Given the following schema:
The zener diode IV-relationship is actually a piecewise-linear approximation to a real diode.
Having the following values:
- RIN = 1 kΩ
- RL = 2 kΩ
- VI = 9.5 V
- vi = 50 mV
Apparently, the zener diode can be replaced by a 5V independent voltage source in series with a 10 \$\Omega\$ resistor. Why is this true?
What I have done
So far, I have used the Thevenin theorem for finding out in which region is the diode operating:
- RL = 2 kΩ
- RIN = 1 kΩ
- VI = 9.5 V
vi = 50 mV
\$R_{TH} = \frac{1}{\frac {1}{R_L} + \frac{1}{R_{IN}}}\$
\$V_t = V_I + v_i\$
\$I_t =\frac {V_t} {R_{IN} + R_L}\$
\$V_{IN} = I_t * R_{IN}\$
\$V_{TH} = V_t - V_{IN}\$
\$print(V_{TH})\$
Result: 6.36666
So, the diode is working in the region that is left to the -5
volts in the iv-characteristics, because its polarity is inverted in the circuit. I get that.
What I don't understand is how can it be inferred that the diode is actually a 5 V voltage source in series with a 10 \$\Omega\$ resistor. Why is the diode augmenting the voltage in 5 V? It is operating in that zone, true, but I don't see why the voltage source is necessary.
In addition, I don't see why the resistor is specifically set to 10 \$\Omega\$ .
Finally, it seems that the resultant voltage source would have the polarity as \$V_I\$ and \$v_i\$, that is, the +
symbol near to \$R_{IN}\$ and the -
symbol near ground. I don't understand this either, since the diode's polarity is inverted.