This is a 0-12V variable power supply circuit at 3 amp. In a normal circuit, the general LM350 regulator has a start voltage of 1.25V.
In the circuit with the components value, makes the output voltage range of approximately 1.25 to 13.5 volts.
But this circuit is special that start of 0-volts. We can adjust the output voltage from 0 volts to 12 volts. We use only one IC and a few other components. So cheap and easy for you.
0 volts for 0-12v variable power supply
We determine the voltage at the output of the regulator, with values of R1 and R2. See LM350 Datasheet! Because these regulators (LM317, LM350, LM338) has a minimum output voltage of 1.25 volts.
Look: Basic circuit of LM350
Ideally, when we use a diode in series. It will make an output voltage lower than the input of 0.7V. With diodes, D3 and D4 will provide a voltage drop of approximately 1.4 volts. Thus, the output voltage of 0 volts
Another advantage of both diodes(D3, D4), will prevent damage to the regulator during certain adverse conditions. Such as the output voltage is higher than the input voltage to the regulator.
They can happen if this type of circuit is variable high voltage power supply, and without the protection of D3 and D4.
We need to use Both diodes that have a current rating of at least 3 amp.
In the circuit, we use diodes D1 and D2 as the feedback blocker. Since the high reverse currents usually flow to the regulator when the devices power down. So we use both diodes to protect any current that is damage to the regulator.
This circuit is a low output voltage but high current. But we need to use a substantial heatsink for IC1-LM350. It will quite a bit of heat while working full load current.
For this reason, this circuit is not recommended for use in handheld throttles.
If the less expensive LM317 regulator is used this circuit makes an excellent test bench power supply just leave out D3 and D4 and add a voltmeter.
Parts you will need
T1 = 12V to 15V @3A primary transformer, Quantity:1.
IC1 = LM350T, 3A Regulator , Quantity:1.
BD1 = 10A Diode bridge, Quantity:1.
R1 = 240Ω 0.25W , Quantity:1.
VR1 = 1K Potentiometer, Quantity:1.
D1,D2 = 1N4001 – 1A 100V – general purpose Diodes, Quantity:2.
D3,D4 = 1N5402 – 3A 100V- Diodes, Quantity:2.
C1 = 3300uF 25V Electrolytic capacitor, Quantity:1
C2,C2 = 10uF 25V Electrolytic capacitor, Quantity:2
What is more? Look LM350 pinout:
Some may not know.
If it does not work
Sound bad, this does not run.
I did not want to see you are disappointed. Please continued to check it again.
Here is an example case that may happen.
First, check all polarity of components. Are they right?
LM350, T1, D1 to D4, Bridge Diodes, C1 to C3, VR1 and more.
You checked them all, right?
Then, if…
High voltage at output— Check D1, may switch polarity.
Zero voltage—Check D3 and D3 may not correct.
Hope you can do it.
Also the 0-12V Variable power supply
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I always try to make Electronics Learning Easy.
How to add voltmeter to this circuit? voltmeter has three wires red, black and yellow?
Can I replace the potentiometer with 10K value?
Hi Ali Mohammed,
Thank you again.
How to add voltmeter.
Did you see this: https://www.eleccircuit.com/diy-digital-voltmeter-panel-meter-0-50v/
Yes, you can use 10K potentiometer. But you may adjust hard the output voltage.
You should add a 1K resistor in parallel with it. Both resistance mix together into about 1K.
I hope you will understand me said.
Have a good day.
Is this working?
I’ve tried it and it doesn’t work.
Hi,
Sound bad, this does not run.
I did not want to see you are disappointed.
Please continued to check it again.
See text.
Should i use the 2 12v terminals of the transformer or the 12 and 0 terminals?
Hi Yo matam,
You may use 12V OV terminal. Use bridge diodes are better than Full-wave rectifier. Keep reading more: https://www.eleccircuit.com/unregulator-power-supply/