what is the "k" in a 100k pot?
Greetings,
Well, I have looked but am still confused. I do not have access to a modern EE textbook & my physics book from college is no help. A pot is a variable resistor, right? If so, is the "k" the for kilo Ohms resistance? My attempt at understanding has left me concluding that the zero resistance would be full volume and max resistance (20, 50, 100, or whatever K) would be the lowest volume. If this is true, then how can some people switch from 50k to 100k--wouldn't the low end change in volume be dramatic?
This is my first foray into building an amp--BrianGT's gainclone group buy. My current amps, 50 wpc from the late 70's, rarely get put above 1 and never above 2 on the volume dial and my speakers are NOT super-efficient. I would rather have the pot set at 3 or 4 for normal listening. Anyone have a suggestion about what "K" to use--I am looking at Nobel pots.
TIA,
Bret Morrow
Greetings,
Well, I have looked but am still confused. I do not have access to a modern EE textbook & my physics book from college is no help. A pot is a variable resistor, right? If so, is the "k" the for kilo Ohms resistance? My attempt at understanding has left me concluding that the zero resistance would be full volume and max resistance (20, 50, 100, or whatever K) would be the lowest volume. If this is true, then how can some people switch from 50k to 100k--wouldn't the low end change in volume be dramatic?
This is my first foray into building an amp--BrianGT's gainclone group buy. My current amps, 50 wpc from the late 70's, rarely get put above 1 and never above 2 on the volume dial and my speakers are NOT super-efficient. I would rather have the pot set at 3 or 4 for normal listening. Anyone have a suggestion about what "K" to use--I am looking at Nobel pots.
TIA,
Bret Morrow
k is indeed 1000. Changing potentiometer value won't do what you fear since the volume is related to the ratio of wiper-to-hot and wiper-to-ground. All it does is change the loading on the circuit driving the pot.
The K stands for kilo ohm.
The input voltage is applied across the resistor track of the pot.
The output voltage is the wiper.
So max resistance is max output and zero is no output.
(relating to wiper to ground resistance)
🙂 sreten.
The input voltage is applied across the resistor track of the pot.
The output voltage is the wiper.
So max resistance is max output and zero is no output.
(relating to wiper to ground resistance)
🙂 sreten.
You are correct that the k, is kilo ohms. So a 100K resistor is 100,000 Ohms, from end to end.
A 100K pot usually has 3 terminals. The two out side ends, will always read 100K ohms. The center terminal, or the wiper as it is sometimes called just moves from end to end. In this case, the resistance from one end to wiper varies from 0 to 100K ohms.
The resistance from the other end to the wiper will be the rest of the 100K ohms.
In short, the resistances from each end to the wiper will add to 100K ohms.
The reason that pot values can be changed, is that it is the ratio of the two resistances that determine voltage gain (or more corretly, probably attenuation). If the wiper is exactly in the middle, the ratio of resistances is 1 to 1. This is true for a 100K, 50K, or any other value potentiometer.
This ratio of resistance forms a voltage divider. If you put 10 volts across the 100K pot, and vary the wiper from end to end, the voltage will vary in proportion the wiper position. In the case of 10 volts across the pot, the voltage will vary from 0 to 10 volts.
...So, you're probably wondering how do you choose a the value of the pot for your circuit. The value you pick is usually determined by other values in the circuit. Namely, input (or supply) and output (or load) resistances. If your source is a low impedance source, you can get away with smaller values for pots. If you have a high impedance source, you will need to match that with a higher value pot.
Do a little research on what a resistive voltage divider is. If you're lucky you should find some information on how the load resistance affects the output voltage.
Hope this helps!
-Dan
A 100K pot usually has 3 terminals. The two out side ends, will always read 100K ohms. The center terminal, or the wiper as it is sometimes called just moves from end to end. In this case, the resistance from one end to wiper varies from 0 to 100K ohms.
The resistance from the other end to the wiper will be the rest of the 100K ohms.
In short, the resistances from each end to the wiper will add to 100K ohms.
The reason that pot values can be changed, is that it is the ratio of the two resistances that determine voltage gain (or more corretly, probably attenuation). If the wiper is exactly in the middle, the ratio of resistances is 1 to 1. This is true for a 100K, 50K, or any other value potentiometer.
This ratio of resistance forms a voltage divider. If you put 10 volts across the 100K pot, and vary the wiper from end to end, the voltage will vary in proportion the wiper position. In the case of 10 volts across the pot, the voltage will vary from 0 to 10 volts.
...So, you're probably wondering how do you choose a the value of the pot for your circuit. The value you pick is usually determined by other values in the circuit. Namely, input (or supply) and output (or load) resistances. If your source is a low impedance source, you can get away with smaller values for pots. If you have a high impedance source, you will need to match that with a higher value pot.
Do a little research on what a resistive voltage divider is. If you're lucky you should find some information on how the load resistance affects the output voltage.
Hope this helps!
-Dan
Ahhhhhhhh!!!
Well, that makes things clearer! I saw the digrams with the proper hookup of a pot, but I was ignoring the grounding line from the pot somehow (after all those unwanted electrons just disappear on their own, right?? ;-)
Thanks all,
Bret Morrow
Well, that makes things clearer! I saw the digrams with the proper hookup of a pot, but I was ignoring the grounding line from the pot somehow (after all those unwanted electrons just disappear on their own, right?? ;-)
Thanks all,
Bret Morrow
- Status
- Not open for further replies.