# In this circuit, please explain why the voltage of Zener diode drops

In this circuit, can you explain why lowering the resistance in right most branch (10 Ω in the first picture) reduce the voltage received by the Zener diode.

I understand that voltage splits proportionally to the resistances in series but it seems that the Zener diode is also in series with the first resistance, so I would expect its voltage to remain at 5 V.

Note: I am learning electronics this question is for pedagogical purpose.

With 10 Ω:

With equal resistor:

• Imagine if you replace your 10 ohm with a short circuit. How much voltage will be across it? Commented Jul 9, 2023 at 20:43
• Intuitively it makes sense but is there a formal way to describe this? Commented Jul 9, 2023 at 20:56
• Yes, but I feel you need to understand it at a more fundamental level before you will be helped with a more formal way. What voltage will you have at zero ohm? Commented Jul 9, 2023 at 21:01
• 0 since the circuit is shorted ? Commented Jul 9, 2023 at 21:35
• Excellent! If you keep increasing, current will commutate towards the Zener diode and you have achieved regulation. Likewise, as long as your load across the Zener (assuming you use it as a voltage regulator), any equivalent load above 70 ohm will work. Lower than that, you loose regulation. Commented Jul 9, 2023 at 22:37

Perhaps if you re-draw the circuit as below, the situation will be clearer:

simulate this circuit – Schematic created using CircuitLab

Here you can see that the two resistors form a voltage divider. Using the voltage divider formula, you will find that the voltage between the resistors is 1.09 V, far below the threshold voltage of the Zener Diode, so no current will flow through the Zener.

When using a Zener diode to regulate voltage, you must remember to include the resistance of any load attached to the circuit when calculating voltages and currents in the circuit.

When you put the 10 ohm resistor across the zener diode the 100 ohm and 10 ohm voltage divider causes the voltage to be less than the zener diode reverse clamping voltage and thus it basically just has minimal leakage current. With the two 100 ohm resistors in the voltage divider that trys to put the voltage for the divider at 6 volts (half the supply) but this is higher than the zener reverse clamping voltage so the zener breaks down and clamps the divider to the 5 volt level.

• At a first order analysis, the Zener diode just prevents the voltage at its anode from being higher than 5v. It's not a buck-boost converter. Commented Jul 9, 2023 at 20:50
• It kinda feels like the the lower resistance branch dictate how the voltage split should be done? Is there a formal way to express this ? Commented Jul 9, 2023 at 20:55
• @vicatcu - Where in the world did you get any idea that I was talking about a buck boost converter?? Commented Jul 9, 2023 at 20:55
• Sorry it wasn't a criticism of your answer at all Commented Jul 9, 2023 at 23:42

Take a look at what happens to the voltage $$\V_{OUT}\$$ across a 5V zener diode (as measured by VM1) configured as follows, as we sweep the supply voltage $$\V_S\$$ from 0V to +10V:

simulate this circuit – Schematic created using CircuitLab

We don't have any load yet, just a voltmeter, but we can use this condition as a "baseline" to get some idea of the diode's behaviour. Here's a plot of $$\V_{OUT}\$$ during the sweep of $$\V_S\$$:

When the supply is below 5V, $$\V_S<+5V\$$, the zener diode behaves as if it's not even there. See the plot of diode current below to see how it passes no current, and can be completely disregarded under the condition $$\V_S<+5V\$$.

When the supply exceeds 5V, though, $$\V_S>+5V\$$, the situation changes suddenly. The voltage across the diode is limited to a maximum of 5V. $$\V_{OUT}\$$ is prevented from rising above 5V, or in other words, the zener diode actively "clamps" its voltage to some maximum, 5V in this case. It does this by becoming suddenly very conductive. Exactly as conductive as it needs to be to develop (almost) exactly 5V. The harder we try to raise $$\V_{OUT}\$$, the harder the diode opposes this rise, the more conductive it becomes to achieve that.

To put this another way, you can apply any potential difference across the zener diode below its rated value, and it will not pass any current, it will not try to conduct, it will not oppose this condition in any way. The moment you try to impose a potential difference across it that exceeds this limit, though, it fights exactly as hard as it needs to in order to prevent this condition, becoming exactly as conductive as it needs to be, to pass exactly as much current as is necessary, to clamp the voltage across it to its rated value.

You can see this onset of conductivity as $$\V_S\$$ rises above 5V in the next graph, showing the diode's current during the sweep of $$\V_S\$$ from 0V to 10V:

Now imagine that the voltage source is not just $$\V_S\$$, but a resistor potential divider formed by R1a and R2a below left, whose output potential $$\V_{OUTa}\$$ is unconstrained (not connected to anything):

simulate this circuit

As you would expect, the potential divider attenuates $$\V_{Sa}\$$ by a factor $$\\frac{R_2}{R_1+R_2}=0.75\$$, meaning that the potential at OUTa will vary between 0V and +7.5V as we sweep:

The addition of a 5V zener diode in parallel with R2, though (as seen above right), imposes a constraint on the maximum potential difference (5V) across R2. This becomes apparent in another sweep; this time we monitor $$\V_{OUTb}\$$ as we sweep $$\V_{Sb}\$$ from 0V to +10V:

As before, diode D1 "clamps" $$\V_{OUTb}\$$ to a maximum of +5V, by becoming conductive. For all $$\V_{Sb}>6.7V\$$ which without D1 would have produced $$\V_{OUTb}>+5V\$$, D1 becomes a "bypass" route around R2b, for current that would otherwise have passed through R2b. In this way, current through R2b is unable to rise above $$\I_{R2b(MAX)} = \frac{5V}{300\Omega}=16.7mA\$$, and the voltage across R2b is unable to rise above 5V.

Your question involves setting R2 to 10Ω. Let's see what happens to the voltage across R2 in that scenario, during another sweep of $$\V_S\$$, this time from from 0V to 100V:

simulate this circuit

The above plot clearly shows that the zener diode is still trying to clamp the output to a maximum of 5V, as it always has, it's just that this time the onset of clamping seems to start when $$\V_S\$$ is huge, well over 50V. This explains why your own circuit doesn't show any clamping to 5V; the source voltage simply isn't high enough, and the voltage across the diode hasn't yet reached that 5V limit!

To understand this curve, first find out the relationship between $$\V_{OUT}\$$ and $$\V_S\$$, if D1 were not there:

\begin{aligned} V_{OUT} &= V_S\frac{R_2}{R_1 + R_2} \\ \\ &= V_S\frac{10\Omega}{10\Omega + 100\Omega} \\ \\ &= 0.909 V_S \\ \\ V_S &= 11 \times V_{OUT} \\ \\ \end{aligned}

For the voltage across R2 to be 5V, the voltage source $$\V_S\$$ would have to be:

\begin{aligned} V_S &= 11 \times V_{OUT} \\ \\ &= 11 \times 5V \\ \\ &= 55V \end{aligned}

That corresponds nicely with the onset of clamping we see above.

The zener diode operates in reverse brakdown mode. This means that it presents a very high resistance (see the blue region in the image) between its terminals until the cathode-to anode voltage exceeds a specified value. The is called the "knee" region where the diode begins to conduct.

For voltages greater than the knee voltage, the resistance (see the red region) between the anode and cathode becomes very low. The zener now dominates any thing in parallel with it. BUT, only if there is enough current through the zener to maintain operation in the zener region.

In your circuit the voltage divider created by the two resistors divides the 12V down to 1.09V. This is less than the knee voltage so the diode is just a high resistance in parallel with the 10 ohms. So the zener will have little effect, the 10 ohm resistor dominates. The zener diode is operating in the blue region.

When the 10 ohm resistor is replaced with a 100 ohm resistor, the voltage would rise to 6V without the zener. This is beyond the knee so the zener starts to conduct moving into the red region. In your case this presents as a constant 5V (5.02 in your example) at any (reasonable) current.

So now the voltage across the lower 100 ohm resistor is 5V and so by Ohm's law, its current is 50mA.

The voltage across the upper 100 ohm resistor is 12-5 = 7V. Again by Ohm's law, its current is 70mA. By Kirchhoff's current law, the current through the zener is the difference, 20mA.

It kinda feels like the the lower resistance branch dictate how the voltage split should be done? Is there a formal way to express this ?

If the upper resistance is decreased to 10ohm, the same "voltage split" would have been obtaned. The currents would be much higher, although 200mA would be too high for real diodes.

A design would call for a nominal load current that may vary between two values. The lower resistor is usually called the load resistor.

For operation as a regulator, the zener diode must operate in the red region.

The current in the upper resistor (call it the supply current) will remain the same as long as the 12V remains the same. So the zener and load currents must add to the supply current. Since it is constant a high load current will result in a low zener current

If the load current is too high the zener current will be too low, moving the operating point into the blue region. So the lower resistor will determine the current split not a voltage split.

The upper resistance is sized so that if the lower resistance is removed (high R) then the max zener current or power is not exceeded. There is a trade off between having enough current for the load and not too much to burn out the zener diode.

You expect the voltage across the Zener diode to stay at 5V in circuit 1. Let's discuss what that would mean, so we see why it can't happen.

As the parallel resistor is connected to the very same two points as the Zener diode, it gets the same voltage of 5V. Being a 10R resistor, it draws a 500mA current (directed downward according to your schematics).

Across the other resistor, 7V are remaining (12V - 5V). For a 100R resistor, that means a downward current of 70mA.

According to Kirchhoff, at every point in a circuit, the sum of incoming currents equals the sum of the outgoing ones. So we still need 430mA of incoming current into the "middle" point. The only connection to this point we haven't yet considered, is the Zener diode. If the 5V were to be true, it should provide 430mA into the point, with a direction against its voltage drop. That means, it wouldn't consume power, but produce it.

Zener diodes don't function as power generators, no more than any other passive component. So, having 5V across the Zener diode in this situation isn't possible.

You can understand the Zener diode as:

A Zener diode will not allow the voltage across itself to exceed the rated voltage, and does so by drawing the current necessary for this purpose (of course, within physical limitations, and only approximately). For voltages below the rated one, it draws no current and thus does not interfere with the rest of the circuit.

So, for analysing a circuit with a Zener diode, you should independently calculate for two different assumptions:

• The Zener-diode voltage is below its rated value. Analyse a circuit where the Zener diode is not there (simply not connected), and compute the voltage between the two Zener diode points. If the voltage is below the rated one, the assumption was correct, and the results are valid. If it is above, continue with the second assumption.

• The Zener diode is actively limiting its voltage by drawing some current. For this calculation, replace the Zener diode by a constant-voltage element, and calculate for this circuit. The current going into the Zener diode will be greater that zero. (Hopefully, it's not infinite, as that would mean blowing up the Zener diode.)