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From: Tong Narak
Subject: Re: I made 1.2V DC-DC converter to 5V drive 4 whiteLEDs in parallel
80mA. The output drop to 3.4v is correctly ?
Date: Tue, 03 Dec 2002 21:49:16 +0700
Organization: C.S. Communications Co. Ltd.
NNTP-Posting-Date: Tue, 3 Dec 2002 14:49:18 +0000 (UTC)
X-Mailer: Mozilla 4.8 [en] (Windows NT 5.0; U)
Very thanks to edit my poor english. And the last sentense I talked
my circuit ( MAX756 ) or equivalent IC doesn't regulate voltage under
all load conditions.
At load only 80mA , the output voltage drop to ~3.4v
So I don't need put the resistor because at 3.4v is ok for LEDs.
Well, I couldn't calculate the resistor value at another load current.
At 100mA, I don't know how many voltage drop.
At 200mA, I don't know how many voltage drop.
Is these the normal behavier of DC-DC voltage regulator ?
Thanks for advance
> 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! :>)
> - Win
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