This is the 5V 3A power supply circuit for those who want to have the Microcontroller or power digital supply source.
For use in experiments such as logic circuit and/or Microprocessor types or even raspberry-pi. These require 5 volts at high current up to 3A rate.
Due to this circuit is very simple with the LM323K (or LM123 or LM223) only one. But the can supply 5V fixed output voltage, and can be connected load current up to 3 Amp.
If you not can buy this IC or want an easy circuit with LM350 or 7805 more transistor.
Please read down.
There are 3 circuits that interesting.
Using LM323K
It is easiest.
How it works
As shown in the circuit below. We will see that use LM323K only one.
Microprocessor 5V 3A power supply circuit using LM323K
Here is step by step a process.
In the circuit, there are only a few pieces of equipment. Such as. The transformer acts to reduce the voltage from AC main (230V or 117V as your country) to AC 9V.
Then, The bridge diode of 4A 100V converts the AC voltage into a DC voltage. Next, we use C1 as the current filter smoothly. And after, the IC1-LM323K regulator will maintain a constant voltage of 5 volts.
In addition, the internal circuit overload protection. Due to Short Circuit output and the thermal protection system over determine as well.
The C2 is a noise frequency filter. The C3 is the output filter.
Read also:
- Many ideas of 12V , 5V Dual Power Supply Circuit at 3A
- 5V 3A Switching Regulator circuit using LM2576
- 5V 5A Power supply circuit
Figure 2: the PCB layout
The parts list
IC1: LM323K or LM323T (To-220) THREE-TERMINAL 3A-5V POSITIVE VOLTAGE REGULATORS
C1: 4,700uF 16V Electrolytic capacitor
C2: 0.1uF 63V Polyester Capacitor
C3: 470uF 16V Electrolytic capacitor
T1: Transformer primary 9V at 3A.
BD1: Bridge diode 4A 200V.
Building
All equipment according to the circuit. Except the transformer can be mounted on the PCB is shown in Figure 2. In installing the IC1, be careful not to input or an output short circuit with the heat sink is strictly prohibited.
Using LM350 as 5V 3A regulator
Many people say that they can’t find LM323. I think we can use LM350 instead, but only need to modify a few additional devices. See the circuit diagram below.
How it works
This circuit uses more equipment. Working on the unregulated power supply is similar to the circuit above.
But we added 3 filter capacitors in parallel Resulting in combined capacitance is 6,600uF. According to the original principles, choose capacitances 2,200 uF per 1A output current.
We should do this because it is cheaper than buying 10,000 uF and it’s easy to find too.
See LM350 Regulator. Normally, we can adjust the output voltage of 1.25V to 30V. And we can set this voltage by R1 and R2.
When we want 5V output, we should use R1 = 270 ohms, R2 = 820 ohms. Why I know it. Because I read…
Recommended: How to use LM350 Regulator
Diode protection
As datasheet, we should add both diodes to protect the backward voltage from the output and others. It can kill LM350.
Transient noise filter
C4, C7 (0.01uF to 0.1uF) filter out the transient noise which can be induced into the supply by stray magnetic fields. C5 and C6 keep the stable voltage.
Other parts
LED1 is power on display like other circuits.
Parts lists
IC1: LM350 Three-Terminal 3A Positive voltage regulator
Electrolytic capacitor
C1-C3: 2,200uF 16V
C5: 22uF 16V Electrolytic
C6: 100uF 16V Electrolytic
C4,C7: 0.01uF 50V Polyester Capacitor
0.5W Resistor tolerance: 5%
R1: 270 ohms
R2: 820 ohms
Others
T1: Transformer, Primary 9V at 3A.
BD1: Bridge diode 4A 200V.
Using 7805 and PNP transistor (cheapest)
If you cannot by both LM323 and LM350. You can use normal IC-7805 and power transistors to make a 5V 4A power supply circuit.
It may be the cheapest circuit.
How it works
We intend to design this circuit to be smaller for saving.
The circuit is similar to the circuit above. Here is step-by-step a process.
In the unregulated power supply. We use it as a full-wave diode rectifier. In the circuit like this, we need to use a center-tap transformer, 9V-CT-9V.
As usual, we know that 7805 is a 1A 5V regulator IC. It runs great. But too low current for us. We need a helper that be kind.
So, the PNP power transistor, TIP2955 will boost up the current to 4A max. When the current flow R1. At B-E of Q1 get a bias current. It conducts current more than IC.
Overvoltage protection
But this circuit has a disadvantage is without overvoltage protection and short circuit protection.
This problem can be easily and effectively solved with fuse and Zener diode.
Fuse Easy but sure
Normally, we do not connect a short circuit or use an overcurrent. Using just one fuse is economical. Definitely works better than complex electronic circuits.
Zener over-voltage protection
We have learned that When the voltage reaches the VZ of Zener diode, it will have a lot of current flowing through it. If we use a 5.6V Zener diode and the voltage exceeds 5.6V.
Learn: How Zener diode works
This is the safe voltage level of the digital circuit. There will be a lot of current causing the fuse burns to immediately
Is it easy?
Parts lists
IC1: LM7805, 3 Terminal 1A Positive voltage regulator
Q1: TIP2955, 15A 80V PNP transistor
ZD1: 5.6V 1W Zener Diode
Electrolytic capacitor
C1-C3: 2,200uF 16V
C4: 10uF 16V Electrolytic
C5: 100uF 16V Electrolytic
C6: 0.1uF 50V Polyester Capacitor
R1: 3.3 ohms, 1W Resistor tolerance 5%
Others
T1: Transformer, Primary 9V at 3A.
BD1: Bridge diode 4A 200V.
F2: 4A Fuse
F1: 1A Fuse
Keep reading: 5V 3A power supply for digital circuit
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