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zenterix
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- Homework Statement
- A voltage source can be represented as the series connection of an ideal DC voltage source and a resistance ##R_{IN}=\mathrm{1k\Omega}##, as illustrated in figure 1a.
The model of the voltage source also includes a small signal voltage source ##v_i## to represent the noise generated by the source.
Assume ##V_I=\mathrm{10V}## and ##v_i=50\mathrm{mV}##.
In practice, connecting a non-ideal voltage source to a load resistance ##R_L##, as in figure 1b, may result in undesirable effects. This problem studies such effects and how to correct them by introducing a Zener diode into the circuit (figure 2a).
(a) In figure 1b, calculate ##v_o## (ie, output noise) and ##V_O## (ie DC output voltage) for ##R_L=\mathrm{2k\Omega}## and ##R_L=4\mathrm{k\Omega}##. What can you say about ##v_o## and ##V_O## as a function of ##R_L##?
(b) Repeat part (a) for the voltage source in figure 2a. In this setting, how do ##v_o## and ##V_O## change with ##R_L##?
(c) In figure 2a, what is the minimum value of ##R_L## that would guarantee that the circuit operates as in part (b)?
- Relevant Equations
- ##V+RI##
Part (a)
The circuit in figure 1b is linear. It is a simple voltage divider circuit.
The relationship between a voltage source ##V_I## and output voltage ##V_O## is
$$V_O=\frac{R_LR_{IN}}{R_L+R_{IN}}V_I$$
This relationship is true individually and independently for the DC voltage source and the small signal voltage.
If we consider ##V_O## as a function of ##R_L##, we see it is a nonlinear relationship.
For ##R_L=2\mathrm{k\Omega}## we have
$$V_O=\frac{2}{3}V_I$$
and for ##R_L=4\mathrm{k\Omega}##
$$V_O=\frac{4}{5}V_I$$
For these two values of ##R_L## and given the values ##V_I=\mathrm{10V}## and ##v_i=50\mathrm{mV}## we have, respectively, that
$$V_O=\frac{2}{3}\cdot 10\mathrm{V}$$
$$v_o=\frac{2}{3}\cdot 0.050\mathrm{V}$$
and
$$V_O=\frac{4}{5}\cdot\mathrm{10V}$$
$$v_o=\frac{4}{5}\cdot\mathrm{0.050V}$$
Part (b)
Let's consider first ##R_L=\mathrm{2k\Omega}##.
Let's consider the case in which ##\mathrm{-5V\leq V_D\leq 0.6V}##.
Then no current passes through the Zener diode and the circuit becomes the same as in part (a).
The range of input voltage ##V_I## that puts us in this range for ##V_D## is ##\mathrm{-0.9V\leq V_I\leq 7.5V}##.
Next, let's consider the case in which ##\mathrm{V_D>0.6V}##.
When this happens, we can see from figure 2b that the relationship between ##i_D## and ##v_D## is
$$i_D=\mathrm{-600mA}+10^3\mathrm{\frac{mA}{V}}\cdot v_D$$
$$=\mathrm{-0.6A+1\frac{A}{V}}v_D$$
Using the node method on the red node above we have
$$\frac{V_I+v_D}{R_{IN}}+(-0.6+v_D)-\frac{(-v_D)}{R_L}=0$$
After some algebra we reach
$$v_D=-V_O=\frac{0.6R_{IN}-V_I}{\left ( 1+\frac{R_{IN}}{R_L}+R_{IN} \right )}$$
If we now input the given values of the variables we get
For ##R_L=\mathrm{2k\Omega}##
$$v_D=\frac{600-V_I}{1001.5}$$
Note that to achieve ##v_D>0.6## we need to have ##V_I<-0.9\mathrm{V}##.
Thus,
$$V_O=-v_D=-\frac{600-10}{1001.5}$$
$$=-\frac{590}{1001.5}$$
and
$$v_o=-v_D=-\frac{600-0.050}{1001.5}=-\frac{599.950}{1001.5}$$
Let's quickly do the calculations for ##R_L=\mathrm{4k\Omega}##.
For ##\mathrm{-5V}\leq v_D\leq \mathrm{0.6V}## the input voltage is ##\mathrm{-0.75V}\leq V_I\leq \mathrm{6.25V}##.
For ##v_D>\mathrm{0.6V}## the input voltage is below ##\mathrm{-0.75V}##.
The relationship between ##v_D## and ##V_I## is
$$v_D=\frac{600-V_I}{1001.25}$$
which is not much different from the relationship when ##R_L=\mathrm{2k\Omega}##.
At this point my impression is that the presence of the Zener diode makes the relationship between input and output voltages less sensitive to ##R_L##.
I won't do the calculations now for the case in which ##v_D<\mathrm{-5V}##. Perhaps in a next post so this one isn't too long. I expect a similar conclusion.
However, I don't quite understand part (c).
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