# Zener Diode usage in Voltage Divider

Need to sense when 12V accessories are switched on for a motorcycle related project.
Testing a voltage divider which seems to be perfect. R1 = 10k ohms, R2 = 4.7k ohms.

The problem is when I add a 5.1 Zener diode. Testing at 18v, the divider outputs 5.7v. When I plug in the Zener, it drops the output voltage to 3.6v.
I tested the Zener (7v, 10mA output from my power supply) and it dropped the voltage down to 5.1, so I know it isn't the problem. What am I doing wrong?

Diode is: Vishay BZX85C5V1-TR

Unable to tell what you are doing, can you draw a schematic?
FYI: zeners are used in reverse biased mode.

LarryD:
Unable to tell what you are doing, can you draw a schematic?
FYI: zeners are used in reverse biased mode.
Zener diode - Wikipedia

yeah, bias is correct. I tried it both ways just to be sure. Not my picture, but this is a diagram:

Your answer is in figure 7 of your datasheet; the breakdown voltage is lower at lower currents. The 5.1V rating is specified at 45ma.

Chagrin:
Your answer is in figure 7 of your datasheet; the breakdown voltage is lower at lower currents. The 5.1V rating is specified at 45ma.

Thanks. Does anyone know how much current an arduino pin will sink to register high? I'm guessing not much, and I probably can't find a Zener diode that will work for me. I think I would have the same problem even with this one:

TZX5V1C-TR Small Signal Zener Diodes

Are there any alternatives? or should I just not worry about it and use voltage divider without any protection?

You can gain some protection by changing your voltage divider to put a lower voltage on the pin. For an Atmel powered between 2.4 and 5.5 V, the minimum guaranteed HIGH level input is 0.6Vcc - for 5 V that's just 3 V. Maximum allowed voltage on the pin is Vcc + 0.5V. If you are just using a pin configured as INPUT (rather than as ADC), it is very high impedance and you might think about adding a resistor between your Vout and the pin as a current limiter - even a fairly high value like 10k won't affect the ability of the pin to sense your signal.
Ciao,
Lenny

jvdb:
Are there any alternatives?

Two plain ole ordinary diodes (second circuit)...
http://www.thebox.myzen.co.uk/Tutorial/Protection.html

The current the input sinks when reading "high" is the Input Leakage. It's "1 microamp" or just generally inconsequential.

But Mike's example in this case has a 22 ohm resistor on the input; OP's is behind a high value resistor voltage divider. Because of that he's going to see less current and his 5V1 zener is going to be in the knee. Given that the input pins of the processor are protected by diodes rated to 1ma, how do you calculate the proper zener to use?

The fact that he has a higher seriese resistor does not matter the input pin is still clamped to the rail.

The zener diode with those 10k/4k7 dividers at analog inputs (for voltage measurement) shall not affect the divider's ratio "too much". I would look for a zener with higher voltage (ie 6v8) as its knee starts at higher voltages, so it may influence the divider's ratio less. The zener is good for environments where large voltage peaks may appear.

The 2 clamping diodes are implemented on the chip already, but I did not find their max currents.

Provided it is 5mA(?), with 10k/4k7 divider the max positive voltage at the divider's input the chip may "survive" is +70V, and max negative voltage is -50V. The assumption is the voltage regulator (5V at Vcc pin) can handle the incoming current (from the clamp diode) properly.
No warranties of any kind

An idea: The "ultimate" protection with Zener in combination with internal chip clamping diodes.

This combines the internal chip clamping diodes, a 10k additional resistor at the analog input pin, and >20V Zener (for example). The Zener diode knee might be now far away from the voltage of interest (0-5V) at the mid of the divider (so a little effect on it), but in combination with R3=10k analog input resistor the ie. 20V maximum at the 20V Zener diode might be safe for the chip. It seems this circuit may protect till many many hundreds of volts at the input of the resistive divider (at least simulation shows that).
Not tested on real hw, no warranties of any kind

The 2 clamping diodes are implemented on the chip already, but I did not find their max currents.

They are not designed for this sort of protection. However if you do use them like this they should not exceed 1mA current. I prefer to have some external ones for robustness.

thanks for all the help!