a 2N3055 or two of them 
(you could also use BD130 or 2N3771-3773).
Replace the 1k resistor at the gate with a link. Connect the Collector 
(case) of the 2N3055 to where the Drain was connected (rectifier 
output). Connect the Base to the emitter of the BD139. Connect the 
Emitter of the 2N3055 to where the Source was. If more than one 2N3055 
is used , wire a 0.22Ohm/5W resistor in series of Emitters.

from rectifier                    to PS output
>----+----------\___/--|===|--+
     |         C     E  0.22  |
     +----------\___/--|===|--+----> 
               C     E

Tie bases together and connect them to emitter of BD139

[matzetronics (Guest)]

Another power transistor would be BD249 in the TO3P package.

[Mathias (Guest)]

Hello matzetronics,

I'm going to rebuild it that way now.

Thanks you very much!
Vielen Dank!

Mathias

[MaWin (Guest)]

Mathias wrote:
> Would it be possible to use an alternative for 4 power transistors in
> the TO220 package?

The power supply is 30V/2A and needs to dissipate about 80 Watts.
A single TO3 insulated on a good heat sink may dissipate 80W (not the 
115 a 2N3055 is advertised, that requires water cooling)
A single TO220 insulated on a good heat sink may only dissipate about 
40W.

But why do we recommend two 2N3055 (not a single 2N3772 as they have a 
terrible hFE current amplification) ?

Because the heat sink may be much smaller if 2 are used in parallel 
allowing for cheaper construction. So you will have to use 4 in TO220 to 
get the same benefit. Clamp them on a heat sink, not using the screw on 
the tab as this will bend the metal tab away from the heat sink).

Each of them will carry only 0.5A, and should get about 0.5V on emitter 
resistor, so 1 ohms should be used and may replace the 0.22 resistor in 
the original. BD243B or C should do, or TIP41B or C.

> Because these holes would already be there.

Each of them does require a resistor in the emitter wire.

If you add more transistors in parallel, this will not require more 
drive current and a larger drive transistor, as the output current 
remains the same and hFE will not get worse.

A good construction will calculate the allowable heat sink temperature 
that the transistors can survive at full load and place a thermo switch 
on the heat sink to prevent damage on hot days or bad ventilation 
(overcurrent may not happen as this is limited in a lab supply).

[Mathias (Guest)]

Hello MaWin,

thanks! This is also a very suitable solution with the 4 TIP41C. As it 
does not need to drill additions holes on the heat sink. I can then use 
the existing. To correct understand: Does it then need 4 x 1 ohms 
resistors for each ones emitter, or 4x 0,22 ohms or only replace the 
existing 0,22 ohms against a 1 ohms resistor (how many watts)? The 4 x 
TIP41C are then paralleled like in the sample with the 2N3055 above?

[hinz (Guest)]

Mathias wrote:
> Does it then need 4 x 1 ohms
> resistors for each ones emitter, or 4x 0,22 ohms or only replace the
> existing 0,22 ohms against a 1 ohms resistor (how many watts)?

Strike a happy medium and use 0.47 ohms, 0.5 W is sufficent.


> The 4 x
> TIP41C are then paralleled like in the sample with the 2N3055 above?

Exactly.

[MaWin (Guest)]

Mathias wrote:
> Does it then need 4 x 1 ohms resistors for each ones emitter, or 4x 0,22
> ohms or only replace the existing 0,22 ohms against a 1 ohms resistor
> (how many watts)? The 4 x TIP41C are then paralleled like in the sample
> with the 2N3055 above?

As the supply needs a 0.22 Ohm measuring shunt for the current that 
increases its output impedance, it is not neccessary to increase that by 
additional emitter resistors.
Just use them to replace the 0.22 ohm shunt.
4 parallel resistors to get 0.22 Ohm would need to have 0.88 ohm. As 
they are not available I said 1 ohm (at 0.5A they only need to have 
0.25W so the usual 0.6W metal film will do fine).
Then either connect a single emitter to the opamp below, or sum up all 4 
emitter lines using 1k resistors, to get to the opamp. The latter one 
measures more exactly if the summing resistors have 1% and not 10% 
tolerance. With 4 x 0.82 ohms you are nearer at the specs, but you 
probably have to buy them from Mouser. I doubt that Reichelt will stock 
0.82 ohm 1% metal film.

--+--CBE--+-----------1R--+-- out
  |       |               |
  +--CBE--(--+--------1R--+
  |       |  |            |
  +--CBE--(--(--+----1R--+
  |       |  |  |        |
 +--CBE--(--(--(--+--1R--+
          |  |  |  |
         1k 1k 1k 1k
          |  |  |  |
          +--+--+--+-- opamp

As a 4k7 resistor as  load is fitted to the joined source connectikn, 
and you have 4 emitters now that are not joined, we need 4 18k resistors 
to replace this.


Your circuit is a copy of 
https://www.mikrocontroller.net/attachment/129845/0_35v_3a_labor_netzgeraet_elektor82_773.pdf

Only with non-working MOSFET and a poorer reference.

[Mathias (Guest)]

Hello, MaWin, thanks a lot! Could you please attached the PDF again it 
states is not existing attachment, even in logged in status.

[hinz (Guest)]

Beitrag "Elektor 12/82 Netzteil Layout"

[Mathias (Guest)]

I could get 0,82 Ohms 1W but with 2% instead 1% from Reichelt. Will this 
fit?

[Mathias (Guest)]

... but (?) should one or all 4 emitters get connected direct to the 
opamp, not to the BD139?

[matzetronics (Guest)]

Mathias wrote:
> but (?) should one or all 4 emitters get connected direct to the
> opamp, not to the BD139?

Neither this nor that. Have a look at the Elektro design. There is no 
connection between the emitters of the power transistors to the BD139. 
Don't get confused. The Opamp comes later and is designed to measure the 
current across the shunt resistor.

MaWin posted a shortcut where the equalizing resistors provide the 
current measurement as well as equalizing, but your original desing uses 
a dedicated resistor for this (the 0R22 in the output rail) and theres 
no need to change that. The equalizing resistors can well be seperated 
from this function.

[matzetronics (Guest)]

matzetronics wrote:
> Have a look at the Elektro design

that should write 'Elektor design' of course

[A-Freak (Guest)]

A little additional note:
The IRF540N is a modernized successor where they could get the same 
switching performance out of a smaller die than the original IRF540 
without "N"
In analog mode it breaks at lower power

If you really want to go with MOSFETs i suggest something like the 
IRF250

[matzetronics (Guest)]

A-Freak wrote:
> If you really want to go with MOSFETs i suggest something like the
> IRF250

Note that MOSFets in this kind of circuit are still one of the wrongest 
things one could do. MOSFets have no fixed relation between Source- and 
Gatevoltage, which is essential in a follower configuration used in this 
power supply. Instead the relation is subject to large variations. This 
even  differs from item to item.
MOSFets used as sourcefollowers are in most cases a wrong choice.

A bipolar transistor on the contrary has a fixed voltage of approx. 0.6V 
between Base and Emitter which also makes it possible to connect them in 
parallel as (emitter-)followers without much effort. Best results are 
achieved with items from the same charge, if this is not possible one 
can use the equalizer resistors as described in this thread.

[MaWin (Guest)]

Mathias wrote:
> Hello, MaWin, thanks a lot! Could you please attached the PDF
> again it states is not existing attachment, even in logged in status.

That's sad, maybe they deleted it for copyright infringement.
But there was already help.

[Gufi36 (Guest)]

A (very!) long time ago I have developed a regulated power supply using 
MOSFET (aka BUZ...).
In order to balance the currents, I have made a circuit using LM324, 
that simply mirrors the source current of MOSFET 1 to MOSFET 2-4. The 
control circuit only controls MOSFET1.
Each OpAmp compares the shunt voltage of MOSFET1 and sets the one of its 
own MOSFET identical. (sorry for poetry instead of schematic!)

This has worked well. All Currents were identical, so was the power 
dissipation. Not sure about the dynamical response. Those days I did not 
care too much...