# Purpose of Zener Diode in Circuit

Can anyone please explain what the purpose of the zener diode is in the circuit shown? The circuit is taken from Figure 3 in the datasheet for the STN1170, an OBDII interpreter IC.

simulate this circuit – Schematic created using CircuitLab

DLC_RAW is the unfiltered, raw +12V from an OBDII bus (i.e. the car battery voltage, so should be in the order of +12.6V). It is also therefore going to be full of transients etc.

The op amp is a non-inverting buffer followed by a LP filter, and the overall purpose of the circuit seems to be to measure the voltage of the car battery.

I'm not sure what the zener is for. What would be a suitable value for the zener diode? And does this not mean that in the event of a large spike coming from the battery (such that the voltage divider output is greater than 3.3V) then the transients will find their way onto the 3.3 supply line?

There is a slightly simpler circuit shown which also achieves the same aim, shown below. I don't know what the zeners do - clamp the voltage to within 0V and 3.3V, perhaps? I have the same question about whether transients then appear on the 3.3 supply line.

simulate this circuit

• I don't believe they are zeners at all, but schottky diodes. They are protecting the inputs from transients as you suggest. Jul 22, 2013 at 22:06
• Indeed, the data sheet makes no mention of the 1N4733A part number, which is a 5.1V Zener. The symbols are clearly Schottky symbols, with the square-brackety-s-shaped cathode, and not Zener symbols.
– Kaz
Jul 22, 2013 at 22:11
• double click a component, then select "edit individual parameters" to get rid of the part numbers. Jul 23, 2013 at 2:09
• Sorry everyone, I hadn't zoomed in enough to see that they are schottky diodes, not zeners. Phil - thanks for the tip!
– John
Jul 23, 2013 at 9:28

Those are not Zeners. They are Schottky diodes.

Shottkies have a lower voltage drop than regular Si junction diodes, about 0.3V instead of 0.7V. They are also generally faster.

Their purpose is to keep the inputs within the supply rails.

• Perhaps you are right about the second circuit (added after I wrote my answer), but not the first. Look at it carefully. Not only is the symbol clearly for a Zener, but a Zener makes sense when you consider clipping the inevitable spikes on the 12 V line. Jul 22, 2013 at 22:10
• They look like Schottky symbols to me: poor ones, that require a very high zoom level to be unambiguous. Jul 23, 2013 at 2:14
• @OlinLathrop, check the data sheet posted in the question for a more clear diagram. They are Shottkey diodes. Jul 23, 2013 at 5:06
• Thanks very much, sorry, I hadn't zoomed in enough to see that they are schottky diodes. In which case, the circuit makes more sense to me - please would someone confirm the following: by using schottky diodes (with a low threshold voltage and fast switching action), whenever the voltage exceeds 3.3v or 0V (in a negative direction) the diode will switch, thereby keeping the input within 0V < input < 3.3V. Will I see a transient on the 3.3V voltage line, however?
– John
Jul 23, 2013 at 9:29
• In the second schematic, the inputs will be clamped between -0.3 and 3.6 volts because of the voltage drop over the diodes themselves. There could be transients on the 3.3 V rail, but any capacitance on the rail, and the load of the circuit itself, will absorb some transients. Also note that the series resistance will limit the amount of rail disturbance. Jul 23, 2013 at 19:17

The purpose of the zener appears to be to clip the input signal to the valid range. The zener should not conduct when the power voltage is in a reasonable range. It will clip the power voltage after the divider to safe levels for the opamp and the analog input in both directions when there are glitches on the supply. These kind of glitches happen regularly on "12V" car power.

Note that the zener is conducting in the forwards direction as a regular diode to clip positive spikes, and as a zener to clip negative spikes.