Would you like to learn about the working of the power supply circuit? I would love to share it. Do you the same interests as me?  I will be telling you about the transistor series voltage regulator with short circuit protection.

I believe that you understand the basics. You must read the relevant content shared previously.

How do filters ripple work in Transistor Series Voltage Regulator?

Imagine we have available 20 volts DC unregulated power supply sources but our load requires 12V DC regulated voltage.
So we must use a transistors series voltage regulator. See the block diagram.

We know that there must not be any ripple voltage on the power supply.

From previous articles, we have found that if any ripple appears on the 20V input, it will cause a rise in the input voltage of about 20.1V or 20.4V.

This increased voltage does not affect the Zener diode because it has a fixed breakdown voltage of 12V. Hence the Zener will be our base reference voltage.

Here the error transistor-Q2 will act as the sensor to this increased voltage.

How it works?

I feel upset because there is no easy way to explain it to you but it’s still worth a try. Here is the step by step process. Observe the circuit diagram above.

Suppose that 0.1V or 0.4V ripple passes through Q1. The output voltage is 12.1V to 12.4V because of 12.0V + 0.1V or 12.0V + 0.4V.

After that, R2-potentiometer will sense the output voltage and it will be set to partially turn on Q2. This will make the circuit give the original balance output of 12.0V.

The output appears at 0.1V and produces a bigger ripple. The voltage on the collector will reduce and this will turn Q2 very little. The emitter will follow the base voltage but it is about 0.6V lesser.

This works very rapidly and can cause a ripple of fairly high frequency. So it will reduce the smoothing output ripple lesser.

If there are some ripples like 2V p-p they may improve it to 20mV at full load. It sounds quite good, right? What is more to learn?

If we make a new circuit, it will look similar to an emitter-follower form. Look at the diagram below:

The emitter of Q1 will always be about 0.6V less than the base.

In this circuit layout, it is clear to know how Q2 runs like a variable resistor between collector and emitter. That way it provides a base voltage for Q1.

When Q2 acts like a low-value resistor, it causes the base of Q1 to connect to the cathode of the 12V Zener.

It then provides an output voltage of 12V-0.6V = 11.4V. When Q2 has a high resistance, the base of Q1 gets connected to the 20V input and the output becomes 19.4V.

Practically, Q2 does not run this wide a range.

This circuit has one small disadvantage. It does not provide overload protection or short-circuit protection.

In the next circuit, we have tried to fulfill these conditions.

Series Regulator with short-circuit protection

This circuit provides short-circuit protection. If the output is connected to ground, the regulator will shut down leaving R4 as the only source of supply.

In this state, there will be no current flowing through R1, R2, and R3. So, there will be no voltage across them and no current goes to the base of Q3 transistor. It does not conduct any current. And finally, Q2 and Q1 do not run too.

For this reason, we must use R4 with a 5 watts wire-wound resistor. Though it does not supply any current in the power supply, it still operates smoothly.

Another time when the R4 is needed for starting up, it must provide about 1V into the output to start-up the circuit.

The set output pot (R2) detects about 50% of the output voltage to begin so that it can turn on Q3.

This will turn on the power driver transistor Q2 and it will also turn on the power regulator transistor Q1.

This condition increases and stabilizes the output at 12V and load current to about 100mA.

Next, when the load increases to 1Amp, the output voltage will reduce to 11.9V. This effectively increases the base-emitter voltage of Q1 to 0.7V so that it is turned on with great force.

The input voltage will reduce to 2V or so, but the base voltage Q1 will remain stable at 12.6V.

The parts you will need

• Q1: 2N3055, 100V 15A, NPN transistor
• Q2,Q3: BC548 or equivalent, 45V 100mA NPN Transistor
• R1,R5: 1K, 0.5W Resistors
• R2: 3.3K to 5K POT(potentiometer)
• R3: 6.8K, 0.5W Resistors
• R4: 1K 5watts Resistors
• R6: 18K 0.5W Resistors

The two circuits explained above have advantages of about two-third of a good power supply.

They provide a smooth and regulating circuit and the 2nd circuit also provides the protection from short-circuit.

Let’s take a look at the 3rd circuit. It has an important feature to include a power supply for overload protection.

Now, we will learn the feature which limits the maximum current rating of the power transformer and the power transistor.

In a short time, the output will be maximum of up to 10 times the normal current. This will cause overheating and damage to many circuit components.

The overload protection is always provided by power supplies that have more than 1 Amp current. Not only that it prevents possible fire risk but it also reduces further damage to the equipment being supplied.

We may put them into two different forms.

• Fuse and circuit breaker
The simplest way to deal with it is using a circuit breaker or fuse (very cheap) in the output. This will blow up when the current rise is approx 30% over the recommended maximum.

When it starts to run, the circuit will not reset itself. We need a manual reset or we will have to replace the fuse.

The only disadvantage here is the inconvenience of physically changing the fuse.

If we use an electronic overload in some other way, it would be better.

To protect a series regulator from excessive current overload we can add the 3 components as shown:

The 1 ohm resistor is connected in series with the output so that all the current will flow through it. As the current increases, a voltage change will occur across the resistor as described by ohm’s law.

Turning the circuit around will make the overload protection easy to understand. It is described  as follows:

• The output current is 100mA and the voltage is 11.99V.
• The voltage at the base of the transistor is 12.6V.
• And the Emitter voltage is 12V.

• Look at the A-B point. The most important point to note here is that a diode is connected  in a forward bias situation so it will drop a maximum of 0.6V.

Do you understand? You must go through one more example as it will help you more to get a better idea.

Let’s increase the current to 500mA.

It looks like the previous circuit. The output voltage is lower than the 200mA load. It is 11.5V.

The voltage drop across the resistor of 1ohm changes.

VR1 = I x R
= 0.5A x 1 ohms
= 0.5V

Look in the circuit diagram. The output current here increases at a slightly reduced voltage.

At 500mA the voltage between points A and B increases to 12.6-11.5=1.1 volts.

This voltage is still below 1.2 volts and thus the diodes do not affect this current flow.

550mA Current to lower output voltage

When we use 550mA current of the load, the voltage levels obtained are shown in the diagram below.

The transistor turns on with a bit of force to deliver the current and this will produce a voltage difference of 12.7-11.45 = 1.25 volts.

Now, these diodes act to limit the increasing of current which will lower the output automatically. You see the diodes are working as the voltage across them does not exceed over 1.2 volts.

The voltage difference between two points (A-B) can never exceed from 1.2 volts because of the clamping effect of the two diodes (D1, D2).

Any increase in current will cause the lowering of output voltage (Due to the voltage drop across the 1ohm resistor). This voltage will go right back to the base via the diodes. This makes the  transistor turn off. This will give a lower output voltage.

The net result will reduce the output voltage so the maximum current passing will not exceed 550mA.

As you already know the ohms law which says that a lower voltage applied to the load will result in a lesser current flow.

Short-circuit protection using Diodes

Let’s learn the case of a short-circuit.

Do you know the main disadvantage of overload protection? Yes, it’s quite clear. Some voltage drop is needed to operate the sensing of the circuit. From no-load to full-load, voltage must be reduced to 0.5V.

Look at the circuit. The point A has dropped down to 1.2V and the emitter will appear at 0.6V. At point B the voltage is 0V. The current in the short-circuit will be 550mA.

We can see that this is very simple but its not stable enough for a few devices.

For example, a TV set. Some sets require a supply voltage of only 11 volts. Even a change of 0.2 volts, will cause a lack of width or height. So, imagine the effect of 0.5 volts. How will it affect?

When the brightness level varies, this circuit will require the change in current and change in voltage too.

How can we improve it?

Have you ever heard about a current limiting circuit?
Yes, I have shared about it with you.

We can place this type of a circuit on the input side of the pass transistor. Here it will have very  little effect on the output voltage.

Look at the following circuit which shows this setup:

This is 13.8V 2A voltage regulator circuit using transistor

• Q1 is the series pass transistor.
• Q2 is a sensing amplifier
• Q3 is the overload detector.

Even if the circuit looks simple, understanding it might be a challenge.

First, the input passes to R1 and then to Zener diode, reference voltage. After that the current bleeds (see the previous presentation for this).

Then, The Q2 works to amplify the voltage. It detects the change on its base to supply Q1 with the required current. Q1 will need at least 20mA base current or even upto 50mA.

This makes the Zener diode to have a lower bleed current.

Under normal circumstances, Q3 turns off and plays no part in the operation of the circuit.

Look at basic circuit zoom at the Q3 circuit.

The four 1ohm resistors in parallel combine to form a 0.25 ohms resistor.

And when the current reaches 2 amps, a voltage of 0.25 x 2 = 0.5 volts will develop across the combination.

This will start to turn on Q3 and if the current increases further Q3 will turn on completely. It will short out most of the Zener diode voltage.

Then, the base of Q2 will detect a voltage as low as 3 to 4 volts. The emitter of Q2 will follow this voltage drop with a voltage on its emitter of 2.4 to 3.4 volts. The output of Q1 will reduce to 1.8 to 2.6 volts.

If this low voltage mode keeps it in a short-circuit condition, the circuit will remain in this shut-down mode. This will in turn protect the equipment.

Of course, this circuit is not perfect. We can improve it by learning more about it.

Parts you will need

0.25W Resistors, tolerance: 5%

• R1: 470 ohms
• R2: 150 ohms
• R3 to R6: 1 ohm 1 watt. Resistors

Q1: 2N3055, 100V 15A, NPN transistor
Q1-Q2: BC5487 or equivalent, 45V 100mA NPN Transistor
ZD1: 15V 1W Zener Diode

Power-Loss in a series regulator

In the series pass transistor regulator, the Q1 power transistor Suffers a lot. We can help it to make it better. This involves much more detail. If you are interested in it, read more.

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