potentiometer

[Bibio]

if i was to take a 50k pot and attach a 20k resistor to the input and wiper pins would i then have a pot that went from 20k to 50k instead of 0 to 50k?

 

[AndrewT]

No.

At top of travel the 20K has not effect the Input impedance is 50K and the output impedance is the Rs that feeds the vol pot.

At bottom of travel the attenuation is maximum, i.e just the same as without the 20K, but the input impedance is now 20k||50k = 14.3
Output impedance is zero, exactly the same as without the 20k.

At other settings you can calculate all the input and output impedances abd attenuations.

I don't see how having a variable input impedance can help.

 

[Pyramid]

What is it you are trying to do? Are you wiring the device as a potentiometer or a variable resistor (rheostat)?

 

[rayma]

if i was to take a 50k pot and attach a 20k resistor to the input and wiper pins would i then have a pot
that went from 20k to 50k instead of 0 to 50k?

That connection would vary from 50k to 14k, since the 20k is in parallel with the 50k at one extreme.
It's simplest to add a series resistor to a pot if you want more than some minimum value.
This also expands the useful range of adjustment.

 

[Bibio]

thanks everyone.

i'm trying to control the Vdc going into an LED so it wont fully dim or indeed go fully bright but still be adjustable in a narrow range via a potentiometer. the LED is inside a Resistive Optocoupler (LDR) so i can control the range of say 10k to 30k of the total 40k of the RO.

i think that it would have a few uses with two being to match impedance of MC cartridges without having to resort to jumper pins and another would be to give a narrow band of gain on amps. yes i understand that i could use trim pots but using RO's would do away with any mechanical devices in the signal path and also make it easier as the potentiometer can be remote on back/front of case.

 

[DF96]

i think that it would have a few uses with two being to match impedance of MC cartridges without having to resort to jumper pins and another would be to give a narrow band of gain on amps.

I'm not sure that a 10's of k resistance will help with impedance matching of an MC at a few ohms. Personally, I would rather have a mechanical device in the signal path than a slightly nonlinear device like an LDR.

Note that LEDs respond to current rather than voltage.

If you add a 20k in series with the two end terminals of a 50k pot (but not the slider) then you have a 90k pot which you can adjust from 22% to 78%. Maybe that is what you want?

 

[Bibio]

Thanks DF96. it was just an example.

ahhh so i need to vary the mA rather than the actual V . so i need to reduce or increase the mA being supplied to the LED and the 5v remains the same. which means that its the value of the pots resistance that will determine how dim/bright the LED will glow.

 

[Bibio]

ok i think i now know the problem with using RO for controlling resistance at 10k to 20k level. the mA is that small a difference that to control it would need more sophisticated electronics than i can deliver. looking at the spec sheet looks like its 0.15Ma to 0.2mA

 

[AndrewT]

..................

i'm trying to control the Vdc going into an LED so it wont fully dim or indeed go fully bright but still be adjustable in a narrow range via a potentiometer. ....................

The LED inside an LED/LDR needs some current to change the light level.
The LED drops some voltage Vf and this is somewhere in the range 1.4Vf to 2Vf.
You can measure this after you have assembled a breadboard test circuit.

If you supply 5Vdc through a variable resistor a current will flow.
Assumming the LED voltage is 2Vf, then there is 3V left across the variable resistor.
Resistor (and LED) current will be (5V-3Vf)/ VR
you want 10mA then VR = 5-3/0.01 = 200ohms, or 200r.
you want 1uA then VR = 5-3/1*10^-6 = 2M0