Variable Switch Mode Power Supply circuit, 0-50V 5A

Today I suggest examples as the variable voltage power supply circuit that can adjust output from 0-50volts at 5A. Which many people interested it a lot. Because is a switch mode type that is Technology available high performance. than normal linear type are they are a small, energy saving and low heat.

I apologize for that. This circuit is just an idea. Ideal for those have basis of this nature circuit. But it is not suitable for beginners. You should also do their homework too. To understand and be able creating the project out.
The heart part that is TL494. Widely used in the computer power supply circuit and the motor control circuits The advantage is high efficiency. easy to use and affordable. And We still use the MJ15004-power transistor to amplify currents up to 5A.

How this circuit works

From circuit in figure below. the AC 220 volts when through a T1-transformer then will be AC voltage 36V. The F1-fuse size of 2 Amperes. The SW1 is ON/OFF switch of this projects.

This 36 volts is Average the voltage. When through the D1-bridge rectifier then will be C1 and C2 are filter to smooth a fluctuating signal(AC Volts).

When push the S1-power switch ON, Both capacitors C1 and C2 will discharges through the R1-resistor.

The input voltage of this regulated circuit is volts across C1 and C2 will equal to 50 volts. And is entered at emitter of Q2. To output next. The circuit in section R2, C3 and ZD1 are the regulated circuit of +15 volts to the circuit of IC1, IC2 and IC3.

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The circuit of IC1 all will be the pulse circuit. wide mod as say above, both C4 and R4 will be RC that create frequency that have value about 20 kHz. In character at pin 13 to ground. The working of both transistor within TL 494 will works in the Single-end Operation type.

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The both transistor in TL494 can pull current up to 400 mA which can drive base of Q2. But will see that VC1 and VC2 have value of 40 volts only.
Which this circuit has voltage high up to 50 volts, if connect directly to them Must damaged certainly.

Therefore, the need for security, Q1-transistor (BD139) to also drive the Q2 . Both R12, R13 that parallel together, as limiter base of Q2 will not over than 250 mA decisively.

View IC1 back again. They R8, R9 and C5 between pin 2 and pin 3 that will be network that be editor Q2 work with narrow pulse before (are Q2 partially ON”) so can reduce pull of the output current.

When push switch SW1 normally to working. then at pin 3 of TL494 will be have a low voltage. So, cause output of the within comparator circuit are “High”. The Q2 will be on. But if the voltage at pin 3 have a higher value in first time. (because C5) The Q2 will be partially no. Then voltage at pin 3 will reduce slowly normal state. (because they will be discharge through R8)

In the primary circuit of the switching section that coil L1 is 0.5 mH. Both C6 and C7 (or C7 will Consists of 330uF 75V – two capacitor connect parallel. Both C8 and C9 acts connected via a high-frequency signal. (because switching) to ground. The D2 use as the fast recovery diode of Toshiba number 6GB 11 (D2 must have special properties than plain diode is able to flow and stop flow quickly. The common diode rectifier will not work. because will be cause the output short circuit when Q2 “ON” cause Q2 damaged immediately.)

LM338 5A Variable regulator Better life with high current in same voltage 1.25V to 30V.

The R14 560 ohm 5 watt serves primarily, 1. are current draw of the circuit or load, can maintain regulated when no load, 2. will reduce undulation of Overshoot voltage. and 3. discharge to C6 and C7 when switch- SW1 “OFF” or No load.

The M1-meter are ammeter that will measure the current through the load circuit. Then M2 is a voltmeter to measure the voltage across load or output voltage of circuit. which can adjust by rotate VR1-2K that divide voltage on R16 are 120ohm, R15 is 4.7K that will ????? input current of error amp circuit (pin 1 of TL494)

The C10-capacitor will reduce ripple at outpu. both VR1 and R16 Changes may be important is the ratio of VR1/R16 must be 100/6 to take 100K to 5.6K, or 50K and 3K time.

The R24 0.1 ohm 5 watts will be the load current detection. The current of all load will flow through R24-resistor and will cause voltage drop across R24.

When current through R24 of 5.6 A will cause the voltage is 0.56 volts at pin 16 of TL494. The comparators limit current will work, make pulse that go to control the Q1 narrow. We will see that if use R24 error, will cause the current limiting slow or fast. But will adjust R11 too. For example : If R24 More value, R11 was increased. (may be 180 ohm or 200 ohm)

Another part of the circuit that we have not mentioned. are the indicator regulation circuit or The size of the ripple at output. This circuit will consist of both IC2 and IC3. We are familiar IC2 is LM741-op-amp IC, they act as an amplifier of AC has the gain is 56 (R20/R19) ripple (Typically less than 10 mV.) Will enter through C14 to pin 3 of IC2. Which will bias to 15/2 = 7.5 volts. So output voltage will be in range 7.5 +- 56 x 10mV or in range 6.94 to 8.06 volts. The C15 10uF 25V will acts as the output monostable circuit at pin 3 will be “Low” state all time. If voltage at pin 2 also not lower than Vcc/3 or 5 volts. Once the voltage at pin 2 is below 5 volts. At pin 3 will be “High” is less than 1.1 R12, C17 or 0.1 second.

We observe that ripple will be expanded to the size 7.5-5 volts is 2.5 volts, before D4 glow. So ripple be must size 2.5×2/56 = 90 mV(peak-to-peak). The part D3 that will be LED show power on of all circuit which include R23 acts as limits current of D3

What’s more? You can look other power supply circuits: Click Here

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