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Topic: Voltage Divider & Voltmeter help (Read 7538 times) previous topic - next topic

scottydb4b

#15
Oct 29, 2011, 08:13 am Last Edit: Oct 29, 2011, 08:18 am by scottydb4b Reason: 1
The 4.555 is the voltage I got when I measured the analogue input variable power supply and multimeter, so 1023 is reached at roughly 4.555 volts not 5 volts exactly, unless its the multimeter telling me different.

and I think i just added the *2 in there just to make the amps close to what it is ment to be. But yes i got a 0.5 ohm shunt on the ground rail.

Im no coding expert and just improvising where i can.

thanks for the help though.

Grumpy_Mike

What does the +5V pin measure? The 4.555V is a bit low, maybe your meter is off? All measurements are relitave to the voltage on the +5V pin.

dc42


So I will change out the resistors to be 4.7k & 10k respectively.. and lets me keep the draw form the 'meter' low...

but why @ a stable 7.4v from the PSU am I getting readings of 1023 from +5v - +7.4v??


My guess is that either the resistors you are using are not really 470 ohms and 1K (e.g. the one you think is 1K is actually 10k, or the one you think is 470 ohms is actually 47 ohms), or you are using one of those breadboards that has a break in the middle of the 5v and gnd rails and you haven't taken account of that break (so the 1k resistor is not connected to ground).
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

scottydb4b

Yeh the 4.55 might be a bit low but at the end of the day i have to calibrate it to something and thats the reading i get @ 1023, its really no big deal i can change it to 5 but is less accurate, according to my meter. But its doesn't have anything to do with the voltage drop issue, or lack of it should i say.

xl97

ack!.. you thread stealer!  =)

anyways..


@ CR-

5V = (8.4V x 10000)/(x + 10000)

should be:

5v = (8.4v x 10000) / (6800 + 10000)

right?



Im still a bit unclear that even with the other resistors in place for a the 7.4v setting.. 'why' it didnt represent a 7.4v 'value' in the serial print?

I mean it was steady 1023 from +5v up to 7.4v..??

I understand why change the resistors.. (since in the end I'll be using 2 x 3.7v li-ions)..

but what was wrong with it currently.. at the 7.4v??  


In my mind.. changing out these resistors will do NOTHING for me?  (but are only to be put in place because in the end I'll be using a 2 x li-ions)

the question (for me) still remains and is unanswered..

WHY with the current set-up of resistors/voltage divder.. and a 7.4v source am I NOT getting any changes in my analogRea() from 5v to 7.4v?

xl97



So I will change out the resistors to be 4.7k & 10k respectively.. and lets me keep the draw form the 'meter' low...

but why @ a stable 7.4v from the PSU am I getting readings of 1023 from +5v - +7.4v??


My guess is that either the resistors you are using are not really 470 ohms and 1K (e.g. the one you think is 1K is actually 10k, or the one you think is 470 ohms is actually 47 ohms), or you are using one of those breadboards that has a break in the middle of the 5v and gnd rails and you haven't taken account of that break (so the 1k resistor is not connected to ground).



sorry I missed this post.. (DOH)..


is there a way I can meter my resistors? to accurately find out the values?

or I suppose I could also try to look up the color bands?  (always hate that)..


any better way?


xl97

ok..

I took BRAND NEW 4.7k Ohm and a 10k Ohm resistor (fresh out of the pack)...  and put them in place

4.7k in between the +Vin and +Vout

the 10k between + and the -   (same as illustrated in the picks.. (but the values changed)

and I 'again' get the same results?  once I turn the PSU up.. and I get to +5v.. Im already capped at 1023 analog value??

what the 'heck' am I missing here? lol..

I dont even have the LEDS in the mix or part of the project until I can fix the analog read values to be 'somewhat' near my ranges..??

I mean damn.. I only have 2 wires going to the Arduino!!!  LOL  GND and A5

some resistors..and two leads going to my variable PSU..




dc42

It sounds to me that you do not have the 10K resistor connected as you intend. Can you post a photo of your wiring?

If you disconnect the 4k7 resistor, do you get a stable reading of zero from the analog pin?
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

Runaway Pancake

#23
Oct 29, 2011, 04:19 pm Last Edit: Oct 29, 2011, 04:28 pm by runaway_pancake Reason: 1
How about borrowing that 10K pot from your LM386 project.
Since you're just experiencing, using a pot will make things a little easier.

Edit -- corrected drawing
"Hello, I must be going..."
"You gotta fight -- for your right -- to party!"
Don't react - Read.
"Who is like unto the beast? who is able to make war with him?"

xl97


It sounds to me that you do not have the 10K resistor connected as you intend. Can you post a photo of your wiring?

If you disconnect the 4k7 resistor, do you get a stable reading of zero from the analog pin?



hi-

her is the image I made of it.. (using a breadboard)



I'll see if the camera is charged (usually isnt)...  and post one up of a 'real pic'..if it'll help.


@runaway_pancake-

I'll grab another 10k post I have laying around to use..

and look at your diagrams.

thanks

xl97

@runaway_pancake-

I did as your diagram outlined..and it works 'perfect'...

I even left the meter included when I hooked up the Arduino just to keep an eye-on readings.

SO goes back to..  I must be doing something wrong on the breadboard set-up?

here is a pic:



thanks for the help guys!.




dc42

Your 4k7 resistor is shorted out because you have both ends in the same row of the breadboard.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

xl97

got ya!.   :)
thanks.
I have also corrected my diagram to show my error/fix




I am now going to go back and change out the resistors to match for 8.4v input.

Thanks to everyone for your help.   :D

CrossRoads

So you've had 7.4V going into your analog pin all this time?  That's not good.
Does the chip feel warm when powered up now? If so, you likely blew the pin.
And if not, it may be partly damaged and fail on you eventually.

"is there a way I can meter my resistors? to accurately find out the values?"
Yes, just measure them with your multimeter. You have gold bands on the resistors? I think that's 10% tolerance (maybe 5%, I'm  not sure which).
If you know the actual values, you can use that in your math for more accurate results.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

Runaway Pancake


I would just add: Ohmmeter your resistors out of circuit,
Never* put an ohmmeter on a live/powered circuit.

(* For your meter's sake.)
"Hello, I must be going..."
"You gotta fight -- for your right -- to party!"
Don't react - Read.
"Who is like unto the beast? who is able to make war with him?"

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