This is 12V 5A linear power supply circuit. Also, how to find resistors onside. I like to use a series of 78xx regulator ICs. Because it is so easy with a few parts. If we want 12V output. So, use 7812. But Maximum current at output does not exceed 1A. What can we do?
Basic 12V positive regulator using 7812
First, look at work only one alone. In a circuit is a basic using 7812. It is so easy. When Vin comes into IC-7812. It will keep the fixed output voltage, 12V.
From my experiment, I use the lowest voltage level. It causes a low output current. But if using high voltage. It causes the output current to rise. But It is too hot.
So, I choose a medium voltage level is 20V.
Inside IC has many components. In general use, We do not need to understand those devices. We just use it by understanding its features and limitations.
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Look at circuit again. How do both capacitors work?
Cin (input capacitor) — It is between Input pin and ground pin of IC. We need to use it when the distance between input and IC exceeds 5 cm.
It will temporarily store power from the input. The IC is able to pull energy to use immediately.
Inputs and IC are close together. It is not necessary to use it. Because IC can get the energy immediately.
If necessary use it. we should choose size 0.33μF to 1μF. Most importantly choose a capacitor type that responds to high frequencies as well
Cout (output capacitor)—It improves transient response. Which does not affect the stability of the circuit
12V 1A power supply using 7812
Meet a nice 12V regulator. It is suitable for a mini load that uses current less than 0.8A. You can build it easily on a small PCB layout.
Above we need 5A linear power supply. There are many ways to do it. Let me tell you this way. Using transistor current booster is suitable. Because it is cheap and easy to build with normal components in a lot of local stores.
Look at in the circuit, many components have a different function as follows.
First, both transistors Q1 and Q2 act as increase current up to 5A.
Second, both fuse resistors RS will protect the saturation of transistor Q1 and Q2.
If both fuse resistors blow or have high resistances
Third, resistor R1 sets a bias current to IC-7812 and output current of Q1. Which it must not exceed its saturation current.
Then, meet the current.
Still, look at the circuit again. The output current is a total Current Source, IT. Which it gets from mixing between IIC and IQ1.
- IIC is the output current of IC-7812.
- IQ1 is the output current of power transistor Q1.
Look back on the path of the electricity.
- IIC get the current from Vin through R1 into pin 1 to Pin 3 of 7812. Which IC1 keeps a fixed regulated voltage, 12V.
- IQ1 comes from working together of both transistors Q1 and Q2.
First, a base current of Q2 flows to its emitter. And, out of a base of Q1 through pin 1 of 7812 to ground as compete for a circuit.
The base bias current of Q2 also causes the base current of Q1 can flow.
So, resistance and voltage across emitter and collector of both Q1 and Q2 are low.
Second, there is a collector of both transistors. Because…
The output current of Q2 flows from Vin through an emitter to collector. Via pin 1 of 7812 to complete cycle circuit to the ground.
Third, the output current of Q1 flows from Vin through a fuse resistor RS. With working of Q1, the current flows through pin 3 of IC1 to power to the load.
How to design 5A linear supply
We have details on the design of the components as shown above as follows.
1. Q1 Power PNP transistor on shape TO-3, should choose…
MJ29555—ICmax = 15A, CEvoltage = 60V, hFE = 20-70
2N3792—ICmax = 10A, CEvoltage = 80V, hFE = 50 min
BDX18—ICmax = 15A, CEvoltage = 60V, hFE = 60 min
2. Q2-PNP transistor on shape TO-66 should choose…
2N6049—ICmax = 10A, CEvoltage = 55V, hFE = 25 min
2N5956–ICmax = 6A, CEvoltage = 40V, hFE = 20 min
BDX78–ICmax = 8A, CEvoltage = 80V, hFE = 30 min
3. The output current is a total Current Source, called IT. Which it gets from mixing between IIC and IQ1.
IT = IIC + IQ1
4. RS- Fuse resistor is a ratio between 0.6V per IC of Q1(IQ1).
RS = 0.6V/IQ1
5. R1 always has resistance to more than 0Ω. But it must not exceed a ratio between (VbeQ1) 0.6V per bias current of 7812(IbiasIC1).
0 Ω ≤ R1 ≤ VbeQ1/ IbiasIC1
R1 ≤ VbeQ1/ IbiasIC1
Value on circuit
We know the value on the circuit below.
- Vout = 12V—voltage to load.
- IT = 5A—total output current to load.
- Vin = 20V—input voltage to this circuit
- IbiasIC1 = 12mA —the current to cause IC1 run
Transistor booster circuit design steps
Look at on the 12V 5A Linear Power supply circuit diagram.
There are component detail as follows.
1. I choose Q1 is 2N3792 and Q2 is 2N6049 respectively. Because they are so easy to buy and cheap.
2. The Q1 must power an output current up to 5A. Also, IQ1 is 5A.
RS = 0.6V / IQ1
= 0.6V / 5A
And the power of RS can find with…
PRS = IQ1² x RS
= 5A x 5A x 0.12Ω
= 3 watts
Thus, I choose resistor is 0.12Ω at 5W.
In normally, a regulator IC-7812 can power current to 1A max. But we set a bias current of 12mA. Or
IbiasIC1 = 12mA = 0.012A
We can find R1 with a formula.
R1 = VbeQ1/IbiasIC1
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