I have a battery-powered wireless sensor system that sends sensor information periodically. One requirement is that it has to be functional at very low temperatures. Powering this system is a Tadiran TL-5903 lithium battery, which are batteries made to power systems at very low temperatures and supplies 3.6V.

When the sensor sends out a transmission by LoRa radio, it can consume a peak current of around 100mA+. At low temperatures, the battery cannot provide enough current to supply the load during transmissions and the battery voltage drops from normal operation of 3V to below the minimum operational voltage of the microcontroller (2.2V).

The current power supply IC is a buck-boost chip TPS63802. It takes in the battery's voltage range (2.0V to 3.6V) to an operating voltage of 3.0V. However, the IC doesn't have an input current limiter that can limit the current so that it doesn't take too much peak current from the battery. At low temperatures, we can't take in too much current from the battery or else the voltage drops substantially.


simulate this circuit – Schematic created using CircuitLab

We tried putting a super cap at the output of the switching regulator but it doesn't seem to make much difference in the voltage drop when doing radio transmissions.

Is there an IC or a simple circuit that can limit the current coming into the regulator without dropping the voltage at the input of of the switching regulator?

  • 4
    \$\begingroup\$ You appear to be asking to violate the laws of physics. You need a substantial energy store : but you could make better use of the one you have if you put the cap on the TPS63802 input, where the voltage variation is, not its output, where as soon as the voltage dips a little the TPS63802 responds by draining the battery. \$\endgroup\$
    – user16324
    Commented Sep 15, 2020 at 18:19
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    \$\begingroup\$ Your diagram shows the LoRa radio's "VDD" sourced from an MCU "GPIO" - that is extremely wrong. LoRa chips should have good low power shutdown and not need power switching, but if they did, you'd need to use something like an external PFET and also make sure to take all I/O lines to the chip low when depowering, to avoid wasting energy partially repowering it through the I/Os. \$\endgroup\$ Commented Sep 15, 2020 at 18:24
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    \$\begingroup\$ Powering the TXCO via a GPIO is reasonable, and if you use it, yes you need to do that to avoid wasting power in between usage. But powering the general VDD as you actually drew would not be workable for higher power transmission - the ~100 mA you speak of is outside the current sourcing capabilities of MCU GPIOs. \$\endgroup\$ Commented Sep 15, 2020 at 18:31
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    \$\begingroup\$ Does the radio have other VDD pins? How are they connected? Why didn't you put TCXO_VDD on the schematic? \$\endgroup\$ Commented Sep 15, 2020 at 18:33
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    \$\begingroup\$ You can filter the current spikes with a cap, but even your average power may not be possible with a single cell. What is your transmit duration (packet size/rate) and interval? \$\endgroup\$
    – mbedded
    Commented Sep 15, 2020 at 19:05

3 Answers 3


As others have said, a current limiter is not what you need here. In fact, this is the very cause of your problem: a current limiter will decrease/block the voltage as it approaches the current limit, which is exactly what the battery is doing already.

The Tadiran TL-5903 indicates it can supply up to 200mA pulses, but it will not maintain the rated voltage for this. The discharge plot shows that the output sags to 3.4V@5mA and 3.1V@31mA. Working backward from your 100mA transmit, a transmit duty cycle beyond about 30% will be entirely impossible.

This all hinges on the average current, which you have already started to attack with the supercap. A cap will filter the current spikes, and ideally it will prevent the spikes from affecting the battery at all. This is not possible at the regulator output, though; the capacitor delivers current in response to a change in voltage, but the regulator's job is to make sure the output does not change. By the time the capacitor engages, the regulator will have already given up after bringing the battery to its knees.

Moving to the regulator input, the cap will engage as soon as the battery starts to sag. The question is now: will the supercap have enough charge to hold this for the entire transmit?

Wireless transmit will typically be short, but LoRa is not particularly helpful here. Since you emphasized using the maximum transmit power, I assume you also use the minimum rate to maximize your range; this may hurt your power even more than the 100mA. At 0.3kbps, just the packet preamble and sync symbols take 160ms; adding 16 bytes of user data, your transmit pulse will last over half a second at the slowest rate.

Since \$I = C\frac{dV}{dt}\$, we can see that your voltage will drop by \$dV=\frac{I}{C}dt\$, or approximately (100mA / 470mF * 0.5s) = 0.11V.

So. If you keep the transmit duty cycle below 30%, limit your transmit length to half a second (16 usable bytes at 0.3kbps), and move the supercap to the regulator input, this might actually be physically possible... at room temperature, with a fresh battery.

You may want to add a second cell.

  • \$\begingroup\$ In many regions LoRaWAN is limited to quite low duty cycles anyway (1% on most channels in the EU868 region for instance). So there’s really just a need to sustain the required voltage and current for a few hundred milliseconds, after that I would expect the device to go to sleep for quite a while (minutes, hours, possibly days). \$\endgroup\$
    – jcaron
    Commented Sep 15, 2020 at 20:56

A current limiter simply won't work - you can't get energy from nowhere, and causing the supply to "brown out" during transmit is likely to at best yield poor on-air results; more likely cause a crash and hopefully restart of the entire system.

To the best of my knowledge, all LoRa radio chips have programmable output power, so if you want to limit the power required from the supply, configure the radio to use a lower transmit power under conditions where the battery will be weak.

A sufficiently large capacitor could in theory be a solution if used with care; you'd probably want to charge it in advance of a transmit or receive operation. Bear in mind that at the slower (larger) spreading factors a LoRa transmission may take several hundred milliseconds to (in places where duration limits don't preclude it) well more than a second.

  • \$\begingroup\$ Using a lower transmit power is not an option. We absolutely need the maximum power output for our application. \$\endgroup\$
    – Alex C
    Commented Sep 15, 2020 at 18:20
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    \$\begingroup\$ @AlexC then thermodynamics is against you; there is no "free energy". Your only option is to find a power source which can deliver the needed power for the desired radio output power, or get the capacitor idea to actually work. Try watching the capacitor voltage with a scope during an actual transmit attempt. \$\endgroup\$ Commented Sep 15, 2020 at 18:21
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    \$\begingroup\$ @AlexC - "it should take it from the supercap in priority" - how would that help? If both sources together can't power the transmitter, using only the capacitor would be even worse. Putting the capacitor before the converter would allow more of the energy in the capacitor to be extracted. \$\endgroup\$ Commented Sep 15, 2020 at 18:58
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    \$\begingroup\$ @AlexC when a battery fails high-impedance, energy would be mostly pulled from the capacitor, not the battery. In reality you're probably draining the capacitor, which is why I suggested observing the capacitor voltage with a scope during the long transmit operation. If you really wanted to, you could rig up something with a silicon power switch (or even on the bench, a relay) to disconnect the battery and attempt the transmission entirely on the capacitor; you'll probably find you didn't store enough energy. \$\endgroup\$ Commented Sep 15, 2020 at 19:05
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    \$\begingroup\$ @AlexC Then use a bigger supercap. Also measure the voltage across the supercap and see whether it finishes recharging up to the battery voltage. If it doesn't finish charging before it starts discharging again, then you need a bigger battery. If it does, but it still drops too far, then you need a bigger supercap. \$\endgroup\$ Commented Sep 15, 2020 at 19:05

In order to utilize the energy in the capacitor better you need to allow it to swing over a wider voltage range. (be careful about the maximum voltage capability of the capacitor)

Putting it before the converter (i.e. across the battery) may be better as there will be a significant voltage change there allowing the capacitor to provide its energy to the load. The regulator will provide constant voltage to the load even though the capacitor voltage is changing.

Or you could use an additional converter to keep the capacitor charged up to its maximum voltage then convert down from there to the radio supply voltage. There would be an additional loss of efficiency due to there being two converters in series but it may be worth it to take care of this peak power requirement at low temperatures.

Capacitors have a different discharge curve from a battery; A battery is essentially an almost constant voltage source (except for changes in temperature, SOC etc). With a capacitor, the energy stored is proportional to the square of the voltage, in order to extract the most energy the voltage needs to be able to swing over a wide range with more energy being available at the higher voltages.


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