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Analog Volts

 
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ZZX
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Analog Volts
PostPosted: Tue Jun 13, 2006 11:26 pm     Reply with quote

Hi forum,
can the analog port read more than 5v, say 12v signal. If not, i can scale the voltage down using the divider circuit to read it. problem is that i have some resistive network further down the line and the current drops due to this divider circuitry.
Any suggestions? How can i read it without this problem?
asmallri



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Re: Analog Volts
PostPosted: Wed Jun 14, 2006 1:32 am     Reply with quote

ZZX wrote:
Hi forum,
can the analog port read more than 5v, say 12v signal. If not, i can scale the voltage down using the divider circuit to read it. problem is that i have some resistive network further down the line and the current drops due to this divider circuitry.
Any suggestions? How can i read it without this problem?


Use a high impedance voltager divider and feed the it to the PIC a/d input via an op amp configured as a voltage follower.
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rnielsen



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PostPosted: Wed Jun 14, 2006 7:51 am     Reply with quote

The specs usually state that the input can reach .3V above VDD which the max is around 6.5v - 7.5v, depending on the PIC you are using. Anything above this and you will, most likely, fry the PIC and let out all of the magic smoke.

Ronald
asmallri



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PostPosted: Wed Jun 14, 2006 8:08 am     Reply with quote

rnielsen wrote:
The specs usually state that the input can reach .3V above VDD which the max is around 6.5v - 7.5v, depending on the PIC you are using. Anything above this and you will, most likely, fry the PIC and let out all of the magic smoke.

Ronald


By using a CMOS opamp and running the opamp from the 5 volt rail, this problem is avoided.
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kender



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Re: Analog Volts
PostPosted: Wed Jun 14, 2006 10:49 am     Reply with quote

asmallri wrote:
ZZX wrote:
Hi forum,
can the analog port read more than 5v, say 12v signal. If not, i can scale the voltage down using the divider circuit to read it. problem is that i have some resistive network further down the line and the current drops due to this divider circuitry.
Any suggestions? How can i read it without this problem?


Use a high impedance voltager divider and feed the it to the PIC a/d input via an op amp configured as a voltage follower.


Or you could use a lower impedance divider with a ground resistor of 2.5k or less. This way you can avoid distortion without using an additional opamp.
Mark



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Re: Analog Volts
PostPosted: Wed Jun 14, 2006 11:58 am     Reply with quote

kender wrote:
asmallri wrote:
ZZX wrote:
Hi forum,
can the analog port read more than 5v, say 12v signal. If not, i can scale the voltage down using the divider circuit to read it. problem is that i have some resistive network further down the line and the current drops due to this divider circuitry.
Any suggestions? How can i read it without this problem?


Use a high impedance voltager divider and feed the it to the PIC a/d input via an op amp configured as a voltage follower.


Or you could use a lower impedance divider with a ground resistor of 2.5k or less. This way you can avoid distortion without using an additional opamp.


Lower impedance wouldn't work. If you reread his post he states that there is some resistance already and using the lower impedance causes a voltage drop. Now if this resistance were fixed, you could probably compensate for it. Depending on the resistance, the best bet might be to use an opamp.
ZZX
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PostPosted: Wed Jun 14, 2006 11:04 pm     Reply with quote

Thank u all for all the valuable suggestions,
Heres precisely something im doing :

14V --------/\/\/\/\/\---------- to load
14A | Rnet |
| |
| |
Div Network1 Div Network2

i.e im reading the voltages on both ends of Rnet(6.25milli ohms) and is meant to calculate the current. Now if my divider network 1 and two both have 3 resistors each of 4.7k, and not high impedance ones, since most current would flow straight to the load. Do u guyz think this solution would work provided i choose apropriate load which will be calculated as:

P= V x I
then R=P/(I)^2
ZZX
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PostPosted: Wed Jun 14, 2006 11:06 pm     Reply with quote

*The dividers in the schematic are at the ends of the Rnet resistors;
kender



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PostPosted: Wed Jun 14, 2006 11:31 pm     Reply with quote

ZZX wrote:
meant to calculate the current.

I would suggest to use a current sense amplifier such as MAX472 or MAX4080. You could also use an instrumentation amplifier, although you'll need a supply rail higher than the the voltage in the line you're measuring.

http://www.maxim-ic.com/appnotes.cfm/appnote_number/746
SherpaDoug



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PostPosted: Thu Jun 15, 2006 10:54 am     Reply with quote

This article describes your exact situation. How to monitor voltage across a small shunt resistor in a +12V bus using an A/D that reads a small voltage relative to GND.
http://www.electronicdesign.com/Articles/ArticleID/12629/12629.html

Look for the "Associated figure" link near the bottom.
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ZZX
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PostPosted: Tue Jun 20, 2006 1:47 am     Reply with quote

Hi folks,
Im back with the issue. I tried reading voltage across a shunt using dividers. Each divider has 3 x 4.7k Resistors, so 11.44 volts at both ends of the shunt gives about 3.8V across each resistor. Now everthings at hand except a small problem. In my divider circuit, one side measures 3.78v while the other 3.79 volts. This creates a problem when i upscale it i.e x3 to know the actual voltage. Hence at the terminals when the actual voltages are 11.44,11.44(across shunt), here i get 11.37 and 11.34v(upscaled calculated). ive used 1% tol resistors here.
Can anybody suggest how to reduce the gap? (0.03 difference )
kender



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PostPosted: Tue Jun 20, 2006 11:34 pm     Reply with quote

ZZX wrote:
ive used 1% tol resistors here.
Can anybody suggest how to reduce the gap? (0.03 difference )

- You can measure the exact value of all of your resistors with a meter. Meters typically give 4 digits, which might be equivalent 0.1% presicision (there's probably an exact number for the precision in the meter's manual). Store the resistor values in thwe EEPROM and apply corrections to your readings.
- You can buy 0.1% resistors. You can also hand-pick resistors from the batch.
- Or (best of all), you can use one of the proposed differential/instrumentation/current-sense amplifier topologies.
- Hall effect current sensors are often used for high current. They are made by Allegro http://www.allegromicro.com/hall/currentsensor.asp and Zetex


Last edited by kender on Tue Jun 20, 2006 11:53 pm; edited 1 time in total
Eugeneo



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PostPosted: Tue Jun 20, 2006 11:53 pm     Reply with quote

It looks like you have a power supply of 14 volts. Could you use an opamp that accepts inputs of 14 volts then follow and divide?
SherpaDoug



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PostPosted: Wed Jun 21, 2006 7:33 am     Reply with quote

There are dedicated chips that will do this much better and cheaper than hand matching resistors. Look at Maxim MAX471 & MAX472. Maxim even gives free samples at their web site.

If you really want to roll your own try either of these two designs from the article I mentioned before.
http://www.electronicdesign.com/Files/29/12629/Figure_01.jpg
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