Charging a battery and monitoring the current w/ an instrumentation amp

[Weaver]

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 V_out is used to vary V_charge in some way, so that I_charge is always 1.3 A. Then when the battery is charged to 4.2 V, V_charge is held constant (at whatever it's voltage is at that time). As a result, the voltage across R_mon drops due to the battery voltage increasing and I_charge decreases as well. When I_charge drops to below 0.026A, V_charge is switched off, presumably from the out V_out 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 R_mon is 1.3A. But we don't know an initial value for V_charge. From my understanding, R_mon drops a voltage across it so that
V_charge - I_charge.R_mon - V_batt = 4.2V

With two unknowns and one equation(I think the equation can be simplified assuming V_batt 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

 

[Tom.G]

Then when the battery is charged to 4.2 V, Vcharge is held constant (at whatever it's voltage is at that time).

What does the voltage across V_MON do when the charging current changes from1.3A to 26mA?

Cheers,
Tom
(Tricky problem statement. I had to read it three times to keep track!)

 

[Weaver]

What does the voltage across V_MON do when the charging current changes from1.3A to 26mA?

Cheers,
Tom
(Tricky problem statement. I had to read it three times to keep track!)

The voltage across R_mon will also drop.

Oh! Would the voltage across R_Mon at that point 4.2V -1% referred to in the problem statement?

Is the voltage applied to the battery I_charge. R_Mon ?

 

[Tom.G]

Oh! Would the voltage across RMon at that point 4.2V -1% referred to in the problem statement?

Is the voltage applied to the battery Icharge. RMon ?

Umm, did you leave a few words or terms out of those questions? I can't seem to follow what you are getting at. (you may be close to the right track, though)

 

[Weaver]

Umm, did you leave a few words or terms out of those questions? I can't seem to follow what you are getting at. (you may be close to the right track, though)

Sorry that was a bad explanation

The voltage at the battery must never go below 4.2-1%V

When the battery is charged, it's at 4.2V

So would I_charge x R_mon be 0.042V (the 1%)?

 

[Tom.G]

Vcharge - Icharge.Rmon - Vbatt = 4.2V

Well, keep that in the back of your head for now. Something close to it will likely be used later.

Can you write the equation for V_charge as it varies with battery voltage, battery current, and the value of R_MON?
V_charge= {some function of} V_BATT, I_BATT, R_MON

Also sketching a graph for I_BATT and V_chargeversus Time may help you to see what is happening. There isn't enough information given to put values on the Time axis, so don't worry about labelling it. No need to post it at this time if it is inconvenient. If words fail us later I may ask for it though.

Cheers,
Tom

 

[Weaver]

Well, keep that in the back of your head for now. Something close to it will likely be used later.

Can you write the equation for V_charge as it varies with battery voltage, battery current, and the value of R_MON?
V_charge= {some function of} V_BATT, I_BATT, R_MON

Also sketching a graph for I_BATT and V_chargeversus Time may help you to see what is happening. There isn't enough information given to put values on the Time axis, so don't worry about labelling it. No need to post it at this time if it is inconvenient. If words fail us later I may ask for it though.

Cheers,
Tom

V_charge = V_batt + I_charge x R_mon

Here’s that graph

(Thanks for all your help)

 

[Tom.G]

I agree with both the equation and graph.
Looks like you've got a handle on it!

All that is needed now is to yank on the handle enough to answer the problem.

Cheers,
Tom