R
RST Engineering (jw)
Guest
Don ...
I've noticed this effect, also. Yet a hallowed source none other than HP
has a full chapter in AN-1005 describing how a pulsed drive increases the
optical output of the lamp. I either don't understand completely what HP is
telling me or my eyes are deceiving me, which is entirely possible.
Let's take a relatively trivial example. How about we generate a square
wave to drive our lamp. Let's not worry for the moment about how we
generate it, let's simply say that we can have either a square wave or a DC
signal for drive. Let's also make the math easy and say our LED has a
forward voltage of 2 volts at 20 mA and 2.5 volts at 40 mA (real data for
HLMP-1340, typical high brightness red LED).
So in the DC case, we choose a resistor value of 150 ohms, and in the square
wave case we choose a resistor value of 62 ohms. The battery doesn't care.
It is pumping out an average power of 100 mW in either case.
However, in the DC case, the (peak) power delivered to the LED is 40 mW and
in the square wave case the (peak) power delivered to the LED is 100 mW.
And yet, to my admittedly untrained eye, they are either the same brightness
or perhaps the square wave is just the littlest bit DIMMER than the DC
drive. What's happening? My understanding of optics is that the eye is
pretty much a peak detector when the PRF is greater than the flicker rate of
24 Hz. or so, but that isn't the case in the experimental lab. Comments?
Oh, and to the feller that wanted to use a PIC and an inductor at a cost of
$5 or so, you can achieve the same thing with a single CMOS quad NOR gate,
two sections as oscillator and one section as monostable multi plus a single
PNP transistor in saturation driving either the LED directly through a much
smaller resistor (the cheap method) or throught the inductor and a catch
diode (the expensive method) for less than two bits worth of parts (the
cheap method) or a buck's worth (the expensive method).
Jim
I've noticed this effect, also. Yet a hallowed source none other than HP
has a full chapter in AN-1005 describing how a pulsed drive increases the
optical output of the lamp. I either don't understand completely what HP is
telling me or my eyes are deceiving me, which is entirely possible.
Let's take a relatively trivial example. How about we generate a square
wave to drive our lamp. Let's not worry for the moment about how we
generate it, let's simply say that we can have either a square wave or a DC
signal for drive. Let's also make the math easy and say our LED has a
forward voltage of 2 volts at 20 mA and 2.5 volts at 40 mA (real data for
HLMP-1340, typical high brightness red LED).
So in the DC case, we choose a resistor value of 150 ohms, and in the square
wave case we choose a resistor value of 62 ohms. The battery doesn't care.
It is pumping out an average power of 100 mW in either case.
However, in the DC case, the (peak) power delivered to the LED is 40 mW and
in the square wave case the (peak) power delivered to the LED is 100 mW.
And yet, to my admittedly untrained eye, they are either the same brightness
or perhaps the square wave is just the littlest bit DIMMER than the DC
drive. What's happening? My understanding of optics is that the eye is
pretty much a peak detector when the PRF is greater than the flicker rate of
24 Hz. or so, but that isn't the case in the experimental lab. Comments?
Oh, and to the feller that wanted to use a PIC and an inductor at a cost of
$5 or so, you can achieve the same thing with a single CMOS quad NOR gate,
two sections as oscillator and one section as monostable multi plus a single
PNP transistor in saturation driving either the LED directly through a much
smaller resistor (the cheap method) or throught the inductor and a catch
diode (the expensive method) for less than two bits worth of parts (the
cheap method) or a buck's worth (the expensive method).
Jim
You need an inductor in addition, unless you were thinking of applying 5
volts directly to the LED at whatever duty cycle results in average
current of 20 mA. By and large, LEDs produce less light from this than
with 20 mA of steady DC - meaning pulsing at whatever duty cycle achieves
20 mA average current with 5V across the LED is less efficient than using
a resistor from a 5V power supply.
- Don Klipstein (don@misty.com)