piclist 1999\04\16\112035a >
Thread: Help -- need power on circuit for LCD
www.piclist.com/techref/lcds.htm?key=lcd
flavicon
face BY : Barry King email (remove spam text)



Brian,

My LCD only needs < 10 mA at 5 V.  So I just power my LCD from a port
pin directly.  To turn it on, set the port latch to 1 and then set
TRIS bit for that pin to 0 (enable output).  PIC I/O pins can source
the 10 or so mA needed for the LCD with only a 0.1 V or so drop, so
you get 4.9 V on a Vdd of 5 V.

When turning off the LCD, Set TRIS to 1. This turns off the output
driver, turning the LCD off.  Don't just set the port latch to 0,
because that asks the PIC slam the LCD power pin low, as hard as it
can, which is not needed and may cause noise problems.

I recommend that any time you use a PIC pin as a power on / off
supply, you use TRIS to control it, because any reactance of the load
(usually capacitive) will cause trouble when powering off.  You
will get noise on the chip when turning the load on, but that can't
be helped.

BTW, Berhards suggestion might work if the load is more than the PIC
pin can handle.  But, there is another way, using a PNP transistor
(like 2N2906):
                    Power
        |             |
        |             /
        |          ||/_emitter
PIC pin  |-/\/\ ----|
(output  |      base|\  collector
0       |            \---------
for ON) |                     |
        |                    load
                              |
                             ground

In this circuit, the PIC drives the base low and can saturate the
transistor so the power lost in the transistor is low.  The
disadvantages are that you need the resistor, and that the base
current is "wasted", that is, it does not contribute to the load
current.

But if the saturation hfe of the transistor is high enough,
you only need a tiny base current.  The resistor from the PIC pin
sets this current.  Check the load curves for your transistor to be
sure what the minimum hfe is, and the maximum base current needed,
Ib = load current / hfe. Then get the resistor value from (Vdd - 0.7
V)/ Ib.

Example: For a 5 V power supply, to supply 100 mA, if hfe (min) =
100, then Ib = 1mA, so resistor is (5 - 0.7)/1 mA = 4300 Ohms.
Standard values are 4700 or 3300.   It probably best to overkill by
using the smaller resistor, 3300 Ohms.

In the emitter follower, the transistor is in linear operation, it's
dissipating Vce (voltage drop) x Ie (load current).  So calculate for
your load which is better efficiency.

So its a classic engineering tradeoff- do you want low power
dissipation in the transistor, and small voltage drop?  OR do you
want highest efficiency?

Hope this helps.

Barry.
------------
Barry King, KA1NLH
Engineering Manager
NRG Systems "Measuring the Wind's Energy"
Hinesburg, Vermont, USA
spam_OUTbarryspamnrgsystems.com
"The witty saying has been deleted due to limited EPROM space"

<25ED4BB09AE@wind.nrgsystems.inc>

In reply to: <Pine.LNX.4.05.9904131417070.773-100000@oxygen.atoms.dynip.com>
See also: www.piclist.com/techref/lcds.htm?key=lcd
Reply You must be a member of the piclist mailing list (not only a www.piclist.com member) to post to the piclist. This form requires JavaScript and a browser/email client that can handle form mailto: posts.
Subject (change) Help -- need power on circuit for LCD

month overview.

new search...