What does a Zener diode do in a circuit with non-ideal voltage source?

AI Thread Summary
The discussion centers on the role of a Zener diode in a circuit with a non-ideal voltage source. It highlights that the circuit behaves as a linear voltage divider, with output voltage dependent on load resistance (R_L) and input voltage (V_I). When the Zener diode is inactive (V_D ≤ 0.6V), the circuit functions like a standard voltage divider; however, when V_D exceeds 0.6V, the relationship between input and output voltages becomes less sensitive to R_L. The calculations provided demonstrate how varying R_L affects the output voltage, indicating that the Zener diode stabilizes the output under certain conditions. The discussion also expresses confusion regarding the problem set's questions, particularly about the minimum load resistance and the Zener diode's function.
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##
1706856399956.png

1706856415761.png


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

$$i_D=\mathrm{-600mA}+10^3\mathrm{\frac{mA}{V}}\cdot v_D$$

$$=\mathrm{-0.6A+1\frac{A}{V}}v_D$$

1706857527772.png

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).
 
Last edited:
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Wow, that's a confusing set of questions, IMO. Part (a) refers to circuit (b), and part (b) refers to circuit (a). And I don't see what part (c) is asking either, at least not yet.

Are you sure you transcribed the question correctly? And asking for a minimum load resistance seems to be implying that they want the output voltage to be clamped by the Zener diode and not by ##R_L##, but I'm not sure. Is that the way you are reading it now? If so, that is not too hard to calculate...
 
@berkeman I think I did transcribe correctly. The problem is the third on this problem set from MIT’s 6.002 Circuits and Electronics.

As for what they are asking for in (c), I really am not sure. I don't have a good understanding of what this Zener diode is doing.
 
zenterix said:
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.

You have a typo in ##V_O=\dfrac{R_LR_{IN}}{R_L+R_{IN}}V_I ##

It should be : ##\displaystyle V_O=\dfrac{R_L}{R_L+R_{IN}}V_I ##
 
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