Calculation help?

M

MartinM57

Moderator
Assuming the answer is found just by applying equation (4) on page 9, then you can find K from your measurements (0.75 = 1000 x 3.75 / K), and then reapply equation 4 for any R value and max load to see what Vsense you get.

But the easy way would be to assume K is constant and so there is a proportional relationship between Vsense and Iout, so...
if Iout = 3.75A gives 0.75V Vsense
then 5V Vsense will occur when Iout = 5/0.75*3.75 = 25A.

But you also need to read the top of page 10 that says Vsense is limited to (typically) 5.5v under short circuit conditions (when equation (4) can't hold as Iout is tending to infinity).

So actually the exercise is how to deal with Vsense being 5.5v when you have a nominal max Vin on the PICAXE of 5v. This is where your "problematic" R2 comes in to limit the input current in this situation - to cut a long story short, I'd use 10K for that (and ignore the neer'sayers about source impedance :))
 
hippy

hippy

Ex-Staff (retired)
Also, don't forget to test it disconnected from the PICAXE under fault conditions, over-current and in thermal shut-down with maximum Vcc to check the behaviour. Theory is all fine but it's good to have confirmation of that before everything does go up in smoke when being used in anger.
 
R

rmtucker

Member
Martin

Thank you for explaining that in some kind of language i can understand.
Could you maybe explain how to re-arrange the formulae to tell me what size rsense resistor to use to give 5v at 30Amps?

Hippy
I do try it disconnected from the picaxe but what worries me is that in the case of a bulb,the inrush ampage can be 3 or more times the normal ampage,so what happens to the adc in that situation.
Or is the vsense limited by the 10k resistor?

Sorry for the numb basic questions but i am a little rusty at all of this.
 
hippy

hippy

Ex-Staff (retired)
rmtucker said:
I do try it disconnected from the picaxe but what worries me is that in the case of a bulb,the inrush ampage can be 3 or more times the normal ampage,so what happens to the adc in that situation.
Click to expand...
Pass. Slap a load in which causes the same inrush current and see what happens. Otherwise you have a solution based simply on guesswork and hope.
 
M

MartinM57

Moderator
Briefly (BlackBerry keyboard too small)..
Working out K from your observation and formula (4)
K=1000x3.75/0.75=5000

So V=IxR/5000

So 5=35xR/5000, so R=5x5000/35=714 ohms

I'd use 680 and 47 in series (or 820) and try to draw say 10 and 20 amps (will be a challenge and exciting both at the same time) and check Vsense with a voltmeter rather than the PICAXe to make sure all is as expected. Hope this helps...
 
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