Problem 64-8
Source:
Problem elaborated by the author's of the site.
Look at the circuit shown in Figure 64-08.1. Assume that VZ1 = 12 V, VZ2 = 5 V, RL = 200 Ω,
PZ1 = 1 W, PZ2 = 0.5 W, and as input voltage we have two sources in series, where Vi = 22 V and
VS = 5 sin ω t V. On the other hand, RL and the zener Z2 are part of a lithium battery charger,
whose nominal voltage is 3.7 V. When the lithium battery reaches a voltage of 4.0 V, RC must ensure that no current flows
greater than 10 mA (charge maintenance current) by the battery. Determine the values of RS1, RS2 and RC for the circuit to function properly.
Figure 64-08.1
Solution of the Problem 64-8
Let's start solving this problem by analyzing the circuit formed by the zener Z2
and RC. According to the problem statement, RC must ensure that
a current greater than 10 mA does not flow through the battery when it reaches a voltage of
4 V. Then the value of RC will be:
RC = ( VZ2 - Vbat ) / Iman = 5 - 4 / 0.01 = 100 Ω
Therefore, 10 mA is the minimum current in the load ( ILmin ) and the maximum current can be
calculated assuming the worst case that occurs when the battery has 0 V at its terminals.
Therefore, the maximum current is ILmax = 5 - 0 / 100 = 50 mA. After these calculations, we can adopt
a current of 70 mA circulating through RS2.
Thus, when the battery has a voltage of 0 V, a current of 50 mA will flow through it and a current
of 70 - 50 = 20 mA will flow through the zener. When
the battery reaches 4 V, a current of 10 mA will flow through it and a current of 70 - 10 = 60 mA
will flow through the zener. All these values are in accordance with the characteristics of Z2,
as it supports a maximum current of 100 mA. So, the value of RS2 is
RS2 = ( VZ1 - VZ2 ) / 0.07 = 100 Ω
To calculate the value of RS1 we must take into account that the input voltage, due to
VS, varies between a minimum value and a maximum value. Let's determine these values.
Vimin = Vi - VS = 22 - 5 = 17 V
Vimax = Vi + VS = 22 + 5 = 27 V
Note that for the calculation above we used the fact that the function sine varies between - 5
and + 5, in the case of this problem.
The current passing through RL is
IRL = 12 / 200 = 60 mA
With this value we can calculate the load current for Z1, or
IL = IRL + IRS2 = 60 + 70 = 130 mA
So, we conclude that for Z1 we have a constant load current ( IL ) and
an input voltage variable. This falls under CASE 3. Therefore,
to find the value of RS2 we will use the equations eq. 64-12 and eq. 64-13.
First, we must find the minimum and maximum value of the current that can flow through the zener
Z1. Using eq. 64-1, we have:
IZ1max = 1 / 12 = 83.33 mA
And for the minimum current we will use the rule of thumb
IZ1min = 10% IZ1max = 8.33 mA
Therefore, the value of RS1min is
RS1min = 27 - 12 / ( 0.08333 + 0.13 ) = 28.13 Ω
And the value of RS1max is
RS1max = 17 - 12 / ( 0.008333 + 0.13 ) = 36.14 Ω
Thus, taking the arithmetic mean between the two values we have RS1 = 32.14 Ω.
We can choose the commercial value
RS1 = 33 Ω
Addendum
Let's check if the value of RS1 makes the zener work within its characteristics. We know that
the maximum (worst case) potential difference over
RS1 is equal to VR = 27 - 12 = 15 V. Therefore, the maximum current flowing
through this resistor is:
Imax = VR / RS1 = 15 / 33 = 454.5 mA
Then, the current that will flow through Z1 will be:
IZ1 = Imax - IL = 454.5 - 130 = 324.5 mA
In other words, as the zener supports a maximum current of 83.33 mA, we conclude that this zener is not
suitable for the project. To overcome this problem, you must choose
a zener that dissipates a power of 5 W, as this will present a maximum current of
IZ1max = 5 / 12 = 416.66 mA
Thus, we conclude that a 5 W zener will support the current of 324.5 mA.