3V6 Zener Diodes With Very Poor Accuracy

Question

I have a question about the regulation accuracy (or lack of) of a 3V6 Zener diode. I wanted to confirm that I wasn't doing anything stupid.

I set up the standard test circuit shown below.

No load except for the DMM. The two Zeners in question are: BZX55C3V6 (3V6 - 500mW) and 1N4729A (3V6 - 1W). Both Zeners came from separate suppliers. \$V_{CC} = 5\$V and is relatively stable (total variance no more than 40mV). I know that Zeners need a minimum current to regulate properly and I worked out the optimum current through the combination of \$I_{zt}\$ from the datasheet and the following "rule of thumb" calculation: $$I = \frac{(P/V) \times .7}{4}$$

Taking the 1N4729A as an example, the \$I_{zt}\$ is roughly 70mA. Working out the current limiting resistor we have: $$\frac{5 - 3.6}{70\text{mA}} = 20\Omega$$

When I tested the zener with a \$20\Omega\$ resistor I get a \$V_z\$ of just over 4V! In fact, as I increase and decrease the resistor value, the Zener voltage increases and decreases along with it. It's like the Zener resistance isn't decreasing as the current increases, which is not what I'd expect. What's odd is that the BZX55C is showing more-or-less the same behaviour although with different resistors. Is this normal?

Answer 1

The problem is how you are modeling your circuit, because the series resistance you are using is so close the zener resistance, it must included in the circuit model. As shown in the circuit below which models the behavior that you are seeing of a voltage reading of about 4V from the zener diode

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

A "zener" diode with a breakdown voltage below ~5.6V are based on the Zener effect and will have a very soft knee, I personally never use them.

The ones with voltage of 5.6 and above exploit the avalanche effect and have a very much sharper knee.

Even so I expect that the diode you have does match its data sheet.

For low voltage shunt regulation either use a series regulator (LDO) or a device such as the TL431 which has a band gap reference, and an opamp in a 3-poin package to create an almost perfect "Zener diode".

kevin

Answer 3

If you want decent accuracy throw the zeners away and use some low dropout 3.3V linear regulators. Zeners are notoriously power inefficient even at no load. At least at no load or very small load the LDO regulator will be way more efficient.

Answer 4

Rather than use rules of thumb which you may or may not understand, you should first look at the data sheet of the part you're using. From http://www.mouser.com/ds/2/427/bzx55-se-58740.pdf you will find that the BZX55C3V6 is tested at a current of 1-5 mA. Since you are driving more than 10 times that amount of current through it, it is no wonder that your voltage is high.

Answer 5

There are a couple of issues here. First this stated in the question: -

In fact, as I increase and decrease the resistor value, the Zener voltage increases and decreases along with it.

If you increase the resistor the voltage across the zener should go down.

Because it rises (as per the quote) you must have an unstable power source of 5V and this is rising as you increase the load resistance. Double check this supply.

The next problem is that at 70mA and 4 volts, the power is 0.28 watts and the self-heating may be causing the zener voltage to rise. I'm not sure what the temperature coefficient is for this device and the data sheets are lacking in that respect.

Answer 6

If the 1W zeners were purchased from an unofficial channel they may, in fact, be something like 500mW units marked with the JEDEC 1N4729 number.

The 3.6V +/-5% @ 69mA is a JEDEC registered parameter and the voltage across the zener should be in that range (at Tj ~= 30°C) with 69mA flowing. 3.6V zeners typically have a negative temperature coefficient of -1.5 +/-0.5mV/°C, so self-heating will not account for the anomaly.

I suspect that if you retest the "1N4729" parts with a test current of something like 20mA (70 ohms) you may find that they're within spec voltage-wise.

The BZX55C3V6 is rated for 3.6V +/-0.2V at 5mA, so your resistor would be more like 280 ohms. You don't say what you saw there.

If you include the dynamic resistance in the model, you must use a lower voltage for the ideal zener. It's incorrect to add a voltage at the test current because of the dynamic resistance. The dynamic resistance describes the small-signal behavior near the test current.

Typically the dynamic resistance of the 1N4729A is about 5 ohms, so you could approximate it with an ideal 3.255V zener in series with 5\$\Omega\$ but only very near Izt = 69mA.

Zeners with nominal voltages less than 5V or so have very soft 'knees' (poor regulation) and are not really that useful as usually better alternative are available.