I have thought about the below question a lot but i can't understand how to go ahead. The question appears to be wrong to me. Any help or even a comment will be appreciated thank you.

A Zener regulator has an input voltage in the range 15V to 20V and a load current in the range of 5 to 20 mA . If the Zener voltage is 6.8V, the value of the series resistor should be ?enter image description here

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    \$\begingroup\$ If your load is taking 20 mA and the zener diode were removed (thought experiment) what value of Rs serves to keep Vload at 6.8 volts? \$\endgroup\$ – Andy aka May 14 '20 at 9:10
  • \$\begingroup\$ @Andyaka Given the standard zener current of \$37\:\text{mA}\$ (see: zeners), R should be about \$220\:\Omega\$. The \$\Delta V_\text{out}\$ will be horrible. \$\endgroup\$ – jonk May 14 '20 at 17:00


It is hard to know what you really want out of this question. The specifications speak to me in a certain way. But I don't know how you are receiving them. And, perhaps, if you could read my mind about this you might have wanted to re-phrase the question, asking for a different outcome. All I can do is interpret what I see in the light of what I imagine I understand.

That said, here goes:

The question specifies two supply rail voltages, \$15\:\text{V}\$ and \$20\:\text{V}\$. It also specifies two load currents: \$5\:\text{mA}\$ and \$20\:\text{mA}\$. Finally, it specifies the zener voltage: \$6.8\:\text{V}\$.

Hidden behind the scenes, and unstated but still present, is the required zener current. For that, see: zeners. There, for the \$6.8\:\text{V}\$ case, the test current is \$37\:\text{mA}\$ and the zener impedance at that test current is \$3.5\:\Omega\$.

Now, I face a problem. If the zener is assumed to be perfect and you are not supposed to go look at a datasheet, then there really isn't a good answer. This is because I could set \$R=1\:\text{m}\Omega\$ and dump torrents of current into the zener. In doing so, the load itself would hardly have any impact at all. So you'd have solved the problem. And even better might be \$R=1\:\mu\Omega\$. So what's the limit here? I mean, seriously? Does this make any sense? It doesn't.

So, the only reasonable way to analyze this is to include a zener model that you have to go fetch (as above) and apply it, intelligently. Idealistic approaches just don't make any sense. Yet the problem doesn't say, specifically.

This is why I started out saying that I'm not sure you read this problem the way I do. And, if you did, you might have asked it, differently.

An Approach

That said, you can approach the problem if and only if you don't assume an ideal zener but instead assume a more realistic one. In particular, I'll use the model we can extract from above. Let's first look at the revised (model-based) schematic(s):


simulate this circuit – Schematic created using CircuitLab

I've included a series resistor as part of the zener model found in the datasheet, \$R_\text{Z}=Z_\text{Z}\$, along with a hypothetical ideal internal zener voltage, \$V_{\text{Z}^{*}}=V_\text{Z}-I_\text{Z}\cdot Z_\text{Z}=6.6705\:\text{V}\$. This ideal zener voltage is computed from values in the datasheet above.

We can now develop two equations using nodal analysis:

$$\begin{align*} \frac{V\text{o}_{_\text{MIN}}}{R}+\frac{V\text{o}_{_\text{MIN}}}{R_Z}+I_{_\text{MAX}}&=\frac{V\text{i}_{_\text{MIN}}}{R}+\frac{V_{\text{Z}^{*}}}{R_\text{Z}}\\\\ \frac{V\text{o}_{_\text{MAX}}}{R}+\frac{V\text{o}_{_\text{MAX}}}{R_Z}+I_{_\text{MIN}}&=\frac{V\text{i}_{_\text{MAX}}}{R}+\frac{V_{\text{Z}^{*}}}{R_\text{Z}} \end{align*}$$

The problem with the above is two-fold. (1) There are three unknowns: \$R\$, \$V\text{o}_{_\text{MIN}}\$, and \$V\text{o}_{_\text{MAX}}\$. (2) We've not said anything about \$V\text{o}\$, itself, which we know we want to be nominally \$6.8\:\text{V}\$ (by some as yet unstated definition of nominally.)

So let's improve the situation. I propose that we set \$V\text{o}_{_\text{MIN}}=V\text{o}-\Delta V\$ and that we set \$V\text{o}_{_\text{MAX}}=V\text{o}+\Delta V\$, with \$V\text{o}=6.8\:\text{V}\$.

We can now re-phrase, as:

$$\begin{align*} \frac{V\text{o}-\Delta V}{R}+\frac{V\text{o}-\Delta V}{R_Z}+I_{_\text{MAX}}&=\frac{V\text{i}_{_\text{MIN}}}{R}+\frac{V_{\text{Z}^{*}}}{R_\text{Z}}\\\\ \frac{V\text{o}+\Delta V}{R}+\frac{V\text{o}+\Delta V}{R_Z}+I_{_\text{MIN}}&=\frac{V\text{i}_{_\text{MAX}}}{R}+\frac{V_{\text{Z}^{*}}}{R_\text{Z}} \end{align*}$$

Here, we have two equations and now only two unknowns: \$R\$ and \$\Delta V\$.

I'm not saying anything good from this is likely to show up, however. The above nodal equations assume that the zener is a voltage source that can supply as well as absorb power. But it cannot supply any power. So the 2nd term in both equations isn't really workable, if \$\Delta V\$ varies more than about \$130\:\text{mV}\$. And it will vary more. So you already know there's a problem that is highly likely and the equations will solve some fictitious situation and not something workable.

At this point, I'd conclude that the problem has no realizable solution. Not, anyway, as stated.

Another Approach

There is a different assumption we could take from the datasheet: that \$I_\text{Z}=37\:\text{mA}\$. We could, instead, treat the zener as a current sink of that value. This would yield a realistic value for \$R\$ and would provide \$\Delta V\$ values that appear useful and would provide a different check.

Let's proceed with that:


simulate this circuit

Keep in mind this isn't realistic and that we are exploring this to see where it goes.

The new equations are:

$$\begin{align*} \frac{V\text{o}-\Delta V}{R}+I_\text{Z}+I_{_\text{MAX}}&=\frac{V\text{i}_{_\text{MIN}}}{R}+\frac{V_{\text{Z}^{*}}}{R_\text{Z}}\\\\ \frac{V\text{o}+\Delta V}{R}+I_\text{Z}+I_{_\text{MIN}}&=\frac{V\text{i}_{_\text{MAX}}}{R}+\frac{V_{\text{Z}^{*}}}{R_\text{Z}} \end{align*}$$

This will result in a reasonable value for \$R\$. But the \$\Delta V\$ will be large (several volts) and obviously beyond the ability of the zener to remain active.

So this will be yet another way to show that it cannot be reasonably achieved.

A Remaining Problem

The above analysis can lead to results, despite the fact that those results will be unreasonable ones. But there's another statement in the datasheet that I haven't yet mentioned:

$$6.46\:\text{V} \le V_Z\le 7.14\:\text{V}$$

As if there weren't already enough problems, this one shows its face, too.

Hopefully, by now, you at least have some approaches to take. In this case, they won't result in anything useful. But in other cases, they may be useful.

  • \$\begingroup\$ Your answer did help. \$\endgroup\$ – Kashmiri May 23 '20 at 16:46

let's do it like this: from mesh analysis we can write the voltage of the loop as: Vin=Vz+Vrs, so Vrs=Vin-Vz and Vrs=Rs*I from hom's law. then Rs=Vrs/I= (Vin-Vz)/I we have Imin=5mA and Imax=20min and Vin_min=15V and Vin_max=20V. so, consider subtituing the value of Imin and I max with Vin_min and Vin_max to find the Rs_min and Rs_max

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    \$\begingroup\$ Thank you but there appear two resistances from your analysis ie Rmax and Rmin which are 164 and 660 respectively. But the answer specifies that the resistor should be 390ohms .I'm just confused \$\endgroup\$ – Kashmiri May 14 '20 at 6:53
  • \$\begingroup\$ well if the Rs is 390 ohm then if you calculate the max current and min current which are 33mA for Vin=20V and 21mA for Vin=15V which exceed the current range given \$\endgroup\$ – Yaakov May 14 '20 at 8:04

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