Zener diode as voltage regulator

Question

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:

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?

Answer 1

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 V_out 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.

Answer 2

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.

Answer 3

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.

Answer 4

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.