Hello Shotgun,<p>Adding to the other posts...<p>Normally when you start out learning electronic circuits you
start with linear circuits and work your way up to non linears.
This is a non linear circuit so it's a little more complicated
then a linear circuit, but it isnt too bad.<p>
The analysis of this circuit goes something like this...<p>The power in the resistor R1 is low when R1 is low and also
low when R1 is high, so first we need to calculate the max
power point of the resistor. In other words, when R1
dissipates the maximum power what is it's resistance?<p>To accomplish this we start by writing an equation for
the circuit in terms of current and voltage, then solve
it for current and plug that result into the power
equation for R1 which is simply P=I^2*R as you know.
Doing this leads to this result:<p>P=R1*(V-Vd)^2/(R1+R2)^2
where
P is the power in resistor R1
R1 is the variable resistor
V is the source voltage (5v in your circuit)
Vd is the voltage across an LED (2v approximated)
R2 is the fixed resistor<p>To find the max power we take the first derivative of
P with respect to R1 and set the result equal to zero
and solve for the solutions, or simply graph the
equation for P above and look for high point maximums.
Doing this and generalizing the results we find that
max power is dissipated in R1 when R1=R2.<p>Noting that last result, we also note that when R1=R2
we also have half the current flowing through R1,
so this allows us to generalize once again about the
current flowing through R1:
The current flowing through R1 will be half the max
total LED current. With the number of LED's equal to N
and each one drawing 20ma nominal, the current through
R1 max will be
iR1max=0.02*N
where
N is the number of LED's in parallel.<p>Also noting that R2's value depends on the number of
LED's and if we assume a 5v source voltage we find that:
R2=150/N
where
N is again the number of LED's in parallel.<p>Since we now know what R2 is we also know what R1 is when
the max power point occurs:
R1=R2=150/N<p>Since the total resistance will be Rt=R1+R2=2R2 when the max
power point occurs the current will be half or 0.010*N amps.<p>Knowing what the current will be and that the resistance will
be 150/N we can now calculate the power and solve for the
maximum number of LED's for a given pot rating:<p>P=(0.010*N)*(0.010*N)*(150/N)
P=0.0001*N*N*(150/N)
P=0.015*N<p>and now we can see how high N can go assuming we can utilize the
full power rating of the pot of 0.1 watt. Setting P=0.1 watt in
P=0.015*N we get:<p>0.1=0.015*N<p>and solving for N we get:<p>N=0.1/0.015
N=6.667<p>so the max number of LEDs should be about 6.<p>
Take care,
Al
LEDs vs Bulbs, LEDs are winning.
