Question 17.3: Design a voltage reference of 3.6 V capable of driving a loa......
Design a voltage reference of 3.6 V capable of driving a load of 200 Ω. The reference voltage is to be produced from a supply voltage that can vary between 4.5 and 5.5 V.
A suitable circuit would be as shown here.
Clearly a Zener diode with a breakdown voltage V_{Z} of 3.6 V is used, but calculations must be performed to determine the value required for the resistor R and the power ratings required for the resistor and diode.
For power dissipation considerations, we would like R to be as high as possible. The maximum value of R is determined by the requirement that the voltage drop across R must not take the voltage at the output (and across the diode) below the required output voltage. This condition is most critical when the input voltage is at its lowest, so the value of R is determined by calculating the value for which the output would be equal to the required value (3.6 V) when the input voltage is at its lowest permissible value (4.5 V). We can ignore the effects of the Zener diode at this point as all the current flowing through R will be flowing into the load.
The current flowing in the load I_{L} can be calculated from
I_{L} = \frac{V_{Z}}{R_{L}} = \frac{3.6 V}{200 Ω} = 18 mAThe voltage drop across the resistance R caused by I_{L} must be less than the difference between the minimum supply voltage (4.5 V) and the Zener voltage (3.6 V), therefore
I_{L}R \lt 4.5 – 3.6 Vand
R \lt \frac {4.5 – 3.6 V}{I_{L}} \lt \frac{0.9 V}{18 mA} \lt 50 ΩSo a standard value of 47 Ω would probably be chosen.
Maximum power dissipation in the various components is generated when the input voltage is at its maximum value (5.5 V). As the output voltage is fixed, the voltage across the resistance is easy to calculate – it is simply V – 3.6 V. Therefore, the maximum power dissipation in the resistance is simply
P_{R(max)} = \frac{V^{2}}{R} = \frac{(5.5 – 3.6)^{2}}{47} W = 77 mWThe power dissipated in the Zener diode P_{Z} is also simple to calculate. The voltage across it is fixed (3.6 V) and the current through it is simply the current through the resistance I_{R} minus the current taken by the load I_{L}. This is also at a maximum when the supply voltage has its maximum value.
Thus
P_{Z(max)} = V_{Z}I_{Z(max)} = V_{Z}(I_{R(max)} – I_{L}) = 3.6 \left(\frac{5.5 – 3.6}{47} – 0.018 \right) = 81 mWThus, the Zener diode must have a power rating of greater than 81 mW.
