Lecture - 29: Voltage Regulator

Example –1

The following data apply to a voltage regulator circuit shown in fig. 1
Unregulated input: Vi = 15 to 20V
Desired output: VO = 10V
Load current requirements: IO = 0 to 0.1 A
Zener diode voltage = 6.8V
hFE min = 15
hFE max = 50

Both transistors are the silicon planer type with negligible ICBO. Determine

  1. Extreme values of RL
  2. Maximum value of IB1
  3. Suitable value for R1
  4. Extreme values of I1
  5. Value of R2 if D1 requires a minimum of 5 mA
  6. Worst-case power dissipation for Q1 under normal operating conditions
  7. Worst-case power dissipation for Q2 under normal operating conditions

Fig. 1

Solution

(a). Since RL = VO / IO, we can write

The maximum load current is zero; therefore the load resistance is anywhere from 100Ω to an open circuit.

(b). The base current of Q1 is IE1 (hFE1 + 1), where IE1 is the sum of load current IO, current through the sampling network, and current through R2. Usually IO is the largest component involved; therefore

(c). R1 should supply the maximum current required by the base of Q1. This is 6.66 mA; however, it is a good practice to increase it by about 50 percent to provide a safety margin. We therefore select R1 on the basis that its current must never drop below 10 mA:

(d). The extreme values of I1 are determined using the extreme values of Vi:

(e).        

(f). Worst-case power dissipation for Q1 occurs when both the voltage and current are maximum. Normally the collector-base junction dissipates the most power, since its voltage is much greater than VBE; however, the total transistor dissipation involves both junctions. This can be approximated as follows:

Voltage across Q1: VCE1 = Vi – VO
Current through Q1: IC1 ≈ IE1 ≈ IO
PE1 ≈ (Vi – VO) (IO)
PD1≈ (20 – 10) (0.1) = 1 W

(g). Q2 does not handle large currents as Q1 does; therefore its power dissipation is relatively low:

PD2 = (VCE2) (IC2)

Both VA and Vr are relatively constant; therefore VCE2 is also constant:

VCE2 = VA – Vr = VO + VEB1 – Vr
VCE2 = 10 + 0.6 – 6.8 = 3.8 V

The maximum value of IC2 is approximately equal to I1 = 21.4 mA; we, therefore, have

PD2 ≈ VCE2 IC2
PD2 ≈ 4(21.4 x 10-3) = 81.5 mW

Note that the current through R1 in this example does not remain constant; in fact, it can vary between 10 and 21.4 mA, a total range of 11.4 mA. The regulator would perform much better if I1 were regulated.

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