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Switching battery through load

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Frongach:
Hi there,
I'm working on a circuit based on Adam Welch's 18650 capacity tester : ; where a logic-level mosfet switches the positive terminal of an 18650 through a power resistor to ground. I am trying to switch the mosfet out for a darlington npn transistor (TIP102), but can't get it to fully switch. I have a feeling the battery polarity is somehow creating a high side switch type situation, but just can't figure it out.
I've tried it with lower value base resistors too. I'm using an ATtiny85, it puts out 5.5v and can source 40 mA.
I've a attached a rough schematic, I'd really appreciate any help. I was thinking to put a pnp there but I'd like to understand what's going on also...
Thanks a million,
Brian

Ian.M:
It depends on the definition of 'fully switch'.  A Darlington can never have a Vce voltage drop of much less than one volt across it (compared to 0.2V to 0.3V Vce for a saturated single transistor), because  the output transistor of the pair cant turn on harder than that without reducing the collector voltage of its driver transistor to the point the output transistor starves itself of base drive.

A MOSFET can get right down to 0V Vds, as there are no junctions in its channel.  In practice Vds wont be zero as it will have *some* on resistance so if any significant current is flowing there will be some voltage drop, but it can be pretty close to zero at lo to moderate currents, with some MOSFETs having on resistances down in the few tens of milliohms region when their gate is driven hard enough.

spec:
Hi Frongach

The 18650 LiIon battery is shown on your schematic as having a terminal voltage of 4V, but the terminal voltage for LiIon batteries when not on charge is 3V6.

You have a 10R load resistor so the maximum current that could ever bee drawn from the battery (if you replaced the TIP100 darlington with a switch) is 3V6/10 = 360mA.

A TIP100 will have a VCE saturation voltage of around 0.8V at around 360mA, so that means that there would be 3V6- 0.8= 2V8 across the 10R resistor, so the current flowing through the 10R resistor will be 2V8/10R = 280mA.

In summary the VCE of the TIP100 will be 0.8V and the current through the 10R resistor will be 280mA. I bet that is what you have got. :)

What are you aiming at. If you give us a specification we can probably advise on a solution.

https://www.onsemi.com/pub/Collateral/TIP100-D.PDF

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