This ciruit Top Linear power supply 5V voltage regulator 5A.
It is good power supply circuit than IC 7805+MJ2955( regulator 5V 5A ) .
Use IC 7812 and LM723 regulator IC, Thansistor TIP142 for boost current to 8A max.
P1 for Control Volt output 2.5-7V,Transformer 10A min.
Detail other see in picture citcuit.
When you need a power supply often missed the 3 legs Regulator ICs. But some jobs that apply a high current over 1 Ampere up there, is very difficult. Even if it is 5 amp and 10 amp, but the price is quite high.
See circuit and pcb below!
This project is designed with the concept of the modular regular circuit is using the output. Which is composed of two or more transistors, to current as shown in the circuit. And a control section is not sensitive to noise, which we would have to use IC-723. Although the present may be perhaps was overshadowed with the 3-pin regulator IC. However, with good characteristics, makes us choose to use it for Supply output voltage from 2 to 7 volts.
The voltage for provide IC1-LM723 get from the increasing the voltage and then filtered to smooth. then, through the control the voltage regulator using 3-pin. This method Is good for power transistors, because we make the output voltage and the before into transistor voltage, both difference as little as possible. without prejudice to the power supply voltage of the IC.
– While running two transistors T2 and T2 may be hot we should hold the adequate heat sink and
All resistors R4 to R6 should used many to parallel to the desired value to average Power Dissipation down.
-Resistors: R4 and R5 use 0.33 ohm 5 watt are 2 pcs.
-And resistor R6, we use 0.22 ohm 5 watt are 2 pcs at the output current on 6 Ampere or 0.33 ohm 5 watt are 2 pcs at current 8 ampere.
In putting, you should be free spacing of the each resistors and the PCB to Cooling.
This circuits’ output voltage can fine up to 14Volt which must be changed a few parts following: the transformer, resistor R1, R2 and capacitor C5, C6. But do not use the boost up voltage (C1, C2, D1, D2). The anode of D3 connect to the rectifier and filter circuits.
Note to TIP142 is although it looks like the common transistor, But within there are structure Darlington Compound. So cannot replace with the normal power transistors.
The electronic parts
Resistors 1/4W +/-5%
R3_________________100 Ohm 1W
R4, R5______________0.15 Ohm 5W
R6_________________0.1 Ohm 10W
P1__________________5 K POT
C1, C2_______________470uF 50V
C5, C6________________1000Vu 25V
BD1 = Diode bridge 10A 40V
D1-D3 = 1N4001
T1 = BD139 – the midterm power transistor
T2, T3 = TIP142 (the Darlington Compound)
IC1 = IC-7812- Fixed voltage Regulator IC DC12V
IC2 = LM723 Adjustable voltage regulator
Tr = Toroid transformer 10V 10A
S1 = switch on/off 2 set.
The testing and apply
In the testing we use the resistor 0.68 ohm to output, then adjust voltage as 5.5 Volt (there is current 8 Amp) The results showed that the voltage drop across the 5.32 volt. Show that the drop to 3.3 percent to 7.8 amperes and measure ripple voltage less than 25 mV (RMS).
7805+Mj2955 power supply 5V 5A for digital circuit
This is circuit dc regulator power supply 5V at current 5A, I used it for my digital production. The mian part electronics is IC-7805 and MJ2955 transistor .
You must use the the 9VAC 5A transformer.
Circuit power supply 5V 5A by 7805+Mj2955
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