5V -1A Power Supply using LM2575

A Power Transistor which is having a drop of 4 Volts across it and passing 3 amps thru it, may dissipate around 12 Watts of Heat, This is the problem in Series Regulators. While a Saturated Transistor or Mosfet with 1 Volts across and 3 Amps Thru will be just 3 Watts. But then a fully on transistor or mosfet cannot be controlled or regulated, for that we turn it ON and OFF very fast so that the right amount of current or voltage is delivered.

Power Electronic Circuits  

5V -1A Power Supply using LM2575

The way this is done is PWM – Pulse Width Modulation. In this the mosfet or transistor is switched ON-OFF at say 100 kHz, but the ON duration is varied to control the output. The longer the duration of ON time more energy or punch is transferred. Switching losses will be present depending on how fast the rise and fall times of the pulses are.

The Pulsed AC or Chopped DC can be smoothed to the Average with Inductors and Capacitors. The reactive pulses of the Inductor has to be absorbed by a Schottky Rectifier 1N5817 — 20V-1A fast switching diode with low switching losses.

This circuit is derived from an application note of LM2575, It is a Power Switching Regulator from National Semiconductor The details are here LM2575

Regulated High Voltage Power Supply

The Circuit below is a paper design and not tested. It can be used for education and information, this can help you make your own design. Please do not just wire it up and expect it to work.

Regulated High Voltage Power Supply

Now let me see if i can explain the circuit, This is a regulated AC power supply. This circuit uses the Mosfet to turn off when voltage goes beyond a reference point. That means it just chops the Sine wave above a point, that also implies that the output may not be pure sine and may have harmonics. The Transformer if well designed may smoothen the chops. Even a Series Inductor or Resonant Circuits may reduce harmonics.

The opto coupler 4N50 Provides isolation and good Current Transfer Ratio. That may mean you may not get a shock and that even a small current signal in Opto-LED will give a saturated or Low Impedance in Opto-Transistor. The Mosfet is used like a Impedance Control switch turned On-Off by Opto. The Optocoupler diode is controlled by the Opamps which work Closed loop. The transformer output is compared with reference to drive opto-led.

This Circuit is based on Teledyne Solid State data book application note. They may not be making these parts anymore but they are available from others.

Regulated High Voltage Power Supply – del20032

Single Polarity Power Supply

This circuit uses a PNP Power Transistor TIP2955, you can use any other according to your current and voltage requirement.

Look at R2 a 10 Ohm resistor, when the current in your load to the power supply is less than 70mA the voltage across R2 is less than 10E * 70mA = 700mV right. The base emitter junction of Q1 will be biased or turned on around 700mV, less than 700mV the transistor just does nothing.

Single Polarity Power Supply

When the current in your load goes over 70mA the voltage across R2 goes above 700mV and a small base current Ib flows from emitter to base of Q1 turning on the transistor. Now a collector current Ic flows from emitter to collector and then to your load supplying the excess demand. The Ic = Ib * hfe where hfe or beta is the DC gain value.

From my Power Electronic Circuits

Some transistors will have only AC gain specified which is lower than DC gain. TIP2955 has a gain of 20 so for an Ib of 50mA the Ic will be 1 Amp which saves the regulator from heating up or shutting down as the main current flows thru the transistor. Q1 should be provided with a good heatsink.

Pulse width modulation using 555

IC1 astable gives a fixed square wave at pin 3, C1 and R1 derive uS trigger pulses from IC1 and this will trigger IC2 monostable or single shot, the voltage at pin 5 of IC2 will change the pulse width output of IC2, to get it working all the three RC combinations have to be figured out.

Optical Obstacle Switch.

You can even build a small SMPS with this or even control the temperature of your soldering iron using the SSR solid state relay circuits in power section, then you need to think and design the cycle time of a soldering iron heat control system, it will be in seconds but then above circuit is running at audio frequencies, then you have to work that out yourself..

Pulse width modulation using 555

Isolated dual power supply from 5V

This is a unregulated supply for low power circuits. You may be able to regulate the outputs with zeners or small regulators like 78L05.

The transformer can be hand wound in a mini ferrite pot core. you can use 2N2222 or any other fast transistor. The transformer should have 1KV isolation. The dot polarity of TR1 should be properly observed, else it may fail to oscillate or give output.

Simple WorkBench Dual Power Supply – del20033

Diode should be fast recovery type, for less than 100mA use 1N4148. transformer, pri-20-20, sec-60-60, a SWG-AWG to suit the current you

design for, any fast switching transistor would work, no regulation, use regulators like 78L12 if you want, circuit like multivibrator used for flashing LED lights.

Isolated dual power supply from 5V

The Source file in CadSoft EAGLE format is here del00010.zip

Dual Polarity Power Supply

This supply gives both positive and negative outputs. Appropriate Fuses should be used to protect from fire hazard and overload of transformer.

Dual Polarity Power Supply

The Filter capacitor C1 4700uF has an impedance of Xc = 1 / (2 * 3.14 * f * C) which comes to 0.6 ohms at 50 Hz. The impedance of the load at 2A for 24V is R = V / I that is 12 Ohms which is more than 20 times the impedance of the capacitor at 50 Hz. That means less than 1 / 20 of ripple current will flow thru the load. The Regulator also reduces the ripple a little.

Simple WorkBench Dual Power Supply

Battery Backup Supply

This is a 9V power supply which will work even on power failure. It uses a rechargeable battery and regulators. A transformer with 15-0-15 AC volts output is required.

From my Power Electronic Circuits

Battery Backup Supply

In the first regulator U1 the output is lifted up by 1.4V and in the second regulator U2 by a resistor divider. In the second regulator the voltage across resistor R3 is 5V, so the current is 5V / 1K = 5mA this adds to the quiescent current of 5mA from the regulators ground terminal and flows into the resistors R1 and R2 in parallel which form 404 ohms, 10mA thru 404 ohms is 4V. So the output will be 5 + 4 = 9V. Note that the charge and discharge paths of the battery are separated with diodes.