Limiting battery input current to avoid voltage drop when activating LoRa radio

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.