# Low-cost, low-current solution for dropping 1V from a battery

I have a 3 x 1.5V NiMH battery pack giving a supply voltage in the range [3.0V ... 4.5V] depending on the remaining capacity. In order to prolong battery life I need to cut power from the entire PCB except for the MCU sleep module that consumes a few uA. The catch is that the sleep module is rated at [1.8V ... 3.6V] and I cannot afford an LDO for this purpose.

Is there a way to achieve a voltage drop between 0.9V and 1.2V without any significant current draw?

• 3 * NiMH gives 3.6V. You can use it directly. Mar 8, 2012 at 20:32
• Yes, but when charging each cell may reach 1.5V. Mar 8, 2012 at 21:03
• @PerE - What precisely do you mean by 'cannot afford an LDO' - Is an LDO unacceptable due to financial cost, space, or power constraints? What cost, space, or power constraints are acceptable? Mar 8, 2012 at 21:27
• @KevinVermeer: I'm sorry for being unclear. Cost is the real issue and considering that I'm already using an LDO for VDD it feels like an 'overkill' solution to use a secondary LDO for the sleep supply. Mar 8, 2012 at 21:41
• Is this for volumes of 1/ 10 / 100 / 1000 / 100,000 ...? Mar 9, 2012 at 1:05

This microchip appnote suggests in section 8-3 to use some rectifier diodes in series as shown in Figure 3-1, which converts a 5V supply into an approximately 3.9V supply.

Each diode will give you a voltage drop, depending on the type of the diode and the current through the diode. Note the inclusion of R1, which the appnote mentions

is present to keep the voltage at the PIC MCUs $V_{DD}$ pin from exceeding the maximum $V_{DD}$ at minimum loads (typically when the PIC MCU is in Reset or sleeping). Depending on the other circuitry connected to $V_{DD}$, this resistor may have its value increased or possibly even eliminated entirely. Diodes D1-D3 must be selected so that at maximum load, typically when the PIC is running and is driving its outputs high, the voltage drop across D1-D3 is low enough to meet the PIC MCUs minimum $V_{DD}$ requirements.

The downside of this approach would be that you will lower the voltage even when you are below the 3.6V, so you will not have the full operating range.

Be extremely careful when selecting diodes; $V_F$ is not constant with respect to current or temperature. Here's an example relationship from the Fairchild 1N414 datasheet:

The forward voltage is proportional to the log of the current until about 100mA, when it begins to increase more rapidly due to carrier saturation. You mention that your device has a minimum current of a few μA, for which you probably paid dearly. You could increase this quiescent current to a little more than 20 μA with a 220 kΩ resistor and put 2 diodes in series to get 450mV drop per diode and a safe output voltage of 3.6V when your batteries are at 4.5V.

Of course, when the MCU wakes up and draws more current (assuming ~10mA) the voltage drop will go up to about 2⋅700mV=1.4V. Instead of operating from your batteries in their full charge range, you'd only be able to discharge your batteries to 1.8V + 1.4V / 3 = 3.2V / 3 = 1.06 V/cell. There's some optimization to be had in increasing or decreasing the resistor value and diode count, but it's difficult to get good results as you could with an LDO.

Conclusion: Don't use this approach unless your application is extremely cost-sensitive!

• This could be OK, given that the forward voltage drop at a few uA is high enough. When I scan through different datasheets I always find that VF is undefined in these regions. Do you have an example diode? Regarding the operating range, it is perfectly OK :) Mar 8, 2012 at 20:47
• I'd add a note that most projects can afford an LDO, and that's a much better solution than series rectifier diodes. Using the $V_F$ drop of a diode is a rather imprecise and wasteful hack. As Per E points out, $V_F$ at low currents is often hugely different than $V_F$ at high currents. It's definitely not a constant $0.7\mbox{ }V$, despite what you may have been taught. The Tips 'n Tricks appnote should be used with caution! Mar 8, 2012 at 20:50
• I hold nothing against this AN (I've actually used a couple of their tips previously). Back to the question - is there really no other solution? Mar 8, 2012 at 21:00
• @Kevin Vermeer: I took a look at the 1N4148 Datasheet. For 1uA the voltage drop is 275mV and at 5uA the voltage drop is about 350mV. Kinda too low. I guess it was a bad idea and I will delete this answer soon. Mar 8, 2012 at 21:06
• @PetPaulsen: I cannot find any information about sub-mA VF, where exactly are you looking? Mar 8, 2012 at 21:21

A common diode in series is not a good idea, like PetPaulsen also showed: the voltage drop varies too much with current.

A voltage reference diode like the LM285 in series with your battery is a better choice if your current is maximum 20mA. Voltage drop is pretty much a constant 1.235V from about 3$\mu$A to the maximum of 20mA.

• Thanks Steven! Not sure I can afford to loose 3 μA standby current in this particular project but I'll keep it for future reference. May 4, 2012 at 19:15

NimH at low current will provide say 1.1 - 1.35 Volt out. Voltage will drop below 1.3V/cell early in discharge range. At low Iout 1.1V is close to exhausted if very good cell life wanted, but you can go substantially lower. If you design 3.3-3.9V or say 3.0 - 3.9V you will recover the large majority of available energy. You do not say what operating current drain is or acceptable vsupply when operating, and these will influence the acceptable design.

Charging voltage max per cell depends on max current charge rate. At moderate charge rates Vmax_chg is typically 1.45V (3 = 4.35V) and 1.4V can be used with minimal capacity loss.

If a reduction of battery voltage range by a fixed amount or ratio was acceptable then an FET with exceptionally high value gate resistors could accomplish this. You can buy "zero gate voltage" FETS which used isolated charge storage to bias the gates to "just not on" at 0 volts $V_{GS}$ so that the source voltage then equals the applied gate voltage. "A bit of playing" would yield a solution similar in performance to the above IC solution iQ wise but utterly inferior voltage accuracy wise. The IC provides an immensely sophisticated and accurate result and is well worth considering.

• I mean \$. Also, I don't follow your reasoning about the supply voltage. The battery configuration is fixed at 3 NiMH in series, and when charging the potential will reach 4.5V. This is above the maximum rating of the sleep supply pin. In running mode, a 3V LDO is used but this is not related to the question. Mar 8, 2012 at 21:27

One 1N4007 will drop about 0.85 volts in the forward direction. It costs watt-hours, but not current.

Just because you supply 4.5V into the batteries doesn't mean you actually get 4.5V across the batteries. They will keep their charge state voltage, plus some voltage across the resistance of the hook-up. The rest of the voltage will be lost across the circuitry going into the batteries. The lower the battery charge, the higher the current they want, and thus the more losses you will have in the circuitry around the battery. Thus, the batteries won't get to 3*1.5V until they are completely full -- and you probably want to cut off before then, say at 3*1.4V. It's likely your regulator will buckle and provide less than 4.5V way before the batteries would start saturating and push the voltage up (but I don't recommend pushing it that hard.)

The only thing that may cost less watt-hours than a diode to drop the voltage is a highly efficient switching power regulator, which is going to both cost a lot of money, and take a lot of board space.

• As explained in PetPaulsen's answer, $V_F$ in diodes is not constant. A rectifier like the 1N4007 will drop .85 volts at 1A at 25C. At the level of a few &mu;A drawn by the sleep module, this will be much lower. Extrapolating from the log-linear response described in this Vishay datasheet below 200mA, I'd estimate a $V_F$ of around 0.4V (though it's not guaranteed, as it's not in the datasheet). Mar 9, 2012 at 4:30