From: Winfield Hill
Subject: Re: I made 1.2V DC-DC converter to 5V drive 4 whiteLEDs in parallel 80mA.
Date: 3 Dec 2002 06:30:16 -0800
Organization: Rowland Institute
X-Newsreader: Direct Read News 4.20
Tong Narak wrote...
> I made 1.2V DC-DC converter to 5V drive 4 whiteLEDs in parallel
> 20mA x 4 = 80mA at output. The output 5v is drop to 3.4v that
> enought for 4 whiteLEDs 80mA.
Thanks for writing, Tong. Let's have a quick English checkup as
part of your answer - it's always good to polish one's skill in
another language, so I fixed your sentences to look more natural.
> I made a 1.2V DC-DC converter to 5V drive 4 white LEDs in parallel,
> 20mA x 4 = 80mA at output. The output 5v drops to 3.4v, which is
> enough for 4 white LEDs taking 80mA.
> With no load, the output is correctly at 5V. With full 80mA load,
> the output is 3.4V. eff. ~ 38- 50%
> P.S. I used a MAX756 in a DIP300 package and an inductor from Wilco,
> 0.070 ohms D.C.R. The input is one ni-mh cell, 1.3v dropping to 1v
> when at full load.
> Does 1.6v output drop out when full load is correctly?
I can't correct your last sentence without guessing at your question.
Are you asking if one should expect to get a 1.6V drop? I doubt that
you can dependably predict the voltage-sag and know if this sag holds
true for various states of your battery. A better approach would be
to make a DC-DC converter that's more efficient and can maintain its
output voltage regulation under all load conditions.
You could set the voltage output for 5.0V, etc., and drop the extra
1.6V with a resistor to determine the LED current. Or even better,
you could make a current-regulating rather than voltage-regulating
circuit, to reduce the wasted power in the current-limiting resistor.
Many adjustable dc-dc converter ICs use a 1.25V reference voltage at
the feedback pin, and you can place a current-sensing resistor in the
ground path, dropping 1.25 volts. That's better than wasting 1.6V,
but not much!
One simple way to reduce the sense-resistor voltage is to use two
resistors to raise the sensed voltage: one in series and the other to
the top of the LED. For example, let's choose 0.5V drop on our sense
resistor at 80mA (Rs = 5 ohms), and add back the missing 1.25 - 0.5 =
0.75V with a 1.0k series resistor. We'll get this 0.75V using a
(3.4/0.75 -1) = 3.6k resistor to the LED's anode, taking advantage of
the LED's more-or-less predictable 3.4V voltage drop.
A more accurate solution is to amplify the sense voltage with a low-
power opamp. This approach allows you to use very low voltage drops,
like 176 mV with a 2.2-ohm resistor. Use a non-inverting stage with
G = 7.2 from 62k and 10k resistors, etc. You can power the opamp from
the LED's anode.
It's too bad more adjustable dc-dc converters don't have low-voltage
feedback pins to begin with! :>)