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I've been trying to make a voltage regulator using a zener diode.

I used the BZX79 3.3V Zener diode. I connected the diode as the following:

Schematic

Data Sheet. I assumed that as as most of them it operates at around 1mA to 10mA safely in the break down region.

I connected my supply which was 5V. The resistance was chosen to be 220 ohms. This will make the max current of the zener around 7mA. Which should be safe.

Now, when the zener was unloaded, I got a smooth 3.3V as an output.

But, when I loaded the zener diode, the voltage across the diode dropped to 2.4V.

Why did that happen?

My assumption is the following:

The load which should operate at 3.3V, is consuming way too much current that the zener current is below the min current. This caused the zener to get out of the break down region.

Is my deduction correct? And what are available solutions, taking in note that I've used probably the smallest resistance possible that is 220 ohms. (At least of what I have).

I also have a supply of 8.4V. Will it work if I connected it directly to the supply? At the end I will have more current available to the zener even after loading it.

Edit:

The data sheet states that the zener is fine for 6 Amps. That is a huge and weird number isn't it?

Datasheet excerpt

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    \$\begingroup\$ You should use a linear regulator to drop the 5V to 3.3 and throw the zener in the garbage. In my opinion. 3.3V zeners (anything under about 5V) have a very soft 'knee' and are practically worthless. \$\endgroup\$ – Spehro Pefhany May 8 '14 at 21:28
  • \$\begingroup\$ The problem that these regulators are not available in my country, and I don't have time to order one. too bad, I'm stuck with the the zener. \$\endgroup\$ – Adel Bibi May 8 '14 at 21:33
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    \$\begingroup\$ You could make a crude one with a couple diodes, a blue or green LED an NPN BJT, and a couple resistors. Or use a TL431 (should be available anywhere- take one out of a duff PC power supply) and a couple resistors in place of the Zener. \$\endgroup\$ – Spehro Pefhany May 8 '14 at 21:39
  • \$\begingroup\$ What you want to power from that circuit? \$\endgroup\$ – Kamil May 8 '14 at 23:16
  • \$\begingroup\$ RF module that I pulled out of an old game. That being said, I have exactly no idea about the module except that it works on 3.3v. @Kamil \$\endgroup\$ – Adel Bibi May 9 '14 at 6:28
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I connected my supply which was 5v. The resistance was chosen to be 220ohms. This will make the max current of the zenor around 7mA.

This means that if your load requires more than 7 mA (I actually calculate 7.7 mA, not 7 mA), the resistor will drop Vout below 3.3 V. In that case the zener will no longer be regulating.

But, when I loaded the zener diode, the voltage across the diode dropped to 2.4v.

Why did that happen?

Your load was taking more than 7.7 mA.

Given that the output voltage dropped to 2.4, we can calculate that the load current was (5-2.4)/220 = 11.8 mA.

You could make this work by reducing your series resistor. If the load is fairly stiff, reducing the resistor to 110 ohms or so should bring you back into regulation. This assumes the load current doesn't increase to more than 15 mA when the supply voltage is raised to 3.3 V.

If it's possible to have an open-circuit condition at the load, you need to choose a zener that can take all the current that would normally go to the load.

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  • \$\begingroup\$ If I used an 8.4 v supply. That will provide me with a max current of 23.18 mAp for 2200 ohms. The load consumes 11.8 mA that will leave me with 11.3 mA for the zener. Will this be fine? \$\endgroup\$ – Adel Bibi May 8 '14 at 21:41
  • \$\begingroup\$ It's usually cheaper to change a resistor than change the supply voltage. But it's up to you. \$\endgroup\$ – The Photon May 8 '14 at 21:43
  • \$\begingroup\$ I'll try connecting two 220 ohms in parallel and see what happens. Thanks. \$\endgroup\$ – Adel Bibi May 8 '14 at 21:48
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    \$\begingroup\$ Accepted this as an answer because I have replaced the 220 ohms with 51 ohms providing the circuit with a max current of 30mA instead of 7mA. It is working like a charm now! Thanks! \$\endgroup\$ – Adel Bibi May 11 '14 at 0:55
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When designing a simple zener regulator circuit like this, there are two important criteria: (1) the maximum load current and (2) the minimum load current.

By KCL, the minimum diode current occurs in case (1) while the maximum diode current occurs in case (2). The KCL equation is:

$$I_D = I_R - I_L = \frac{V - V_Z}{R} - I_L$$

Clearly, one must pick \$R\$ small enough such that the minimum diode current is at or above the zener holding current.

Also, one must pick \$R\$ large enough such that the maximum diode current does not cause excessive diode power dissipation.

Now, with the values picked, the upper bound on load current is

$$I_{L,UB} = \frac{5V - 3.3V}{220 \Omega} = 7.72mA$$

This is an upper bound because (a) the output voltage must be less than 3.3V for any larger load current and (b) the diode current is zero for this load current.

Since there must be some diode current for the diode to have 3.3V across, the actual maximum load current should be less than this.

Thus, if your load requires more current than this, you must either decrease \$R\$ or increase the source voltage.

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No, at high load, you can ignore your zener. You then just have a voltage divider between your 220 ohm resistor and your load.

A zener can be used successfully as a voltage reference, but if you want it to actually regulate voltage to power a load, you need a very light load. Instead use a zener voltage to set the voltage level of a common drain amplifier to get ample current while still regulating voltage.

The 6 amps is only to be done for very short periods of time 100 microseconds to be exact. And only if the ambient temperature is 25 degrees C. That means for very short periods of time, you can pulse it with 6 amps of current without frying it. This amount isn't useful for your application because you want a constant current on it which it doesn't really mention from what you've shown.

The circuit I speak of would look pretty much like this except the BJT would be replaced with a mosfet (although a BJT would work too). http://i.stack.imgur.com/qlBYs.png, here, if BJT used, the output will be 3v3 - vbe, ~ 0.6v. Also instead of 10K use 330Ohm for 5mA zener holding current.

A better solution is a linear regulator. They come cheap and are meant for exactly what you're after.

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  • \$\begingroup\$ So, at high loads my zener doesn't regulate the output anymore? \$\endgroup\$ – Adel Bibi May 8 '14 at 21:16
  • \$\begingroup\$ Precisely. The voltage divider drops the voltage below the zener voltage and the diode conducts in neither direction. It looks like an open to the circuit when the voltage falls below that zener level. \$\endgroup\$ – horta May 8 '14 at 21:19
  • \$\begingroup\$ What's the solution then? Does that mean I can't use this zener as a regulator for the load? Although I'm using a simple RF tranciver. It shouldn't be taking that much load! \$\endgroup\$ – Adel Bibi May 8 '14 at 21:22
  • \$\begingroup\$ I doubt an RF tranceiver is running off of less than 10 mA. If the voltage drops to 2.4 V then you know it at least wants (5-2.4)/220 = 11.8 mA. A solution as I said is to use a common drain amplifier (1 mosfet) along with your zener voltage generator. \$\endgroup\$ – horta May 8 '14 at 21:25
  • \$\begingroup\$ Could you provide me with a schematic of the connection? \$\endgroup\$ – Adel Bibi May 8 '14 at 21:26
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You have to choose the resistor so that the load current plus some current for the zener diode will still leave 3.3 volts across the zener diode.

The BZX79 series of zeners are rated at 500 mW maximum power dissipation, so your 3.3 volt diode should safely handle up to 150 mA.

If your maximum load current is 20 mA, and you want to allow 10 mA minimum in the zener, the resistor should be about (5V - 3.3V)/(20mA + 10 mA) = 56 ohms. Even if you disconnect the load, the zener will only have to carry 30 mA, which is well within its capabilities.

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