# Calculating the voltage drop of two Zener diodes in series? (reverse conducting)

I'm building a circuit powered from a 12V 55mAh (A23 battery). Needless to say that power consumption is a primary concern. The circuit is an ATtiny85 which is driving 13 LED's (in 5 groups). I initially used a voltage divider circuit to reduce the 12V down to 5.5V to supply the ATtiny85. This worked but I found the voltage divider was eating up too much current, 5mA, even when the microcontroller is put to sleep and all LED's are turned off. When active it consumes 5mA to 7mA. I replaced the voltage divider with two 1N4733A 5.1V Zener Diodes in series:

simulate this circuit – Schematic created using CircuitLab

This circuit works beautifully for saving power. Power consumption with this circuit is in the micro-amp range when the ATtiny is sleeping (too low for me to measure) and consumes only 0.1 mA to 0.3 mA when the ATtiny is awake and pulsing the LED's. A FIFTY fold decrease in power consumption.

The trouble with the new circuit is that it is very sensitive to the supply voltage. At 12V input (using a bench power supply) the circuit fires up and works well. I get around 5.1V at Vcc. At 11.5V supply voltage the circuit will not start.

When I was using a voltage divider instead of the diodes the circuit worked well down to about 8V supply voltage. With the diode setup it needs 12V to start and will keep running down to about 11.5V before it shuts off.

Is there a way to modify this circuit so that it works over a wider range of supply voltages? Ideally down to around 6V (as the battery discharges). The datasheet's I've looked at for the Zener diodes don't mention anything about the expected voltage drop across the diode when it's conducting at the breakdown zener voltage. If I replace D1 and D2 with 3.3V Zener's should I get the same voltage drop? I'm not sure how to calculate the expected voltage drop at Vcc.

• Zener diode is usually used in parallel to the load in order to regulate the voltage. – Eugene Sh. Dec 10 '18 at 18:13
• Yes, I've seen Zener's being used to regulate voltage using a resistor to limit current but this has the same disadvantage as a voltage divider. The limiting resistor will burn current. – Jeff Wahaus Dec 10 '18 at 18:37
• Everything will burn current, with passive dividers, Zener parametric stabilizers, or LDO regulators. If you really need to save power, use switching mode power regulator. Period. – Ale..chenski Dec 10 '18 at 18:42

Since no-one has tackled your basic question, I'll attempt to explain:

You cannot calculate the voltage across a Zener at low current. The reverse voltage is a characteristic of the device and depends on the construction/architecture.
There is no better reference I've found than the OnSemi Theory of Zener booklet and you should start by reading this.

Zeners come typically in one of two distinct architectures, Zener effect and Avalanche effect, You are using Zeners down in the Zener effect region, which can be characterized down to very low currents and the voltage across them will vary somewhat predictably.

From the OnSemi document, here is one very pertinent graph of the Vz characteristic:

Notice that Zeners below 6V Vr have quite different Vr curves. Those below 6V have a very large slope and as you lower the current toward the leakage limit the Vr voltage varies. The leakage current is normally stable (at a given temp) for voltages over 1V.

## Now to your problem space, putting 3 Zeners in series.

The leakage current for each of the Zeners will be different. The Zener device acts as a constant current source once it has sufficient Vr and until tunneling begins. Leakage current will be typically in the 10-200uA range and you will find the maximum specified in the datasheet for some devices.

For your devices (1N728A and 1N729A) the leakage current will be below 100uA @1V and will begin Zener effect between about 1V and the rated voltage.

For your configuration ONE of the Zeners will have the lowest leakage current which will restrict the other two below their leakage current. This means that at very low currents (whatever that leakage value is) the voltage across the two of the Zeners will be much less than their rating, and may fall below 1V.

## First bump in the road is an ATTiny with much lower current when asleep

In your configuration the ATTiny85 will draw a very low current when in sleep modes and from the datasheet you can see that when asleep with the Watchdog timer used for wakeup you can expect less than 10uA.

This very low current for the ATTiny85 will restrict the current through the Zeners to below the leakage current of any of them.

This may result in the VCC for the ATTiny rising to above the Absolute maximum rating of 6V and result in device failure. And indeed you already showed that the VCC rises to 5.1V.

simulate this circuit – Schematic created using CircuitLab

This low voltage drop across the Zeners explains why your circuit works down to 7VDC on the battery, but it is not a good mode of operation and you could easily damage the MCU. My assumption here is that you actually have a capacitor across the MCU you have not shown, and you are effectively running the ATTiny from that when it wakes up.

## Viable Solution based on a regulator

Your best bet would be to invest in a linear series regulator. There are many variants available that have very low quiescent current such as the MCP1703 which would provide a single chip solution at low cost (

It's doubtful that you will find a switching regulator for the same price and low power.

• Thanks for the detailed explanation of how zener diodes work. I found the voltage drop I was looking for by trial and error but I was wondering if there was a good way to calculate the zener voltage drop. I know I'm running this circuit in a very non-standard way and the ATtiny is getting voltage swings but it doesn't seem to mind. I wouldn't design a pace-maker this way but this is for a Christmas tree ornament and is working well without a voltage regulator. – Jeff Wahaus Dec 12 '18 at 16:03
• @JeffWahaus. It's up to you of course, and since you have built and tested only one and it works maybe you are OK. Build more than one and you may start destroying MCUs. – Jack Creasey Dec 12 '18 at 16:45
• Understood. I'm building only one. I was getting by with the parts I have on hand and the only voltage regulators I have are L7805's which have 4.3mA quiescent current. Thanks for the link to the MCP1703 BTW. I wonder if it would run longer with such a regulator? I doubt it, the additional current draw is minimal and it keeps the voltage within range longer as the battery dies but since I'm driving 3 LED's at a time wired in series they will only light if the voltage of the battery is fairly high. The micro might keep running but the LED's will stop lighting at around 7V batt voltage. – Jeff Wahaus Dec 12 '18 at 17:41
• Your power requirements Chart for the ATtiny are high when active. I'm running at 1MHz and the current draw is more like 1 ~ 2 mA . Also, the cap was 100uF. I wonder what a simulation would spit out if given such circuit? I'm betting its answer would not reflect reality very closely. – Jeff Wahaus Dec 13 '18 at 12:46
• As I expected, the simulation I ran didn't reflect reality very well. It uses the zener value as the voltage drop value. The actual voltage drop is device dependant and is closer to 2V for 3.3 and 3.6 zeners and 3V for 5.1 zeners. The simulation also indicated the Vcc changes very little as the resistive load increases between 50uA and 5mA. – Jeff Wahaus Dec 13 '18 at 15:13

If you really need to save power, you must use switching mode power regulator. Everything else will burn current, with passive dividers, Zener parametric stabilizers, or LDO regulators.

To get optimal design, my advice is to go to TI.com, enter your design parameters (DC 6V-13V Vin, 5V 0.02A out), and launch their free WEBBENCH Power Designer. Here is the top of their 162 variants that meet your parameters:

If you don't want to mess with 7 components and inductor, here is a single-piece design using their module with integrated inductor:

But the latter is more expensive. Both designs will fit into 7x7mm area.

You might want to consider alternatives as LTC3642 ot ADP2360, they need only one inductor. I am sure there are many other options depending on market locality.

• Given the power consumed by the ATtiny, it's not really worth worrying about the efficiency of the regulator, the efficiency penalty of a simple linear LDO with a low quiescent current is tolerable and is smaller and cheaper. – Phil G Dec 10 '18 at 19:19

Fundamentally, the problem you're having is that you have an input voltage that varies over a wide range, and your circuit demands a much smaller voltage variance. Diodes drop a fixed amount of voltage, so as you've seen, while they work well to drop your maximum input voltage down to where you want it to be, they will shut off when your voltage drops.

What you need is a Voltage Regulator. A simple linear regulator should work okay, given your low current draw, but for maximum efficiency you might consider a switching voltage regulator.

• At extremely low output currents, the circuitry to drive a switching regulator can take a significant fraction of the input current. In this situation, linear regulators can actually be more efficient; you'd need hard numbers to determine if that's the case here though. – Hearth Dec 10 '18 at 18:25
• The ATtiny85 works over a fairly wide Vcc voltage. I'm running it at 1MHz so anything between 1.8V and 6V is fine. A voltage regulator will burn 5mA just like the voltage divider did. A switched dc buck converter would be better than a simple voltage regulator but it's too bulky for this project. – Jeff Wahaus Dec 10 '18 at 18:31
• True, good point @Hearth. A linear regulator will likely be better during standby, and although I think 0.3mA might still be better done by a switcher, it depends on the duty cycle whether that would be worth it. If it were me I'd probably slap a 7205 on there and call it a day. – Nate S. Dec 10 '18 at 18:31
• @JeffWahaus, not every voltage regulator will burn 5mA, and certainly will not be as bad as a voltage divider. Look for "low quiescent current" voltage regulators. – Nate S. Dec 10 '18 at 18:33
• @JeffWahaus: A LM7805 will burn 5mA. But it is a 40 year old design. There are voltage regulators with much lower quiescent currents. All the major manufacturers have voltage regulators with quiescent currents in the microamps. – TimWescott Dec 10 '18 at 18:34

This configuration does not save any power and provides unstable voltage. It might even contribute to higher sleep current because voltage grows when current falls. There is nothing than can be done to improve power efficiency, because in this configuration power is always current multiplied by supply voltage and all voltage overhead is wasted as heat on the diodes. Any other linear solution would only add extra consumption to power the regulating circuitry. Thus, the only way to reduce total power consumption is to use a switching regulator.

You should use a regulator. It's possible to make a crude regulator with a micropower shunt reference and a BJT but I would suggest not doing that.

LT3970 is a switching regulator that uses 2.5uA quiescent. Efficiency will be about 71% at 1mA out, 3.3V.

TPS709 is a linear regulator that uses 1uA quiescent current. Efficiency will be about 27% at 1mA out, 3.3V.

Both are very compact and easy to use. The better choice (of these or many other similar parts) will be determined by the exact requirements of your application.

Another solution is change your LED strings to <3V shunt strings then use CR123A cell 3V 1500mAh Li Ion vs. your A23 12V 55mAh.

The ATtiny85 works down to 1.85V at lower rates.

By changing the zener diodes to lower values and using 3 in series as well as moving the position of the capacitor the following circuit improves the usable supply voltage range to between 12V and 9V.

simulate this circuit – Schematic created using CircuitLab

Once started, the circuit continues functioning until the voltage reaches about 7.5. This is a big improvement over the original circuit's range and keeps the low power feature of the dc to dc voltage conversion. This circuit drops the battery input voltage by a more or less constant 6V.

The voltage drop introduced by each diode varies by the Zenier value. The circuit above uses one 3.6V and two 3.3V zener diodes. Zener diodes with higher breakdown voltages have larger voltage drops when conducting in reverse.

The ATtiny85 works over a wide voltage range (1.8V - 6V) so a regulated voltage source is not needed. With an average current draw of 0.2mA the A23 battery should last for around 275 hours.

Here is the complete schematic, it works well despite the voltage swings at Vcc which is between 6V when idle to as low as 3.5V when the LED's are pulsing rapidly.