I notice it has over-current protection and under-voltage lockout. Are these features “ok” to rely on?
More specifically: I’m thinking of using it to step up voltage from a single 3400mAh 18650 lithium ion battery (which has voltage protected in the range of 4.2V down to 2.75V), up to a regulated voltage somewhere in 6 to 9V range.
At the lower end, 2.75V is below the suggested minimum of 2.9V. What would happen when voltage is in that range?
In my project the regulator would drive two HP micro-metal motors which will typically require less than 600mA in total, which the efficiency graphs suggest the regulator would easily handle. But the motors may very rarely stall, and require 1.6A each, or 3.2A total. I’ll have encoders detecting for this so should be able to stop it quickly (but am not sure yet exactly how quickly). May an attempt at drawing this current damage the regulator, or can I rely on the over-current protection?
(Previously I had been planning to use two of these batteries, but if I can get away with one that would be great for weight reasons).
Edit to post: to clarify, my question isn’t about whether or not the motors will work perfectly (since clearly there won’t be enough current available under some circumstances) but whether it will damage the regulator trying.
The under-voltage lockout for that regulator does not activate until at least 2.7V, and might not activate until the voltage drops to 2.5V or lower, so it would not be appropriate for protecting your battery from over discharging. A search for “Li-Ion battery protection circuit” should turn up a lot of useful hits that discuss how to protect your battery.
The over current protection of the regulator can activate with input currents of anywhere from 5.25A to 7.75A, so depending on whether both of your motors stall at the same time and what the threshold is for your particular regulator, you might not trigger the over current protection. Even if you don’t trigger the over current protection though, 1.6A is still near the limit of what the regulator can do when converting from around 3V to 6V or 9V, so I expect that any prolonged current draw will just cause the output voltage to drop and/or the regulator to go into thermal shutdown until the current decreases. I don’t expect it to damage the regulator.
Thanks very much for your answer. The batteries have a built in protection circuit (at 2.75V) so I’m not worried about the batteries, rather just going below the stated minimum for the regulator. So - can I assume the voltage regulator won’t be damaged with an input voltage between 2.75 and 2.9V? Would the likely outcome just be that it can’t meet the targeted output voltage?
And regarding the higher current, since any bursts of higher current will be very short (so I presume less likely to trigger thermal shutdown), it doesn’t sound like I have much to worry about.
