# Pots, and how to understand ...

Discussion in 'General Electronics Chat' started by SlowCoder, Apr 1, 2012.

1. ### SlowCoder Thread Starter New Member

Mar 25, 2012
26
0
In the learning projects I'm doing, some of the schematics that use pots will only connect the input and output pins. But some schematics will connect the 3rd pin to ground.

In the learning schematic I've attached, the pot is connected to ground (#39). The circuit is intended to show how a zener diode works. While connected to ground the circuit works as expected. I disconnected #39 and found that the circuit allowed more power to the circuit.

I'd like to understand what is happening here. What is the purpose in connecting the pot to ground? Why only use in some projects?

File size:
23.2 KB
Views:
49
2. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
Current can cause hot spots in something like this arrangement, possibly killing the pot.

The reason they did it this way is simple. You are creating a variable voltage, from +9VDC all the way to ground (- lead).

That is why a pot is called a pot, which is short for potentiometer, (potentio = voltage).

So if you are making a power supply a true variable voltage is absolutely necessary, which is the heart of this circuit.

Side note, powering LEDs can be done so many better ways. I prefer to use a simple LM317 IC (which is cheap) and make a precision current regulator. That circuit sucks for that application.

As far as LEDs go, I wrote this article as a tutorial, but then it took on a life of it's own.

LEDs, 555s, Flashers, and Light Chasers

3. ### SlowCoder Thread Starter New Member

Mar 25, 2012
26
0
Just so we're on the same page, this is part of the electronics lab from Radio Shack I'm working from. I can understand if you got confused that I was attempting to build a variable power supply, as I did inquire about it previously. This particular project is a demonstration of the voltage threshold of the zener diode to active the LED. I just chose it because I'd previously wondered about the use of ground in a pot.

Going to your comment about the possibility of hot spots on the pot, I understand what you're saying there. Too much current in a single spot on the resistive material can burn and possibly short the pot.

Is the ground of the pot essentially dividing the voltage, as the outcome in the experiment was different between using the ground and not? I still don't understand.

4. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
Actually it would probably burn the pot open, but the point is the pot dies.

A pot in the configuration you show is a voltage divider. The wiper of the pot (which physically moves over the resistive material) is a tap point for the pot, and is a variable voltage source.

Pots are one of those items I have found hard to get, so I have ordered about 10 of each kind over time. 1KΩ, 10KΩ, 100KΩ, and 1MΩ. With the right parts you can compensate for other parts when you design like I do.

5. ### SlowCoder Thread Starter New Member

Mar 25, 2012
26
0
Ok, so in this case, they wired the pot to ground to divert some of the current to ground so it won't burn out?

6. ### kubeek AAC Fanatic!

Sep 20, 2005
4,631
787
No, they wired it so so that you get the full range of voltage between 0 and 9V going to the LED. If you wire it without ground it will be a variable resistor, so the LED will see 9V all the time and the pot will just vary the current.

7. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
92
Simplified.

Both Ends connected is Voltage divider = Potentiometer = vary voltage level.

One End connected is Variable resistance = use as Rheostat = vary the Current available.

8. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
The LED will never (in this circuit) see 9V. If it did (and assuming it is a normal, run-of-the-mill diode) it would be transformed into a DED (dark emitting diode).

You didn't say what the LED forward voltage was (I'll assume 2V), nor did you say what the Zener voltage was for the Zener diode (I'll assume 2.7V). Under these assumptions, then no current will flow if the bottom of the 1kohm resistor is anything less than 4.7V. As others have noted, without the third terminal grounded, you have a variable resistor (a rheostat) that can be varied from zero ohm to 10kohm. Combined with the 1kohm resistor that it is in series with, you then have a resistance range from 1kohm to 11kohm and this resistance will always have about 4.3V (9V-4.7V) across is, so you will be able to vary the diode current from about 400uA up to about 4.3mA. You will NOT be able to turn of the LED, though it may or may not be bright enough to see, under normal room light, at the 400uA end of the range.

With the third terminal on the potentiometer grounded, imagine disconnecting the wiper and then adjusting the wiper until you measure 4.7V on it. To achieve this you would have to move the wiper about 52% of the way from the ground to the 9V terminal. Below that, you will not have enough voltage at the wiper to result in any current flow into the 1kohm resistor (and, hence, the Zener and the LED), so they might as well be disconnected. But once you reach this point, you will have enough voltage and so some (not all) of the current will flow out of the wiper to the LED. The maximum current is the same as the earlier configuration, namely about 4.3mA, because with the wiper all the way to the top, the current flowing out the wiper is unaffected by any current flowing between the fixed ends of the pot.

To see if you understand this, consider the following question:

The pot is adjusted so that it is 75% of the way toward the 9V end of the pot. What will the current be in the LED? What will the current be in the battery?

Now answer the same questions if the grounded end of the pot is disconneted from ground?

To approach the problem, replace the potentiometer with two normal resistors, one that is 2500ohms between the wiper and the 9V terminal and the other that is 7500ohms between the wiper and ground.

Good luck!