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#1
04-27-2011, 01:21 PM
 and1_hotsauce Junior Member Join Date: Apr 2011 Posts: 19
how to connect Pot in series with resistor

Hi,

I'm trying to connect the circuit pictured to the left. Both resistors are going into pin 2. How would I be able to make 5k a variable resistor?

I have connected leg 3 of POT to resistor. But would that not also make the 10k a variable resistor??

#2
04-27-2011, 01:37 PM
 DerStrom8 Senior Member Join Date: Feb 2011 Location: Vermont, U.S.A. (GMT-5) Posts: 1,883 Blog Entries: 12

Quote:
 Originally Posted by and1_hotsauce I have connected leg 3 of POT to resistor. But would that not also make the 10k a variable resistor??
No, if you connect it the way it is drawn, the 5K pot will work independently of the 10K fixed resistor. It appears that you have it connected correctly.
Der Strom
#3
04-27-2011, 01:39 PM
 Adjuster Senior Member Join Date: Dec 2010 Location: London UK (GMT) Posts: 2,147

Your sketch on the left side of the picture shows one fixed resistor and one variable resistor connected to pin 2 of an IC. The breadboard seems to show two resistors connected to pin 2, one of which is a 5k variable, connected to another pot of unspecified value.

You then ask if connecting the pot would make another resistor variable. I would think not, although the total resistance would be variable, within a limited range. Could you explain more clearly what you are trying to do?
#4
04-27-2011, 01:58 PM
 and1_hotsauce Junior Member Join Date: Apr 2011 Posts: 19

sorry let me be a bit more specific

Assume I dont have a 30.6K and 510 ohm resistor(fig below). Instead I have a 28K and 500 ohm resistors instead. I want to connect TWO pots in series with each resistor in order to bring the value up to 30.k and 510 ohm respectively.

The dilemma is, if I connect a pot in series with R1 does it not automatically make R2 a variable resistor as well?

In the circuit above I have connecetd the pot to 30k pot to the 5k resistor FOR EXAMPLE, to bring it to 30.6 k. By twisting the knob, does it effect both resistors?

#5
04-27-2011, 02:02 PM
 and1_hotsauce Junior Member Join Date: Apr 2011 Posts: 19

Quote:
 Originally Posted by DerStrom8 No, if you connect it the way it is drawn, the 5K pot will work independently of the 10K fixed resistor. It appears that you have it connected correctly. Der Strom
What you mean? Both resistors are connected to pin 2 which is aht I have drawn (ms paint). How do I connect the pot to control only one of the resistors?
#6
04-27-2011, 02:29 PM
 Adjuster Senior Member Join Date: Dec 2010 Location: London UK (GMT) Posts: 2,147

If the 10kΩ is a fixed resistor, then it will not change just because another resistor connected to the same point is variable.

If the two resistors are connected in series or parallel, then the total equivalent resistance will vary, but this is not the same thing. The 10kΩ remains 10kΩ, only an equivalent resistance changes.
#7
04-27-2011, 02:36 PM
 and1_hotsauce Junior Member Join Date: Apr 2011 Posts: 19

thanks for that! So can I say now that my 5K resistor is effectively a variable resistor? Is this the proper way to connect a pot to a resistor in order to make it variable??

Last edited by and1_hotsauce; 04-27-2011 at 02:56 PM.
#8
04-27-2011, 04:01 PM
 bluemarvin New Member Join Date: Apr 2011 Posts: 8

The 3 pot terminals may be confusing you...you need to use only the wiper and either of the other two terminals...this will vary the resistance across those two points from 0 to 5k or whatever the pot value is. Connect one of those two points to your fixed resistor in SERIES and you will make a variable resistor ranging from the fixed value to the fixed value plus 5k. You could also connect them in parallel but that would only get you 0 to 3.33k if the fixed is 10k and the pot is 5k.
#9
04-28-2011, 12:34 AM
 and1_hotsauce Junior Member Join Date: Apr 2011 Posts: 19

thanks got it now

Last edited by and1_hotsauce; 04-28-2011 at 12:49 AM.

 Tags connect, pot, resistor, series

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