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After the Zener breakdown voltage, a small increase in voltage across the diode causes large increase in current through the diode.

So in a way Zener diode in the breakdown region acts like a resistor with very low resistance (zero for the ideal case.)

So to regulate voltage across an LED for example, why don't we just use a small resistance in parallel instead of the Zener diode?

Here small resistance means small compared to the resistor in series like we use in a Zener voltage regulation circuit.

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  • \$\begingroup\$ LED's are just like 2V and 3V Zeners so what you are suggestion won't work and is not logical. \$\endgroup\$ Aug 31 at 17:27
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    \$\begingroup\$ You don't regulate the voltage across an LED, you regulate the current through it. \$\endgroup\$
    – Finbarr
    Aug 31 at 17:36
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    \$\begingroup\$ why not use lead weights to slow a car instead of brakes? \$\endgroup\$
    – dandavis
    Aug 31 at 18:00
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    \$\begingroup\$ For a start, you don't drive an LED by regulating the voltage across it. \$\endgroup\$
    – Finbarr
    Aug 31 at 18:12
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    \$\begingroup\$ Well, I suppose that the question is more like "How does Zener diode voltage regulator work?" than "How to drive LED?", so no reason to say that this is a wrong way to drive LED, the LED was just a badly chosen example of "some load", right? \$\endgroup\$ Aug 31 at 18:33
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The way an ideal Zener diode is supposed to behave is, that it is non-conductive (in reverse bias) until a certain break-down voltage and above that, it is perfectly conductive.

schematic

simulate this circuit – Schematic created using CircuitLab

This is probably the circuit you meant by "Zener voltage regulation circuit". This is a basic linear voltage regulator. I see no point in using it for LED (you are supposed to regulate current through an LED, not voltage across it; so just use serial resistor), but let's explain how it works:

Example 1) The Zener diode has a break-down voltage of 3V, there is a 2V supply. No current flowing, because the voltage across Zener is too low.

schematic

simulate this circuit

Example 2) The Zener diode has a break-down voltage of 3V, there is a 3V supply. The voltage is just enough to make the diode conductive, so some current is flowing.

schematic

simulate this circuit

Example 3) The Zener diode has a break-down voltage of 3V, there is a 5V supply. The voltage is way higher than the threshold. 3 volts are still used to keep the diode conductive. Where goes the remaining 2V? In this ideal case, nowhere, so current would be infinite. With a real Zener diode, the current is limited by the internal resistance of the diode, but the current will likely overheat the diode.

schematic

simulate this circuit

Example 4) The Zener diode has a break-down voltage of 3V, there is a 5V supply. The voltage is way higher than the threshold. 3 volts are still used to keep the diode conductive. Where goes the remaining 2V? It is "used" to push some current through the resistor that is connected in series with the diode.

schematic

simulate this circuit

The resistor should have high enough resistance to limit the current through Zener, so that it won't overheat. But too high resistance will limit output current, if you connect load parallel to Zener, current draw of the load must not exceed current through the resistor at 2V (otherwise, the resistor would "consume" higher voltage than the 2V, which would mean lower voltage remaining for the load).

Now, if you want to model the Zener diode using basic components, it would look like this:

schematic

simulate this circuit

The resistor is there to simulate behavior in the breakdown region (as you said in the second paragraph). Voltage source simulates the fact that there must be at least 3V across it to keep it in breakdown region (this model only works in the breakdown region, when voltage across it is higher than the 3 V; for lower voltage, it just behaves as open circuit).

Whe you suggest using only small resistor instead of Zener diode, you are completely missing the "voltage across the diode" part:

schematic

simulate this circuit

Here, the "small" resistor is just a short circuit parallel to LED, so there is nearly no voltage across it.

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    \$\begingroup\$ OK, this was a good way to handle the question. Simple and accurate. I think this kind of answer helps the OP much more than closing the question. \$\endgroup\$
    – mkeith
    Aug 31 at 23:55
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You can make a voltage divider with low value resistors. If the load draws very little current compared to the current through the voltage divider, then the voltage will stay (somewhat) stable - provided the input voltage is stable.

It won't be much of a regulator. It will waste at least ten times more current than the load consumes. Any variation in the input voltage will also change the output voltage.

The thing about a Zener diode is that it varies its current depending on the input voltage. That property makes the output voltage of a Zener regulator (somewhat) indifferent to changes in the input voltage.


You don't want to set the voltage across an LED. What you should do is limit the current. If you take care of the current, the voltage across the LED will take care of itself. LEDs are much like Zener diodes in that sense - with, of course, the difference that LEDs work in forward bias while Zeners work in reverse bias.

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LED with a Zener diode.

schematic

simulate this circuit – Schematic created using CircuitLab

The Zener voltage, Vz, is relatively constant.
LED operates by the current flowing through. It has "forward voltage drop, V_led", which has dependency to the current, but relatively constant as well.
Thus, the current through LED is: I_led = (Vz - V_led) / R2
Since, Vz & V_led are "relatively constant", R2 can set the LED current that is "relatively constant", while the Power Source varies the supply Voltage, Vs.
Meantime, when the Vs changes (above Vz), the excess voltage is taken by R1.
Thus, Vs = R1 * I_r1 + Vz

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