# very simplistic potentiometer question

Discussion in 'Electronic Basics' started by Barbara, Jul 1, 2005.

1. ### BarbaraGuest

I am just beginning to tinker with electronics. I've decided to start
very simply and work my way up to the fun stuff. I like the thought
of playing with LEDs for some reason so I'm starting there. I've
figured out already that without a resistor that LEDs last about .2
seconds before they fry. Now that I can keep one going for more than
5 seconds I'd like to add a pot to control the brightness. I know a
pot is a variable resistor that basically adds resistance. I had this
idea that I would simply insert it between my resistor and the
positive lead on my power source (a 9 volt battery in this case.) I
bought an assortment pack of pots and was astonished to find they have
3 leads. This doesn't fit into my notion of how they work so I am
obviously missing the big picture here. I figure the answer is simple
sow I've done a lot of searches on Google and yahoo and I can't find
simple enough circuits to make sense of how they work. I need some
basic info on how to wire a circuit using one led, a resistor, a pot
and a battery.

Barb

2. ### steamerGuest

--FWIW if you get into Basic Stamp (Parallax.com) the first thing
they show you is how to drive an LED. After that it gets really fun, heh.

3. ### Rich WebbGuest

LEDs are (to a first approximation) constant brightness devices.
Garden-variety LEDs (waves hands, points back at "first approximation")
will want to drop 1.5 volts with about 15 milli-amps of current.

The voltage drop is constant and independent of the current flowing
through the LED (see waving hands, above), so you were "shorting" about
7.5 volts of the battery across the very small resistance of a forward
biased diode. That's why the LEDs had a very brief but exciting life.

Given your 9 volt source, and the 1.5 volts dropped across the LED, you
need to limit the current driven by the remaining 7.5 volts to about 15
milli-amps. That works out to (7.5 / 0.015) = 500 ohms.

There's a standard resistor value at 470 ohms that will be close enough.

You should also check that the power dissipation capability of the
resistor is adequate (that's the wattage rating). The resistor is
dropping 7.5 volts and carrying 0.015 amps so it's dissipating about 120
milli-watts. You'd want to use a 1/4 watt resistor to allow yourself
some headroom for things like a fresh battery (that's > 9 V) and other
device variations. If you use a smaller size 1/8 watt resistor then it
may get too hot and, eventually, fail.

LED brightness is typically controlled by pulsing it on and off very
quickly. If it's only on half the time, the perceived brightness is
about half of the 100% duty-cycle brightness.

There are lots of ways to vary the duty cycle. A 555 timer chip is a
good place to start. It has a wide input voltage range (will work with
the 9 volt battery) and there are lots of sample circuits available.
Examples at http://www.kettering.edu/~bguru/EE323/EE323-05.pdf and
http://www.schematica.com/555_Timer_design/555.htm.

Regarding the potentiometers, the two leads on the ends are connected to
a resistive strip that's the nominal value of the pot, say 10 KOhms. The
middle lead is connected to a wiper that moves along the length of the
pot and so it has an output voltage that changes along the length of the
resistive strip. If one of the outer leads is connected to the (-) end
of the battery and the other to the (+) then you could measure a voltage
at the middle lead that varied from the (-) to the (+) value, or 0 to 9
volts relative to the (-) terminal.

You can use a potentiometer as a variable resistor by connecting
together the center and one of the outer leads. The resistance between
the center and the remaining outer lead will vary from nothing (when the
wiper is turned all the way towards the remaining lead) to the full
value of the resistive strip when it's at the other extreme.

4. ### Andrew HolmeGuest

The outside pins connect to the ends of a resistive track. The middle pin
is the wiper. You can just use the middle pin and one outside pin to get a
variable resistor like you imagined. You'll get best results from low-value
pots (try 1K) and, of course, you *must* retain the fixed resistor in series
as well to avoid frying the LED when the wiper reaches the end. With this
rather crude method, you'll find the brightness varies non-linearly along
the track.

5. ### BanGuest

So you have already calculated the resistor: If it is a white LED, it might
have a forward voltage of 3.3V at 20mA current. The 9V -3.3V= 5.7V must be
dissipated. R=Voltage/Current = 5.7V/20mA = 285 ohms. So 270 will also do.
Now you can take a pot of low value (1k) in put it in series with that
resistor and you can vary the current from 4.5mA to 21mA. But 0mA can not be
reached.
If we connect the pot in a different way, we can vary the LED-current from
18mA to 0. (3.3mA run through the pot)
The best is to use a 1W wire-wound pot for this purpose to not burn the
wiper or the carbon trace. Also a linear not log characteristic should be
chosen, they have the best power rating.
.---------.
| |
| .-.
--- | |
- | |270
--- '-'
- |
--- 9V |
- .-.
--- | |<------.
- | |1k |
--- '-' |
- | V white LED
--- | -
- | |
| | |
| | |
'---------o--------'
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
view\fixed font

6. ### Guest

You need some bqsic info on both LEDs and potentiometers. LEDs are
junction diodes that have all the normal properties of any other
diodes. When you apply reverse voltage to them, they conduct only a
small leakage current, till that voltage gets to some breakdown level,
and then the current rises very fast. In the forward direction, the
current rises with an exponential relation to the voltage. Equal
increments of voltage increase raises the curent by equal multiples.
every 60 millivolts or so increases the current by a factor of 10.
That means that once the forward voltage is in the normal working
range, very small additional voltage will drive way too much current
through them, and you see a flash and it is all over, as you have
noticed.

The series resistor acts as a current stabilizer, using up all the
extra voltage as the LED current passes through it, also. Resistors
have a linear relation between current and voltage, so that doubling
the voltage across them simply doubles the current through them. This
linear relationship is called Ohm's law.

Potentiometers are resistors with a connection to each end of a strip
of resistance material and also a sliding contact that can be moved to
any spot along that strip. The sliding contact is almost always the
center of the three terminals.

If you want to use one as a variable resistor (make connections to a
variable length of resistance material) you use one of the end
connections and the sliding contact.

Most LEDs have a maximum current rating of 20 milliamps (.02 amperes)
and pass this current when there is a volt or three across them. The
voltage depends on the color. Red photons have less energy each, so it
takes electrons falling through about 1.5 volts to produce them. Blue
photons are about twice as energetic, so electrons have to fall through
about 3 volts to produce them.

So, based on the color of your led and the total voltage available, you
the LED. Leds say you have a red LED and a 9 volt battery. This means
that you have about 9-1.5=7.5 volts to burn up across the resistor.
7.5volts/.02amperes=350ohms. Ohms is just a label that means volts per
ampere. So if you put something close to that in series with the LED,
you can expect it to shine with about rated light output and still last
a long time.

Then you can connect a pot in series with that pair (using the center
and one end connection) to allow you to turn the current down lower
than the rated current and see how the light varies. I recommend you
get a \$20 digital multi-meter at Walmart and make a table of LED
current versus LED voltage to get familiar with the exponential diode
current voltage relationship I spoke of earlier. You can keep the
meter set on voltage and use the voltage drop across the fixed resistor
(dividing the voltage by the resistance to calculate the curent) to
make this table for various pot settings.

7. ### Byron A JeffGuest

Define fun stuff?
But have you figured out why? I'll take a minute to explain. LEDs required
a minimum voltage to light. Also there is a maximum amount of current that
can flow through is before it burns up. Without the current limiting resistor
the LED sees too high a voltage and allows too much current through and it
blows like a fuse. POOF!

A good discussion of this is here.

http://www.palosverdes.com/pvarc/2002QRO/Sept2002QRO.htm
You can do that, but you're not going to get the expected result.
They all do. The two outer ends are like a regular fixed resistor. The
middle lead is known as the wiper and it's resistance varies vs. the other
the variable resistance you're looking for.
See above. That should be good enough to get started.
Well that's a different kettle. LEDs are highly non-linear with respect
to current and brightness. Also as you've already learned they burn up
if too much current goes through them. And they go out if there is too
little voltage.

What you can do is hook up a low resistance pot (50-100 ohms) as a voltage
divider. You do this by tying the outside two leads to the two leads of
your 9V source and pull the variable voltage from the wiper lead. Be aware
that if you do this with a 9V batter that it'll go out pretty quickly because
the pot will be burning 10-20ma of current the entire time you're
experimenting.

BTW dimming LEDs usually isn't done this way. It's usually done by turning
them fully on/full off very fast. It's a technique known as Pulse Width
Modulation (PWM).

Hope this helps,

BAJ

8. ### tlbsGuest

One other thing to add to all the good information, above.

This table is just a "rule of thumb". You need to read the datasheet
for each LED for the exact current ratings and voltage drops.

RED -- 1.5 V 10 to 20 mA (some as low as 1 mA)
AMBER -- 2.0 V 10 to 20 mA
GREEN -- 2.2 V 10 to 20 mA and even higher
For the following colors it is best to read the datasheet
BLUE -- 3.5 V to 5V(on some) 20 to 50 mA and higher
WHITE -- same as blue

Note that some of the BLUE LEDs have very low reverse voltages, and can
be damaged by connecting them backwards to the 9V battery.

9. ### Bob MyersGuest

To a better approximation, they are devices whose brightness is
current-controlled rather than voltage-controlled (as you noted,
the drop across such things is fairly constant - since it is, after
all, a diode). Control the current, and you control the
brightness, with a *reasonably* linear relationship between the
two. (Well, actually, it's luminous intensity that has a nice
relationship with current, and not "brightness," but now I'm
definitely getting in WAYYYYY too deep....) You DO, of course,
as noted always want to limit the current so as not to exceed the
device's maximum power- handling capabilties (the limit is generally
power, not current, which is why higher peak currents are permissible
for most devices in pulsed applications). Pulsed operation as a means
of varying brightnessis often desirable, because it lets the device
operate in a more efficient manner (relative efficiencies for
LEDs tend to go up with current, at least for a while), but might
be more difficult for the beginner hobbyist to implement than a
simple current-controlled, constant-on mode.

So a variable resistor CAN be used to vary LED brightness, as
long as you ensure that you never can get to the point where there
is so little resistance in the circuit that the maximum permissible
steady-state current (as determined by the LED's power-dissipation
limit) is exceeded. A potentiometer configured as a variable
resistor, plus a fixed resistance in series, will generally do nicely.

So let's figure on about a 2V drop on a red LED operating within
the usual range of currents, and a 9V battery; if we want to be
able to vary the current up to, say, a safe peak of 30 mA for the
particular device in question, that's

(9V - 2V)/ 30 mA = 7/0.030 = 233.3 ohms.

With the nearest standard-value fixed resistor at 240 ohms (for
5% parts), we'll err on the side of caution and choose that. Now,
adding a 470 ohm potentiometer configured as a variable resistor
(center terminal tied to either of the outside two), we can vary the
current from 30 mA down to:

7V/(240 + 470 ohms) = 7/710 = 9.9 mA,

for roughly a 3:1 range in current and therefore intensity. If you
want more range, you can always drop the bottom end by using
a larger potentiometer (at the cost of being able to set the mid-range
and higher brightness levels to as fine a degree of control).

Bob M.