1.2 volts power supply

This switch mode power supply circuit require an input voltage range , between 4.5V to 16V,and will provide a very stable output voltage of 1.2 volt . The LTM4627 supports an output voltage range of 0.6V to 5V, set by a single external resistor , so you can modify the output voltage .


High switching frequency and a current mode architecture enable a very fast transient response to line and load changes without sacrificing stability.
 

1.2 volts power supply

6V to 12V Converter Circuit with BD679 - BC547

6V to 12V Converter Circuit with BD679- BC547

This is a design circuit for converter circuit. This circuit is based on transistor as controller the circuit. There are two types of transistor that is BC547 and BD679. This circuit is a simple design of converter or inverter. This is the figure of the circuit.

This inverter circuit can to 800mA of 12V power supply with a 6V. For example could you 12V Car Accessories (UK turning into a 6V?) Car. The circuit is simple, more than 75% efficiency and very helpful. By changing a few components you, you also change for different voltages.

Electronic Part List

R1, R4 2 .2 K 1/4W Resistor

R2, R3 47K 1/4W Resistor

R5 1K 1/4W Resistor

R6 15K 1/4W Resistor

R7 33K 1/4W Resistor

R8 10K 1/4W Resistor

C1, C2 0.1uF Ceramic Disc Capacitor

C3 470uF 25V electrolytic capacitor

1N914 diode D1

D2 Diode 1N4004

D3 12V 400mW Zener Diode

Q1, Q2, Q4 BC547 NPN transistor

BD679 NPN transistor Q3

L1 See Notes

Notes

1. L1 is a custom inductor wound with about 80 turns 0.5 mm magnet wire a ring around the core with an outer diameter of 40 mm.



2. Different values of D3 can be used to obtain different output voltages from 0.6V to 30V is about. Note that at higher voltages, the circuit could perform just as well and can not produce much electricity. You may need to use a larger C3 for higher voltages and / or higher currents.



3. You can use a larger value for C3, in order to achieve a better filtering.



4. The circuit requires about 2A from the 6V supply to provide the full 800mA at 12V.

18v power supply USING LM350T

18v power supply USING LM350T

This is a classic linear power supply which produces a regulated 18v, rated at about 1 amp.

POWERSUPPLY

18V / 1 AMP POWER SUPPLY

7805 power supply

5V power supply using 7805

7805 is a 5V fixed three terminal positive voltage regulator IC .The IC has features such as safe operating area protection,thermal shut down, internal current limiting which makes the IC very rugged.Out out currents up to 1A can be drawn from the IC provided that there is a proper heat sink.A 9V transformer steps down the main voltage , 1A bridge rectifies it and capacitor C1 filters it and 7805 regulates it to produce a steady 5V DC .

Circuit diagram with Parts list.

powersupply

Notes.

The bridge D1 can be also made by yourself by using four 1N 4007 diodes.

If more than 400mA current is supposed to be taken from the circuit , fit a heat sink with to the 7805 IC.

12vdc power supply

12vdc power supply

12Volt DC Power Supply

D1 = BRDIGE

C1 = 2000UF / 25V

R1 = 200 R

D2 = IN759

R2 = 10K

Q1 =2N3053

0 - 300V Adjustable Power Supply

0 - 300V Adjustable Power Supply

Introduction

To prevent my high voltage experiments to go up in smoke completely, I designed
a simple circuit which can provide an adjustable voltage source of 0 to 330 Volt..
The supply is short-ciruit proof: the current is limited to about 100mA.

Circuit description

TR1 is a 1:1 mains transformer; it is included for safety.

The mains voltage from TR1 is rectified with bridge D1 (1Amp / 500V) and large elcap C1.

T1 is switched as a source follower: the source of T1 will follow the voltage of the
wiper of R3. D2 is included to protect the gate of T1; although in theory not necessary
I strongly recommend to include it!

T2 and shunt resistor R2 build the current limiter. When the output current becomes too high, T2 will discharge
the gate of T1. This will prevent the current to become too high.
The value of R3 has been determined experimentally; it depends also on the Hfe of T2 so you may need to tune the value of R2.

Note that T1 needs a large heatsink: in worst case T1 will dissipate 330V x 100mA = 33Watt!
Instead of a BUZ 326 (400V/10.5Amp) you can also use an IRF740 (400V/10Amp).
The output impedance of the power supply is determined by the beta of T1, so the larger the MOSFET
the lower the output impedance!

power supply schematic - 12 Volt 30 Amp Power Supply

power supply schematic - 12 Volt 30 Amp Power Supply

Description
Using a single 7812 IC voltage regulator and multiple outboard pass transistors, this power supply can deliver output load currents of up to 30 amps. The design is shown below:








Notes
The input transformer is likely to be the most expensive part of the entire project. As an alternative, a couple of 12 Volt car batteries could be used. The input voltage to the regulator must be at least several volts higher than the output voltage (12V) so that the regulator can maintain its output. If a transformer is used, then the rectifier diodes must be capable of passing a very high peak forward current, typically 100amps or more. The 7812 IC will only pass 1 amp or less of the output current, the remainder being supplied by the outboard pass transistors. As the circuit is designed to handle loads of up to 30 amps, then six TIP2955 are wired in parallel to meet this demand. The dissipation in each power transistor is one sixth of the total load, but adequate heat sinking is still required. Maximum load current will generate maximum dissipation, so a very large heat sink is required. In considering a heat sink, it may be a good idea to look for either a fan or water cooled heat sink. In the event that the power transistors should fail, then the regulator would have to supply full load current and would fail with catastrophic results. A 1 amp fuse in the regulators output prevents a safeguard. The 400mohm load is for test purposes only and should not be included in the final circuit. A simulated performance is shown below:

Calculations
This circuit is a fine example of Kirchoff's current and voltage laws. To summarise, the sum of the currents entering a junction, must equal the current leaving the junction, and the voltages around a loop must equal zero. For example, in the diagram above, the input voltage is 24 volts. 4 volts is dropped across R7 and 20 volts across the regulator input, 24 -4 -20 =0. At the output :- the total load current is 30 amps, the regulator supplies 0.866 A and the 6 transistors 4.855 Amp each , 30 = 6 * 4.855 + 0.866. Each power transistor contributes around 4.86 A to the load. The base current is about 138 mA per transistor. A DC current gain of 35 at a collector current of 6 amp is required. This is well within the limits of the TIP2955. Resistors R1 to R6 are included for stability and prevent current swamping as the manufacturing tolerances of dc current gain will be different for each transistor. Resistor R7 is 100 ohms and develops 4 Volts with maximun load. Power dissipation is hence (4^2)/200 or about 160 mW. I recommend using a 0.5 Watt resistor for R7. The input current to the regulator is fed via the emitter resistor and base emitter junctions of the power transistors. Once again using Kirchoff's current laws, the 871 mA regulator input current is derived from the base chain and the 40.3 mA flowing through the 100 Ohm resistor. 871.18 = 40.3 + 830. 88. The current from the regulator itself cannot be greater than the input current. As can be seen the regulator only draws about 5 mA and should run cold.

power supply schematic - Gyrator Circuit

power supply schematic - Gyrator Circuit

Description
An electronic recitification circuit. The use of large, heavy and expensive electrolytic capacitors is avoided, being replaced by an active transistor in this gyrator circuit.




Circuit Notes
To avoid excess ripple output on a power supply feeding a heavy load, usually a large value capacitor is chosen following the rectifier. In this circuit, C1's value is only a 470uF. The gyrator circuit works on the principle that the value of input capacitance at the base-emitter terminals of a transitor is effectively multiplied by the static forward current gain, HFE of the transistor. In this circuit C2, a 100uF capacitor is effectively magnified at the ouput ( Vreg ).

If you assume a dc current gain, HFE of 50 for the 2N3055 power transistor, then the effective value of the smoothing capacitor would be 50x this value; or be the same as using a 5000uF capacitor without the power transistor. The graph below shows the output voltage and current through the load :-



The load draws nearly 400mA. With the output directly from the rectifier there is about 5v pk-pk ripple in the output. Using the output at the emitter of the transistor things are much better. The circuit will take a few hundred milliseconds for the output voltage to stabilize and reach maximum value. The advantages are that a smaller, less costly reservoir capacitor can be used with this circuit to give a high quality