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This circuit is the easiest, simplest and a very accurate voltage regulator circuit. I decided to share this circuit even though I am making a seperate linear regulator series because this design is extremely simple and anybody can replicate it.
The MOSFET you see is modelled to behave like a TL431, which is a very commonly used IC, and this is one of its applications.
This is the configuration of the MOSFET:
Width - 1mm
Length - 1nm
Kp - 10mA/V2
VTO - 2.5V
Lambda - 60m1/V (default)
This is the configuration of the BJT:
Saturation Current - 1fA (default)
Forward Beta - 150A/A
Reverse Beta - 1A/A (default)
Collector Resistance - 100mohm
Base Resistance - 1ohm (default)
Emitter Resistance - 100mohm
This is how to calculate the feedback resistors:
Let Vo be the desired output voltage.
Let the upper feedback resistor be R2
Let the lower feedback resistor be R1
Then, the output voltage is given by:
Vo = 2.5(1 + R2/R1) ...(1)
So for this circuit, I wanted Vo = 5V. A good rule of thumb to calculate the feedback resistors is to choose a high R1, like 10Kohm.
Therefore,
5 = 2.5(1 + R2/10k) ...(2)
On solving for R2, we get R2 = 10k. Therefore, to get an output of 5V, use two 10k feedback resistors. It's is also a very good idea to use a potentiometer, such as a 10k pot, as you may not necessarily get standard resistor values of R1 and R2.
How to calculate the base resistor value:
*A simple rule of thumb is to choose a value of resistance that allows 1-2mA to flow through the TL431 when maximum load is attached. In other words, after establishing the feedback resistors and attaching the maximum load, vary the base resistance until 1-2mA of current flows through the TL431.*
Here is the mathematical procedure-
Let Vb be the voltage at the base of the BJT
Let Ve be the voltage at the emitter of the BJT
Vb = Ve + 0.7 (approx.)
But Ve = Vo, the output voltage. So
Vb = Vo + 0.7 (approx.) ...(3)
Since (for this circuit), Vo = 5V, then Vb = 5.7V.
The TL431 is rated for a current between 1mA - 100mA. Let the minimum TL431 current Ik be 1.5mA. Therefore,
Ik = 1.5mA
Let the base current of the BJT be Ib. We know that
Ib = Ic/beta ...(4)
We also know that Ic, the collector current, is approximately equal to Io, the output current. If the maximum output current is to be 500mA and Beta is 150, then
Ib = Io/beta = 0.5/150 = 3.3mA
Therefore, the total current to be supplied by Rb is
Irb = Ib + Ik ...(5)
Therefore, Irb = 3.3mA + 1.5mA = 4.8mA. Let's round it to 5mA.
To calculate Rb, we need to apply ohms law.
Rb = Vrb/Irb ...(6)
Where Vrb is the voltage across Rb and Irb is the current through Rb.
The voltage across Rb is the difference between the minimum input voltage and Vb.
Vrb = Vin(min) - Vb ...(7)
We know that Vb is 5.7V and the minimum input voltage is 10V, so Vrb turns out to be 4.3V. And since Irb = 5mA, Rb turns out to be
Rb = 4.3V/5mA = 860ohms
Let's round it of to 820ohms.
Thus, all component values have been calculated. Now the last thing to do it's to calculate the power dissipation of the BJT. The power dissipation is given by
Pd = (Vin(max) - Vo) * Io(max) ...(8)
Since my input is a constant, Vin(max) is also 10V. And Io(max) is 500mA. Therefore, Pd turns out to be 2.5W.
For more details on linear regulators, checkout my Understanding Linear Regulators series http://everycircuit.com/circuit/5551163267874816
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