Example –1 (Design)

Design an 11.3V regulated power supply shown in fig. 1 for a load current that varies between 400 mA and 500 mA. Assume an input of 120 V rms at 60 HZ into a 3:1 center - tapped transformer. Use a 12 V Zener with R_Z = 2Ω. The transistor has V_BE = 0.7 V and β = 100. Set C so Δv_s= 30%.

Solution:

The design consists of choosing the values for R_sand for C_F. First V_{S max} is obtained multiplying the rms voltage by √2.

V_{S max} = √2 x 120 = 170 V

The transformer output on either side of the center tap is one-sixth of the input, so V_{S max}is 28.3V.

Since Δv_s= 30%,
Therefore,          V_{S min}= (0.7)(V_{S max}) = 30%
Given that         I_{Load max} = 500 mA
and                      I_{Load min}= 400 mA
                         V_Z= 12 V

Notice that the transistor keeps the value of I_{Z max} quite small since β appears in the denominator.

The source resistance R_S can be calculated from equation (E-1) of previous lecture,

Note that

The capacitor size is estimated from equation (E-5) of previous lecture:

Equation (E-7) can be used to evaluate the percent of regulation at the load.

Example – 2:

For the circuit of fig. 2, determine the following:

1. Nominal output voltage
2. Value of R₁
3. Load current range
4. Maximum transistor power dissipation
5. Value of R_S and its power dissipation

The relevant information is as follows:

V_i = constant 8 V; D: 6.3 V, 200 mW; requires 5mA minimum current

Q: V_EB= 0.2 V,         h_FE = 49,             I_CBO≈ 0

Fig. 2

Solution:

a. The nominal output voltage is the sum of the transistor's V_EB and the Zener voltage:

V O = 0.2 + 6.3 = 6.5 V

b. R₁ must supply 5mA to the Zener:

c. The maximum allowable Zener current is

P / V =0.2 / 6.3 = 31.8 mA

The load current range is the difference between minimum and maximum current through the shunt path provided by the transistor. At junction A, we can write

I_B = I_Z – I₁;
I₁ is a constant 5 mA; therefore

I_B =I_Z – I₁ = (5 x 10^-3 ) – (5 x 10^-3) = 0
I_B =I_Z – I₁ = (31.8 x 10^-3) – (5 x 10^-3) = 26.8 mA

The transistor's emitter current is (h_FE + 1) (I_B). I_B ranges from a minimum of around zero to a maximum of 26.8 mA; therefore the load current range is (h_FE + 1) (26.8 x 10^-3) = 1.34 A. The Zener alone could provide a maximum range of 26.8mA.

d. The maximum transistor power dissipation occurs when the current is maximum. Using I_E ≈ I_C, we have

P_D = V_O I_Z = 6.5 (1.34) =8.7 W

e. R_S must pass 1.34 A to supply current to the transistor and R_L:

The power dissipated by R_S is

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