- #1
Weaver
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Homework Statement
Homework Equations
CMRR = Av/Acm
Acm= Δ/R , Δ = (2 x Tolerance of Resistor).R
The Attempt at a Solution
I have to admit I am a bit confused by the premise of the problem.
My understanding is Vout is used to vary Vcharge in some way, so that Icharge is always 1.3 A. Then when the battery is charged to 4.2 V, Vcharge is held constant (at whatever it's voltage is at that time). As a result, the voltage across Rmon drops due to the battery voltage increasing and Icharge decreases as well. When Icharge drops to below 0.026A, Vcharge is switched off, presumably from the out Vout being at a set value
Is that correct?
If so I am confused by part (i) then. The voltage applied at the battery is supposed to be always above 4.2V - 1%. And the current following through Rmon is 1.3A. But we don't know an initial value for Vcharge. From my understanding, Rmon drops a voltage across it so that
Vcharge - Icharge.Rmon - Vbatt = 4.2V
With two unknowns and one equation(I think the equation can be simplified assuming Vbatt is 0 flat at the start of charging).
I know that assumption must be wrong but I don't really know why? Any help would be appreciated
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