Zener Diode Operation Question - Page 1
Forum section: electronics / projects

author: rcbuck
author_rank: Frequent Contributor
timestamp: October 24, 2019, 08:59:51 pm
content: I am dropping 12V to a maximum of 3.3 volts to drive an input pin on a micro. I have two resistors setup as a voltage divider to create the 3.3 volt level. I wanted to use a zener diode for extra protection in case the lower divider resistor failed (unlikely).With just the two resistors (photo 1), I get 3.56 volts at the resistors junction point. I wrongly assumed that when I put the zener from the junction to ground, the voltage would drop to roughly 3.3 volts. However, the voltage drops to 2.94 volts. The zener is a 2% device and 2.9 volts is outside the tolerance range. If I change the top resistor to 820 ohms, the junction voltage then rises to 3.286 volts. This confirms the zener does work.So why does the voltage drop to 2.94 volts when there is only about 3 mA of current? Decreasing the top resistor value so there is roughly 10 mA of current and the zener works properly.Does the operation I am seeing have something to do with the zener internal resistance? Does it behave like a resistor before it avalanches? The data sheet shows a internal impedance ZZT@IZT of 1600 ohms at a current of 0.25 mA. However, I would think this only applies to an AC signal? The part number of the zener is Mouser 863-MMSZ5226CT1G.Last Edit: Changed ZZT@IZXT to ZZT@IZT

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 24, 2019, 09:49:17 pm
content: Check the V(I) curve of the diode, and check what current I_ZT it's measured at.Tim

author: SiliconWizard
author_rank: Super Contributor
timestamp: October 24, 2019, 09:53:56 pm
content: Oh yeah. Real diodes are not ideal.

author: rcbuck
author_rank: Frequent Contributor
timestamp: October 24, 2019, 10:27:07 pm
content: Tim,Looking at Fig 9 in the data sheet it appears that any current over about 1 mA is a straight line for any zener voltage above 5 volts. Unfortunately there is no curve for the 3.3 volt part. Below 3 volts the curve is not straight line. Maybe I need to select a different zener.

author: Tomorokoshi
author_rank: Super Contributor
timestamp: October 24, 2019, 10:44:16 pm
content: It's better to characterize the Zener and then tune by selecting resistors than to select resistors and tune by selecting Zeners. Figure on +/- 2% to 5% for that part.Then compare the actual range of the circuit with the Zener to the actual High and Low ranges of the processor. 2.94V may be perfectly within the High range.

author: mikerj
author_rank: Super Contributor
timestamp: October 24, 2019, 10:49:27 pm
content: Low voltage zeners are basically useless for this kind of application (and most applications tbh), they have a very soft "knee". Consider using a Schottky diode to clamp the input to the 3.3v rail, but make sure the rail has sufficient load to sink you worst case fault current.

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: October 24, 2019, 11:14:31 pm
content: If you use high value resistors (> 10k) the ESD diodes at the inputs of the MCU will clamp the voltage to its Vcc + 0.5. The ESD diodes typically have low forward drops.

author: rcbuck
author_rank: Frequent Contributor
timestamp: October 25, 2019, 12:17:03 am
content: mikerj and EEEnthusiast, those are both good suggestions. The output is actually coming from a LM555 which has 200 mA source capability. But a high value resistor would work with the Schottky diode. The MCU is an ESP32 and I can't find any indications that there are clamping diodes inside the MCU. The data sheet is mute on that subject. In either case I would need to keep the internal pullup resistors off.I wasn't thinking and it just dawned on me there is a far better way to handle my problem. I am using the 555 in a missing pulse detector circuit. So I only need to know when the output goes low. I don't need the high output signal from the 555. The attached circuit does what I need. I'm using an external 10K resistor to the 3.3 volt line instead of the internal 45K pullup resistor.

author: SiliconWizard
author_rank: Super Contributor
timestamp: October 25, 2019, 01:17:02 am
content: Quote from: mikerj on October 24, 2019, 10:49:27 pmYup.

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: October 25, 2019, 06:23:26 am
content: Each GPIO pin of almost all MCUs have clamping diodes at the input. If not for ESD protection, these are used for continuity testing on the production line when the IC is manufactured. Without these diodes, there is no easy way for the manufacturer to test the continuity of the pin from the internal die.

author: Renate
author_rank: Super Contributor
timestamp: October 26, 2019, 06:06:58 pm
content: Quote from: rcbuck on October 25, 2019, 12:17:03 amSo, why not use pin 7 of the 555?This has the same logic polarity as pin 3, is open collector and there are no diodes or bias to the 555's Vcc.The MCU pullup would be fine, or throw on your own to 3.3V(Of course we can ask why you're using a 555 at 12V.You could use a CMOS 555 at 3.3V.Or even just a CMOS Schmitt trigger.Or do your missing pulse detection in software.)