Select a ‘dummy load’ resistor value that will sink your one amp at about 2.8-3 volts or so.
…Or use an LED as the load… then you have an indicator light.
Don’t ask the 7135 to sink all the current!
Point was there’s got to be something different going on, when the load’s connected or not on startup.
Wellp, 7135s (or any linear regulators, for that matter) work beautifully when the load is almost matched to the source.
Instead of a linear load like a resistor, try a nonlinear load like a hotwire bulb, 3W or so. Resistance is low at low voltage and increases with higher voltage (and wattage) and almost “saturate” at their working voltage/wattage. So something that draws ~1A at ~3V would still draw just a bit more current at 4V vs 3V, vs 33% more.
That should do more heavy-lifting so the 7135 doesn’t have to.
Of course, you’d have to experiment with which bulb would work best, but…
Small bulbs are actually used to stabilise feedback in some oscillators, precisely because the effective resistance would change nonlinearly.
That was a common circuit back in the old National Semiconductor databooks (the dark blue “telephone books”) that were a treasure-trove of incredibly useful circuits.
That’s a pretty good explanation. Thanks.
Ohm’s Law:
V/I=R
$ = (∑ evil)½
∴ $² = ∑ evil
If I'm all evil, shouldn't I get my money squared?
Uhhh, forget all the nonsense, thought this could be a teachable moment here.
Just use a 3ohm 5 watt resistor for your load then. Might want a bit less resistance if you want to drain your battery below 3 volts at 1 amp. A bit of heat sink on the 7135 and they will be fine.
