Question

im having trouble with exercise 2.7. It requires info from 2.6 so I have posted 2.6 and the solution to it. Thank you!

Exercise 2.7. Design a +10 V supply with the same specifica- tions as in Exercise 2.6. Use a zener and emitter follower. Cal- culate worst-case dissipation in transistor and zener. What is the percentage change in zener current from the no-load condition to full load? Compare with your previous circuit.

Answer 1

First we revisit the problem 2.6 whose analysis is as follows :

Note that the Zener diode works in reverse bias during voltage regulation. Minimum Zener diode current to operate in breakdown region maximum Zener diode current to ensure safe operation clearly I_1 = I_2 + I_L therefore (V_in-V_out)/ R = I-1 V_out = I_L. R_L I_2 = I_1- I-L given V_out = + 10 V V_in [20,25](I_1)_min = (I_l)_max + IZ min = 100 MA + 10 mA = 110 mA I_1 = (V_in) - (V_out)/ R I_1_min = 20-10 greater than or equal to 110/R R less than or equal to 90.91 ohm. To analyses f R essay 100 then I_1 would be less than 110 MA 10 MA of which will be consumes by the Zener. Hence load will get less than 100 ma of current which is less than the maximum load current that needs to be delivered. Now for this value of R, we need to find I_z max I_z max = (V_in)_max - V_out /R = 25-10/90.91 = 0.165 A. power rating of the Zener diode must be able to withstand 165 MA of current.

To obtain better voltage regulation in shunt regulator, the zener diode can be connected to the base circuit of a power transistor

as shown in the figure :

This configuration reduces the current flow in the diode. The power transistor used in this configuration is known as pass transistor. The purpose of C_L is to ensure that the variations in one of the regulated power supply loads will not be fed to other loads. That is, the capacitor effectively shorts out high-frequency variations.

Because of the current amplifying property of the transistor, the current in the zener diode is small. Hence there is little voltage drop across the diode resistance, and the zener approximates an ideal constant voltage source.

The current through resistor R is the sum of zener current I_Z and the transistor base current I_B ( = I_L / β ).

I_L = I_Z + I_B

The output voltage across R_L resistance is given by

V_O = V_Z – V_BE

Where V_BE » 0.7 V

Therefore, V_O= constant.

The emitter current is same as load current. The current I_R is assumed to be constant for a given supply voltage. Therefore, if I_L increases, it needs more base currents, to increase base current I_z decreases. The difference in this regulator with zener regulator is that in later case the zener current decreases (increase) by same amount by which the load current increases (decreases). Thus the current range is less, while in the shunt regulators, if I_L increases by ΔI_L then I_B should increase by ΔI_L / β or I_Z should decrease by ΔI_L / β. Therefore the current range control is more for the same rating zener.

NOTE : HERE SOURCE RESISTANCE IS NOT MENTIONED IN THE QUESTION , SO WE NEGLECT IT.

The simplified circuit of the shunt regulator is shown in the above figure .

Now we analyse the modified scheme :