Bench Lab Power Supply 0-50V 0-5A

bench-top-power-supply-project

Every laboratory needs a few critical devices, the most important of which is a power supply for powering the projects. But as demand grows and the projects get bigger a professional and adjustable power supply becomes a necessity. Here is an adjustable 50V/5A power supply with a variable output from 0V to 50V and adjustable current limiting from 0A to 5A. Most simple power supplies cant get the output to come down to exactly 0V or 0A. But in this circuit, the differential amplifiers have a negative power supply rail at (-3V), which can pull the output down to exactly zero.

CIRCUIT DIAGRAM OF BENCH POWER SUPPLY:

bench-top-power-supply-circuit-diagram-project

 

WORKING:

The power supply relies upon two differential amplifiers made from T1 to T6. The first one being responsible for controlling the output current limiting. The second differential amplifier controls the output voltage. They both are driven by the reference voltage created by D5 and D6. The use of zener and a normal diode is to compensate for thermal drift of the reference voltage generator. This is because they both have opposing thermal coefficients.

VOLTAGE CONTROL CIRCUITRY:

The voltage control circuitry is created from T4,T5 and T6. This works by measuring the differential voltage on two base terminals of T4 and T5. One terminal is supplied with the reference voltage, and the other with some of the output voltage. The reference voltage created by D5 and D6 is around 15.4V. Hence the differential amplifier must amplify the voltage difference by 3.4 times to match the 50V output. This is done by the voltage divider (R23,P3) on the inverting terminal of the differential amplifier, setting the gain of the diff amp to 3.4 times.

When tweaking the power supply, you must set P2 to its upper most level. Then fine tune the maximum output voltage to 50V by P3. Since the current sensing resistor (R24) is in a low side configuration, the differential amplifier must correct for the voltage drop it makes when the power supply is loaded.  This is why the reference voltage generator is connected to the (-) terminal of the power supply and not ground terminal. By connecting the reference voltage generator in such way, it allows it to drift up or down by the same amount of voltage the current sensing resistor creates as a voltage drop. Therefore it keeps the output steady through the load.

CURRENT LIMITING CIRCUITRY:

The current limiting circuitry is comprised from T1,T2 and T3. This works by measuring the voltage drop created by the current sensing resistor and comparing it to a given reference voltage created by R11 and P1. I actually suggest replacing R11 with a 220k trimmer and fine tuning the maximum current limit to match your requirements.

As it is, the protection is set to enable at 5.3A. With an adjustable value for R11, you can set the protection at any level up to maybe 6-7A without compensating the circuit for increase in power. When the protection is ON, T3 drops the voltage at its collector, thus creating an appropriate potential difference through the diode D7. At this point it starts stealing some of the biasing voltage of T4. By dropping the base voltage of T4, the output voltage drops sufficiently to keep the current through the load constant.

Hope this project will be useful to you guys 🙂 Do post your questions/ feedback in the comment box below. Happy DIY making 🙂


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Bernard
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Bernard

Hi, I did a lot of simulations. I see that the voltage reference from D4 and D6 are related to the – lead but the feedback form the output is related via R23 and P3 to the -3V (which is a stable voltage), then the voltage drop is not compensated. Change the P3 to the – lead then it is compensated. The differential T4, T5 have then the same references.
Comments?

Bernard
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Bernard

Hi, with the R4 = 470 ohm it gives about -2.1V, is that used in your pcb supply instead of the 240 ohm as other people suggested (datasheet LM337)? I see that the -3V (R4= 240 ohm)as ripple that can be reduced by add 1uF capacitor parallel on R2 and 1uF on the -3V, and make the C1 and C2 higher capacitance. As you suggested, I removed the Cap on the base of T9 (ripple came from -3V). The response is quicker (<2ms in stead of 200ms). It can be more quicker by reducing the output capacitance 100uF to 22uF… Read more »

Bernard
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Bernard

BC546A / BC556A or BC546B / BC556B Simulated in LTSPICE, reduce ripple by capacitor 10..20uF (to gnd) on base T9 Or on the base extra transistor on BC546B on BD139 circuit.
This slows down a little the dynamic current limiter.
For B types a need to set R20 to around 40k to adjust 0V.
Not clear why I see in the simulation much voltage 2V drop above 4A. I simulate further to understand the voltage compensation on the R24 resistor. Something to do with the A or B versions (hfe). Any Idea?

jack
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jack

why are there two r11 resistors, no volts getting to the base of T2.
what should they be.

Gr Tech
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Gr tech

can i use 48V 10A transfo.

Norbert
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Norbert

Of Coarse you can, only…..total power will be 500W , think of the powertransistors if you would adjust the powersupply to give out 5V with 10 A rectified voltage from 48V transformer could get up to 70V, and if you give out 10 A, then you will have to dissipate 65 V x 10 A, that is 650 Watts, than you can NOT do with just 3 output powertransistors, think of beefing it up to 10 transistors!!, next to T7 and T8 make another 7 or 8 tranaistors, and place them on a big big heatsink and/or include a cooling… Read more »

Erik
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Erik

Can you clarify please. The Ground output terminal seems to be at the ground side of C9 but you have a resistor (R24) running from that junction to Ground which doesn’t make sense to me. I do not have any other issues with your schematic & thanks for posting.

Bruno
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Bruno

Hi, this circuit is very interesting, I have to correct some little things:
1-the emitter resistors of the power stage must be written R510, let say half Ohm,
2-the partition resistors of LM337 must be corrected as R4 = 240 Ohm to get -3 V,
3-the reference diodes could be better compensated using Dz5 = 12 V and Dz6 = 3 V, or 3 Zener diodes in series 5,1 V each?
However I am building this circuit, thanks for your project,
Happy New Year

Chris_W
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Chris_W

I’ve noticed a few things about this circuit: R3 and R4 values should be swapped, in their current configuration the output of the negative reg. is about -2V not -3V. -out is in the right place, essentially the supply is floating on the 100mOHM resistor. So regardless of how much current you pull, the potential between the +out and -out will still be your set value. The Voltmeter is directly across the outputs (+out & -out). You can buy displays with programmable gain, so the ammeter is one of these displays connector across the shunt R24. P2 appears to be… Read more »

Anonymous
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Anonymous

where is the display circuit??

maharadga
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maharadga

This circuit is wrong. Out “-” is not from right place.

Youcef
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Youcef

R24 = 0.1 Ohm 10W its correct friend