That’s wrong.
The leds will get his negative from the chip, it needs the get the positive from the source.
It’s possible that it will only work with a led (or any load higher than an ammeter or multimeter)
Think of it like a mosfet. If voltage + is supplied on the VDD pin then current flows threw GND and OUT but is limited to 350ma per chip.
VDD would go to battery + and led +, they are shared. GND would go to battery - , OUT goes to led - .
The VDD pin can be PWM’ed for lower led output but the current is the same just a percentage of off time in a second. You probably already Know that.
Yea, the datasheet suggests it needs at least a 0.2v difference to work with 
The datahseet does indeed label the “IN” pin as “OUT”. Regardless of labeling your schematic is correct.
Are you sure you have enough voltage? 7135s have a minimum dropout voltage. The voltage “4.0” in your case has to cover the voltage over the LED at plus the minimum dropout voltage over the 7135, which is around 0.12V according to the sheet. I’ve never checked myself. If it doesn’t have the 0.12V dropout voltage it can’t do anything.
The 0.2V from the datasheet are between the Out and gnd pins, say that's what the 7135 needs for himself to be able to regulate the current
Example :
if the supply is 3.4V and the led eats 3.0V there is 0.4V left for the 7135 and it's ok but with the same led and a 3.1V supply then there is only 0.1V left for the 7135 and the current will decrease.
If you short the led it should still work (ie limit the current to 350mA) because then the 7135 will have the full supply voltage wich is higher than 0.2V (except that the 7135 will dissipate all the power and will heat fast)
If you share the Oshpark files maybe one of us can help you find the problem.
Or PM me and I’ll see if I can help.
All right, I’m back.
So the project is just to make a 1 amp dis-charger for 18650 cells. I thought (3) 7135 would be a simple solution.
I have learned by starting this thread that yes, my schematic was correct.
But here’s the issue:
Scenario #1
- Power up the (3) 7135 (with a 0.6 ohm load connected).
- Frustration. It is now drawing 135ma and will stay that way.
Scenario #2
- Power up the (3) 7135 with the load disconnected.
- Connect the 0.6 ohm load. This creates a 0.68v drop.
- Satisfaction. It is now drawing 350ma per 7135.
- Things get a little hot.
SO yay for the learning, crap for the project.
Try a bypass cap to VDD. Sometimes the control circuit needs to settle first before hitting it with full load. Same like trying to start a car’s engine if it’s still in-gear.
The state of VDD on boot does not change the results, so idk if that would work or not. But I’ve abandon the 7135 because of the intense heat they create. I expected the heat to be acceptable, it was not.
It’s a linear regulator. All that power has to go somewhere.
Well I figured if they survived inside a hot stick of aluminum, they would be even better on a bare board. But they even seemed to thermal-shutdown when run for a few minutes.
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.