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This is the second part of my Linear Regulators Series, so read part 1 first if you haven't already http://everycircuit.com/circuit/5551163267874816
Now let's talk about this circuit and how it's different from the previous one. Firstly, you'll have noticed that it has a lot more BJTs in it compared to the previous one. And second - this one's not so obvious - it has current limiting, an essential feature of any voltage regulator.
Let's discuss the improvement in the series pass transistor first. It is now a Darlington Pair, as you may have noticed. A Darlington Pair is advantageous in that the op amp is required to supply only a small amount of current to the driver transistor (the NPN closer to the op amp), and the driver transistor intern helps provide an appropriate base current to the main series pass transistor (the NPN further away from the op amp). The main or the series pass transistor is typically a Power BJT. A Power BJT or a Power Transistor in general is a type of transistor that is designed for heavy duty work and is capable of handling very high power. It is different form the driver transistor, which is typically a small signal transistor. Small signal transistors are rated for low current/power applications and are suited for this purpose.
To fully understand the need for a power transistor, let us see how much power is dissipated in the series pass transistor. The formula for power, as we all know, is
P = V * I ...(1)
We need to calculate the power dissipated by the series pass transistor. The voltage appearing across it is equal to the difference between the input voltage and the output voltage, as the component lies between the input and output, and the current through it is equal to the load current. Therefore
Pdiss = (Vin - Vout) * Iout ...(2)
putting Vin = 15V, Vout = 5V and Iout = 500mA, we get Pdiss = 5W (This is true under normal operation only, as we shall see later). A small signal transistor in a TO-92 package is typically rated 0.5W, which is a rating 10 times lesser than what is required. Also, a TO-92 is not heatsink mountable, unlike a TO-220 package. Heatsinking is mandatory in order to keep the series pass transistor cool. A transistor rated 10W or more is a good pick for this particular design. The driver transistor can be a small signal transistor like the 2n3904.
Now it's time to understand the need for the 4.7k resistor and 470R resistor that are placed in parallel to the base-emitter terminals of the two BJTs. All BJTs, as some of you may know, have some base-emitter parasitic capacitance. Due to this capacitance, some amount of charge is stored in the base-emitter junction of the transistor. The stored charges slows down the turnoff time of these BJTs, because of which bleeder resistors are used to help dissipate these charges and turn off the transistors quickly. Although they are not as critical in such linear applications as they are in switching applications, it is still recommended to include gate bleeder resistors in your design.
It's now time to discuss the current limit functionality of the regulator. As I already said, it is an important feature that must be present in all voltage regulators and power supplies. It is what limits the output current in the event of a short circuit and prevents the load from going into flames and prevents damage to the regulator and input voltage source by limiting the short circuit current. Try reducing the load resistance to 1ohm or 0.1ohm and see what happens.
The current limiting circuitry in this design is a very simple and rudimentary one. The lowest BJT in association with the 1ohm resistor helps limit the short circuit current. It does so by the following mechanism. Notice the location of the 1ohm current sense resistor. It is parallel to the base-emitter terminals of the auxiliary BJT (the lowest BJT). The load current, which passes through the 1ohm resistor since it is in series to the load creates a voltage drop in the resistor. Now as long as the voltage drop is less than 0.7V or thereabouts, the auxiliary transistor stays off. This is during normal operation. Now let's say that, for some reason, the load draws an overcurrent. The regulator obviously tries to supply that overcurrent in order to maintain the output voltage at the same level. The results in the drop across the 1ohm resistor to increase and that turns the auxiliary transistor on. This transistor redirects the current that is supposed to go to the Darlington pair through it's collector-emitter path. This results in an insufficient current flowing into the base of the Darlington pair and results in a decrease in the output voltage. This decrease in the output voltage is the reason for the output overcurrent to stay constant and not exceed a predetermined limit. This limit is easily calculated using Ohm's Law. The drop across the 1ohm resistor must be enough to turn on the auxiliary BJT in order to activate current limitation, which means that the value of overcurrent to do so can be calculated using
Isc = Vbe/Rse ...(3)
where Isc stands for short circuit current (or overcurrent), Vbe stands for base-emitter voltage and Rse stands for Current Sense Resistor. Since Rsc = 1ohm and Vbe = 0.75V, Isc turns out to be 750mA or thereabouts. I say thereabouts because, as I mentioned earlier, this is a very rudimentary implementation of this feature and is not very precise. Also, it has a very high temperature dependence. But the reason why it is used anyways is because of the fact that it is a very, very simple design.
Equation 3 can be re-written as
Rse = Vbe/Isc ...(4)
The reason why it is better to write it in this manner is because it is the resistance that has to be determined using Vbe and Isc, which are known quantities.
Let us revisit equation 2, which tells us how much power is dissipated in the series pass transistor. In the event of a dead short, the output voltage drops to zero, which means the power dissipation in the event of a short circuit becomes
Pdiss(sc) = Vin/Isc ...(5)
which is a much higher value. Putting the appropriate values in 5, Pdiss(sc) turns out to be 11.25W. Since a short circuit is possible, this calculation must be considered while choosing a transistor and heatsink. Leaving a large headroom is, as always, very important. So I would choose a BJT rated for at least 20W in this design. I wont go into heatsink calculations in this post, as thermal design is out of the scope of this post.
I hope you made it to the end of this post. I know, it's a very long read, but my objective is to share all my knowledge on Linear Regulators (or a big chunk of it, at least) in this series of posts.
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