Hello steven and audioguru,
It is difficult to measure
the power because a neon bulb are very different and and an LED are not
linear.
A Neon bulb draws no current until the AC voltage rises to about 90V which
is its "striking voltage" then draws current as its forward voltage drops
to about 65V. When the AC voltage drops less than about 60V then the neon
bulb turns off and draws no current until the next half-cycle voltage
reaches about 90V again.
An LED has such a low voltage compared to the very high mains voltage that
the LED conducts for almost the entire half-cycle that it has forward
bias.
That's an interesting observation. I was tempted to think of the neon as
linear myself.
With that information we can model the neon roughly as a device that is
open circuit until the line voltage reaches 90v, and then becomes a 65v DC
source which of course then draws current through the resistor. Knowing
the resistor value, we should then be able to calculate current and thus
the power lost in the resistor and in the neon bulb. It wont be perfect
but it should be good enough for a simple comparison. Ive done this below.
steven:
I have powered LED's at low current from the line. For a little more
brightness i have used 50k, and for more normal use 75k and 82k. 82k works
out very well because the resistor does not get warm even though the line
is 120vac and as audioguru said the LED is on almost the whole time. This
is when using a 1/2 watt resistor for physical durability and for it's
power rating.
All:
The line voltage is:
Vt(t)=170*sin(2*pi*f)
Solving this for Vt=90 we get:
T1=asin(9/17)/(2*pi*f)
Solving Vt(t) for Vt=65 looking for the solution between 90 and 180 degrees
we get:
T2=acos(13/34)/(2*pi*f)+1/(4*f)
Now that we have the turn on and turn off times per half cycle we can
calculate the power in the resistor and the power in the neon bulb.
The instantaneous current in the resistor is:
IR(t)=(170*sin(2*pi*f*t)-65)/R, {T1<=t<=T2}
and so the instantaneous power in the resistor is:
PR(t)=IR(t)^2*R=(170*sin(2*f*pi*t)-65)^2/R, {T1<=t<=T2}
The average power then is:
Pavg=(1/T)*integral PR(t) dt [from t=T1 to t=T2], {T=1/(2*f)}
which for any frequency f comes out to:
Pavg=4249.0894/R
That is the power in the resistor alone.
Next, the voltage across the neon bulb is:
VN=65, {T1<=t<=T2}
and the current (from above) is:
IR(t)=(170*sin(2*pi*f*t)-65)/R, {T1<=t<=T2}
so the instantaneous power in the neon bulb is:
PN(t)=IR(t)*VN=65*(170*sin(2*pi*f*t)-65)/R, {T1<=t<=T2}
Integrating from T1 to T2 and multiplying by 1/T where T=1/(f*2) we get:
Pavg=3286.9903/R
Adding the two we get the total power:
PT=7536.08/R
Calculating this for R=300k we get:
PT=25mw
and for R=100k we get:
PT=75mw
For comparison, my semi high brightness LEDs run at about 100mw but this can be reduced with ultra high brightness LEDs.
Without specifying the peak line voltage Vpk or the turn on voltage of the neon bulb VH or the turn off voltage of the neon bulb VL here is the rather long formula for the total power:
Code:
PT=
((4*VL^2-260*VL+2*Vpk^2)*acos(VL/Vpk)+(2*VL+260)*sqrt(Vpk^2-VL^2)-8*sqrt(Vpk-VL)*VL*
sqrt(VL+Vpk)+(2*pi-4*asin(VH/Vpk))*VL^2+(260*asin(VH/Vpk)-130*pi)*VL-8*sqrt(Vpk-VH)*sqrt(VH+Vpk)*VL-
2*Vpk^2*asin(VH/Vpk)+(2*VH+260)*sqrt(Vpk^2-VH^2)+pi*Vpk^2)/(4*pi*R)
The only variables are:
Vpk the peak line voltage (Vpk=170 for a 120vrms line),
VH the turn on voltage of the neon,
VL the turn off voltage of the neon,
R the series resistance (in Ohms).
The formula may be able to be simplified but that's it for now.