Current booster circuits for IC-78xx

Here is the Current booster circuit using transistors. Normally 78xx regulator series ICs can power max output currents of 1 A. Or in the common usage, it is only up to 0.6A. But we need more current up. You must buy a new one, like a 78H05 voltage regulator, LM350, or LM338, etc.

But it is expensive. How do you do? Read this may be alternative good for you. Don’t worry you can use the current booster circuit for 7805 (78xx).

The current booster circuit for 7805

In this increasing Regulator Current or circuit, the power transistor is used to provide most current to the load, to maintain a constant voltage, by the little current still through the regulator IC intact.

Let’s say, I want a 12V voltage regulator, and the current output voltage at 5A, I must to use an IC No. 7812 and it can take a transistor number MJ2955 or TIP2955. The input voltage is enough I think that about 20V, which voltage drop across the transistor is approximately 8V. And it takes about 40Watt energy, which can be calculated = Vce x Ic. When is this. We need a cooling pad for Q1, which is large enough, it is very hot, maybe cause damage.

If you want to add current to the IC 79xx or negative voltage, you can also easy to make, but simple NPN transistor type 2N3055 is only a number.

Want more brilliant ideas? Here’s how to get them through electronic circuits.

10A fixed  regulated Supply using 78XX – MJ15004

If you have a lot of old pieces of equipment. You may build a 10A DC power supply circuit.

That be simple please, when I sees the equipment that him has had already. Then take this circuit see use 78XX integrated circuits. Have many the number and Transistor power number M15004 that can enhance current. Get tall arrive at 10A.

The important aspect is must use the Transformer that be appropriate give current 10A More than. And use C-Filter 10000uF sizes go up with. The detail is other see in the circuit has leisurely sir.

Increase current of 7805 with protection

But above circuit not have protection over load We need to use the below circuit!!

Sometimes we want to use. high current at a dc regulator 5v 3a. We can do this by using the voltage regulator IC1 positive constant number 7805.

Learn more 7805 Datasheet

The Q1-2N3792 and Q2-2N5956 of the circuit will serve to increase current for the IC regulator.

The R2 acts as the anti-saturation of Q12N3792 and Q2-2N5956. Without R2 will allow the two transistors to do not work. The output current can be from IC1 alone.R1 specify the bias current to IC1. And the current output of you against too much saturated with.

The C1 functions in the input power filter to smooth. And C2 serves to prevent interference with the pressure to payout (Vout).

This circuit can help enlarge the current of the power supply circuit that uses an Ic regulator so well. you can use other IC 7805,7809, IC-7812, and IC7815. Q1 is BFX88 or Equivalent, Q2 is BD132 or Equivalent

Note: The section below inspired me to learn from here. Thanks

Experiment increase current of 7805 with 2N3055 transistors

Imagine you need a 5V 2A power supply to experiment with digital circuits. Only, you have plain components on hand. Is it possible to use those components? to create a circuit according to this problem?

We will find out which circuit layout provides maximum power and efficiency.

Let’s learn by experimenting with circuits.

Create the load

First of all, let’s get a good load. Why? It is very important for power supply circuit experiments.

We think roughly if the power supply is 5V 2A, our load should have a resistance of 2.5Ω. Of course, the most economical way is to use ordinary resistors, with 10 watts or more of wattage. If we use 0.25W-resistor it will burn instantly.

But I don’t have a 2.5Ω 10W resistor lying around. So, I decided to put two 4.7Ω resistors in parallel. They become a new resistor with half the resistances, at 2.35 ohms. More importantly, it can withstand a power of up to 10 watts. It’s close enough to what we need.

Then, connect it to a 5V 3A power supply, it takes around 2.1A of current, close to what we will need. And they seem not to be overheated.

Basic circuit 5V 2A using 7805 and 2N3055

See the circuit below, transistor Series Voltage Regulator. It’s simple and cheap. But it has low efficiency because we use Zener Diode to keep the voltage stable and the transistor can supply up to 2A as needed.

Next, we tried replacing the Zener Diode circuits with a 7805 regulator. Look at the circuit.

It’s much better. Because the 7805 works well, it’s specifically designed as a voltage regulator. But it can only supply about 1A of current.

Put the 2N3055; This method might solve the problem. The output current of 7805 flows to a base of 2N3055. It will allow a much larger current to flow between the collector and the emitter.

This circuit quite worked. The output current is high up, according to the gain of the transistor itself.

However, we try connecting the load. Then, measure the voltage and current of the load. The result is unsatisfactory.

Note: you should read 7805 datasheet/ see pinout before assembling this circuit on a breadboard.

The load voltage is 4.4V. It is too low using a digital circuit. While, it is low current about 1.6A, too.

Why? I measure Vin, it is stable at 12V. Then, measure voltage the 7805’s output, it is quite lower to about 4V. The problem may be caused by 2N3055 or circuit form, unable to fully extract energy.

Oh…we only have 2N3055, don’t despair. Let’s find a solution.

When finding a problem, go back to the beginning. Let’s look at the structure of this circuit again. It makes it easier to understand.

We need a high gain transistor circuit. Which form circuit is best? Experiments are the best way to find answers.

Transistor circuit experiment

Prepare a load

It’s fun to find out the value of the devices. We intend to use the LED as a display device and the power supply is a 9V battery, for ease. It requires a limiting current resistor.

As the red LED 3mm uses 10 mA of current at 2V.

Thus we can calculate R2:
R2 = V_BAT – V_LED / I_LED

= (9V – 2V) ÷ 0.01A = 700Ω
We use 680Ω 0.25W of the resistor.

Common collector NPN transistor circuit

Assemble the components as the circuit below. The LED glows.

Then, measure the voltage and current. The load current is 0.009A, both IC and IE.
The load voltage is 8V. Some voltage across VBE is 0.7V.

In short, we call this a circuit pattern that common collector.

We prefer to use the power supply circuit working with the Zener diode as in the example above: transistor Series Voltage Regulator. It’s the form of the circuit that we use.

But there is a disadvantage, the voltage gain is less than 1 and the power gain is very low. So, can amplify the current less than we need.

See other simple electronic circuits

Common emitter NPN transistor circuit

Change the position of Load to connect the Vin and the collector of the transistor according to the circuit diagram below.

The LED1 lights up as before. Then, measure the current of the load. It is 0.01A exactly as calculated. And it gets almost full voltage as the power supply, 8.9V. It worked well. This circuit pattern is a common emitter.

It has a very high power gain but it is not suitable for this power supply circuit because we always need to connect Vout and Ground( negative).

What should we do?

Common emitter PNP transistor circuit

Use a PNP transistor, to reverse the voltage. I changed the transistor from the original 2N3904 to the 2N3906 Small PNP transistor. See the circuit below.

The load connected to GND and Output is similar to a normal power supply circuit. This circuit works fine, the LED is bright and has a current flowing through it, which is 0.01A exactly as calculated.

The form of this circuit is a Common emitter using a PNP transistor. It has a very high power gain just by changing the position of the R1 bias current to the base of the transistor.

But 2N3055 is an NPN transistor. How do we modify it?

Look at the Figure.

We place a small PNP transistor at B-C of a large NPN transistor. They work like one large PNP transistor. When B of Q1 is bias negative, Q1 conducts current. It allows a higher current from E to C of Q2. Turn the circuit to the other side.

Are you beginning to see ideas clearly?

1. The current flows through R1 to 7805 regulators, it maintains a constant output voltage at 5V.
2. Meanwhile, causing the bias voltage (VB-E) of Q1, it conducts current through B of Q2.
3. This allows both Q1 and Q2 to conduct the current. Thus, a large current can flow through Q2’s C-E fully into the load depending on the gain of 2N3055.

Cheap 5V 5A voltage regulator using 7805 & 2N3055

Let’s start experimenting with the circuit assembly.

Look at the circuit below. The compete cheap 5V 5A voltage Regulator uses 7805 and 2N3055.

Can increase the current of 7805 with 2N3055 for Arduino? Read more

Interested components

The selection of devices to test is something that must be exciting. Sometimes its principle may not match the real devices. Because the quality of it may be different from the datasheet. So it’s a good idea to test it out.

• Q1: PNP transistors like 2N3906(TO-92), BD140(TO-139), BC327(TO-92), C9012(TO-92), etc.
• R1: It can be calculated from a simple formula: R1= 0.7/I_IC; Note: IIC is the Bias current of the 7805 chip. Normally use about 0.012A, but I tried 0.046A with the highest efficiency.
  R1 = 0.7V / 0.046A = 15 Ω
• C1, C2, C3: These capacitors should be installed because they are very effective when the load is high current and reduces noise signals as well.
• All resistors use 0.5 watts.

My daughter measures a load current of 2A and a voltage drop of approximately 5V as needed.

It worked well. What are the results of your experiment? Don’t forget to share some. If you have any comments or problems, don’t be considerate. Let us know quickly. we will learn together.

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