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MrsssSu

Joined Sep 28, 2021
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Hi. above is a charging cut-off circuit. As you can see in my demo. when voltage source is 5V, the load (R9) has 5W power. When voltage drops to 4V, the load (R9) has 0 W (cut off). However, my lithium ion charging pin for my battery is 4.26V. Thus, when battery fully charged at 4.2V, the voltage drop between the 4.26V-4.2V=0.06V. This means I have to assume the voltage source in my demo to be lower than 0.06V to cut-off the circuit.

Question: How do I implement a 0.06V voltage drop in the voltage source when 0.7 is needed to turn on and off the transistors? Of course, you can say that you can just attach a 4V voltage source with the 0.06V but the 4V voltage source maybe coming from a battery will reduce over time because it consumes power to charge the load. What are some suggestions (You can download my LT Spice files below and place it in a new folder) to simulate it? You can attach schematics below :) .


Thank you
 

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ericgibbs

Joined Jan 29, 2010
14,189
hi Mrss,
This is what LTSpice shows, using a ramped 5v down to 4v Vsrc.
E


Update:
Added another sim showing temperature etc....
EG 865.png
 

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