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Stabilised Power Supply with Current Limiting

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mechie

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A simple bench PSU capable of :-
Approx. 5v to 15v output at 1A
Current limited from about 500mA to 1A (adjustable)

Discreet_PSU_V1.gif


Circuit Operation
An error amplifier is formed from Tr1 and Tr2, wired as a differential amplifier (often called a long-tailed pair -- look at those collector leads). One input of this diff. amp is taken from the zener diode which provides a stable 4.7 volts, the other diff. amp input is a fraction of the output. Any difference between the two will cause the current through R3 to alter and so will alter the voltage across it and hence to the base of Tr3.
Tr3 and Tr4 are connected as a darlington pair, this produces a very high gain to aid stability of regulation. This darlington pair can be visualised as a single transistor connected in an 'emitter-follower' configuration, the emitter voltage will follow the base voltage (less the 1.2v required to forward bias the two base-emitter junctions), but with a much greater current capacity.
Current limiting is provided by Tr5 which will be forward biassed by a fraction of the voltage drop across the current sensing resistor, R5, set by Vr1. As the current through R5 increases so does the voltage dropped across it, this begins to bias Tr5 on and in so doing causes Tr3/Tr4 to be deprived of base current and so reduces the output voltage.

Tr4 needs a heatsink of at least 10 degrees C/Watt (10C / W) and R5 is expected to carry (1A * 4R) = 4 Watts, so it will need to be a 5W device or four 1R 1W devices in series, mount them on a peice of metal as they will get hot!
The transformer needs to be selected to suit your local mains voltage and provide between 15 v and 20 v out at 1A (thats 15 or 20VA, respectively).

As there is no current measurement built in to the unit, a useable scale could be marked around the potentiometer Vr1's knob such that a known current could be chosen in advance of connecting a load. To do this, connect an ammeter directly across the power supplie's output - this will instantly overload the supply and cause the current limiting to operate. Adjust Vr1 so that the ammeter displays 500mA and mark the knob's position. Repeat the procedure for 600mA, etc. ## DON'T keep the supply shorted for too long as Tr4 and R5 will soon get hot! ##

Bits List...
R1, R2 1k0
R3 2k2
R4, R7 470R
R5 4R 5W (see text)
R6 47R
Vr1 100R
Vr2 1k0
C1 6800uF 25v (4700uF will probably do)
C2 0u47
D1, D2, D3, D4 1N4001 (or a 1 amp bridge rectifier)
Z1 BZY88C 4V7
Tr1, Tr2, Tr3, Tr5 BC108 (or BC109)
Tr4 BD131
Transformer 110/240v to 15 or 20v, 1A
 
Does anyone know what a sutible altherive for the BD131 transistor would be i could proplerly find it myslef but a bit busy with collage at the moh so any help would be apreshated
 
Does anyone know what a sutible altherive for the BD131 transistor would be i could proplerly find it myslef but a bit busy with collage at the moh so any help would be apreshated

TIP 31, TIP41, BD243, BD535 etc. Almost any transistor with simillar power voltage & current ratings.
 
hi. what is the best way to indicate that the supply has gone into current limit, i.e. using an LED. thanks
 
It's a simple PSU and there is no simple way to indicate current limiting mode.
This could be fixed by changing arrangement of current limiting circuit.
This is nice example of linear regulator and good excercise for sizing but there
are some things to be aware of. For example, for full 15V/1A output, there will be
4V drop on R5 and some more on transistors since Tr3/Tr4 darlington is wired
as common collector. This means that supplied DC voltage should be at least
some 25V. C1 should be rated 40V or higher since 25V is already border line
and would not provide any safety margin for AC line hickups. For linear PSU fed
from 50 or 60Hz circuit it is common to use about 2000uF for each Amp.
This means that even 2200uF cap will be just fine, specially since we have
regulator after it. Bigger caps offer better filtering and lower ripple and this makes
sense only for not regulated power supplies (like for audio amplifiers etc.).
In worst case scenario Tr4 will be overloaded. BD131 is good for up to 3A but
absolute maximum power rating for this transistor is only 15W. If the voltage accross
C1 is 25V for example and output current is 1A, this transistor will have to handle
(25V-9V)*1A=16W. 9V is sum of output voltage of 5V and 4V drop on R5.
As said, it's a good excercise although i would look for something else as permanent
solution (minimum output voltage of 5V and that 4V drop is pretty big sacriface if you ask me).
I like to be able to use supply voltage as low as 1-1.5V to be able to test and repair
circuits running on single battery (...and not just brand new 1.5V battery).
I would also strongly recommend built in voltmeter on the output of adjustable
power supplies, even if it is salvaged VU meter with matching series resisor and
hand marked scale.
 
you are right, it's not there anymore...
 
brihoo2k said:
Why cant I see the attached image
hi no problem i will give u another psu diagrams this is 30V2A bench psu
best regards

ataul
 

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15 million parts instead of a single LM350 3A adjustable voltage regulator IC?
Even an inexpensive LM317 is good for 1A.
 
raybo said:
I go with the guru this is totaly ridiculous lm337 and a 2n3050 is needed
LM337 is a negative regulator. 2N3050 is a matched pair of PNPs.
You're the guy calling people idiots.
When you feel your butt, do you think that it might in fact be a hot rock?
 
Wow what a dumphuc!
 
rowley said:
Does anyone know what a sutible altherive for the BD131 transistor would be i could proplerly find it myslef but a bit busy with collage at the moh so any help would be apreshated
I used BD237.It's even better.
 
try this simple current/voltage adjustable

current/voltage adjustable:)
 

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Last edited:
mechie said:
The transformer needs to be selected to suit your local mains voltage and provide between 15 v and 20 v out at 1A (thats 15 or 20VA, respectively).

Isn't the RMS input current more than the DC output current?

Here's an exmple:

Consider a 12V 50Hz AC source driving a 1A constant current load via a rectifier and a 10,000uF smoothing capacitor.

Vout = 12 * root(2) = 16.97V (ignoring the rectifier losses)
Ignoring the 1V ripple the output power is 16.97W, so the rms input current must be greater than 1A, it must be at least 16.97/12 = 1.414A, hence the transformer should be sized accordingly.
 
Hi Hero,
You are correct. The transformer will be overloaded if it has the same current rating as the project's output current.

The power passing (VA) rating of a transformer should be used to specify it. Then the power wasted by heating the recifiers and regulator will be included.
 
What make the situation even worse is that the power isn't drawn thoughout the AC cycle but in surges at the high parts of the waveform. These surges overload the transformer in extreem cases causing core saturation which increases losses (and therefore heating) even further.
 
I don't think a transformer manufacturer would make a transformer so poor that its core saturates when it has a rectifier and filter capacitor feeding a load and it is operating within its ratings.
 
I've seen many transformers feeling so hot to the touch by just idling without any load connected. This is a sign of inferior core material or some core saturation occurring. Could be all cost related.
 
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