Subject: Re: [Shop-talk] Need a variable 85-285 ohm resistor

From: "Mark Watson" <[email-masked]>
Date: Tue, 12 Mar 2013 20:42:49 -0400 s=q20121106; t=1363135372; bh=eXg8q59n4wECb1vmG2jg0Eu+eaZvSa3fRlWn1E/aAH4=; h=Received:Received:Message-ID:Reply-To:From:To:Subject:Date: MIME-Version:Content-Type; b=SD7nHOMv9ook4J+vrsw2xKKec8aPHbbSYjf06nUDjYrQ5zd1/WUgrE2SIHuAVKvWR I2vdd4cNDfHxyPnHkwssYiJHq8HgZDHv8yTUKrTABbNOT4W1TLfehNjGBlsbDevv3F +1zNftmVFIuNtEhSry7GXRFw8+y03IyxVqZW4/GxCU4GDLkhaeChyd51d3Tc5y2bfo 2WPH1EtgAML4gj0nMXzWkB9gFdNE/pEM/Aj4KcfZcb8AI1q2IL6Ml2bh605Xsn4N+j K8ou7rbsieNb04LCnn/RWpfrQd8Uz87vzS7UXRvHO6EGpDbpgXqH2WL4IuqkkSkUwM I54eqgRMGpjFg==

Hi, Darrell,

    OK, the Electronics teacher in me just had to come out.  I poked around 
on the net a little and couldn't find many details about how an electric 
gauge like this would work.  My concern is about the power that a pot would 
need to dissipate to simulate the sender.  According to one description the 
circuit in the meter applies a constant current through the sensor so the 
meter is just basically an ohmmeter with the meter scaled appropriately.  If 
this is true of yours and the current is low enough you won't have any 
problems.

    Based on my reading on the net I don't think the following is an issue 
but I'm going to mention it anyway:

    On the other extreme if a full 12V is applied to the sender then at the 
lowest resistance it would pull I = V / R = 12V / 85 Ohms = 0.14A.  If you 
use a potentiometer set to 85 Ohms then the pot would be dissipating P = V^2 
/ R = 12^2 / 85 = 144 / 85 = 1.7 Watts.  The two pots mentioned so far would 
fairly quickly fry trying to dissipate this amount of power.

    I did see one mention of 5V being applied to the sender.  In that 
instance the power dissipation would be 5^2 / 85 = 0.3 Watts.  The pot that 
Greg linked to is a 1/2 Watt unit so should would survive this treatment 
undamaged IF you don't go much lower than that resistance (P = V^2 / R so R 
= V^2 / P = 5 ^2 / 0.5 = 50 Ohms minimum).

    Now you've got me curious about how an electric oil pressure system. 
Blast - yet another thing to investigate ;-)

    Hope this helps and doesn't confound the issue.

Mark Watson
1956 Daimler Regency Mk II '104' (stalled restoration :-(
1965 Ford Falcon - brake work
and various other uninspiring transportation pods

-----Original Message----- 
From: Randall
Sent: Tuesday, March 12, 2013 6:43 PM
To: [email-masked] Talk ; DarrellWalker
Subject: Re: [Shop-talk] Need a variable 85-285 ohm resistor

Radio Shack unfortunately doesn't carry much in the way of components any 
more.  You're lucky if they get even close to what you want.

But in this case, I think that
http://www.radioshack.com/product/index.jsp?productId=2062307
will do well enough.  You won't be using its full range, but since it is 15 
turns from 0 to 1K, you'll still get about 3 turns between 85 ohms and 285 
ohms.

Randall
_______________________________________________

[email-masked]
Archive: http://www.team.net/archive

---

From: Darrell Walker <[email-masked]>
Date: Tue, 12 Mar 2013 17:50:53 -0700

Hi Mark,

Thanks for the reply.  I believe you are correct, the gauge is just an ohm
meter, with the resistance values mapped to pressure (in this case, temp
gauges work the same way).

The gauge has three connections, +12V (or whatever the electrical system is
producing, I believe the gauge has its own voltage compensation, unlike the
older gauges), ground, and the sense wire to the sender.  The sensor would
provide resistance between the spade connector and ground.  So I should be
able to connect the gauge up to 12 volts, and then measure voltage between the
sense terminal and ground to see what voltage the gauge uses to read the
resistance, right?

-Darrell

On Mar 12, 2013, at 5:42 PM, Mark Watson <[email-masked]> wrote:

> Hi, Darrell,
>
>   OK, the Electronics teacher in me just had to come out.  I poked around on
the net a little and couldn't find many details about how an electric gauge
like this would work.  My concern is about the power that a pot would need to
dissipate to simulate the sender.  According to one description the circuit in
the meter applies a constant current through the sensor so the meter is just
basically an ohmmeter with the meter scaled appropriately.  If this is true of
yours and the current is low enough you won't have any problems.
>
>   Based on my reading on the net I don't think the following is an issue but
>   On the other extreme if a full 12V is applied to the sender then at the
lowest resistance it would pull I = V / R = 12V / 85 Ohms = 0.14A.  If you use
a potentiometer set to 85 Ohms then the pot would be dissipating P = V^2 / R =
12^2 / 85 = 144 / 85 = 1.7 Watts.  The two pots mentioned so far would fairly
quickly fry trying to dissipate this amount of power.
>
>   I did see one mention of 5V being applied to the sender.  In that instance
the power dissipation would be 5^2 / 85 = 0.3 Watts.  The pot that Greg linked
to is a 1/2 Watt unit so should would survive this treatment undamaged IF you
don't go much lower than that resistance (P = V^2 / R so R = V^2 / P = 5 ^2 /
0.5 = 50 Ohms minimum).
>
>   Now you've got me curious about how an electric oil pressure system. Blast
- yet another thing to investigate ;-)
> -----Original Message----- From: Randall

[email-masked]