LCD backlight control during PIC "sleep"

A

Andy_123

Member

• Mar 19, 2008

• #1

Need some advice:

I am using 3.3V PIC with 5V LCD.
Currently backlight is powered by 5V PS via 100 Ohm resistor (~10mA).
LCD backlight forward voltage ~4V
I need to turn backlight off when PIC goes to sleep mode.
What would be easiest and cheap way to control?
PIC output is 3.3V that I can turn off before PIC goes to "sleep".
I tried few differnt ways with transistors, but like none of them.

Any ideas?

 

K

kchriste

New Member

Forum Supporter

• Mar 19, 2008

• #2

If you can switch the negative side of the backlight, a simple logic level MOSFET may do the trick. Something like a BSS138LT1 or BS108 have a low RDSon of 10hm: for a gate voltage of around 3V.

 

A

Andy_123

Member

• Mar 20, 2008

• #3

Control negative side - geat idea! I did not even think about this and connected K pin to GND
I am sure I can control it.

I probably can control negative side directly from PIC (25mA pin capacity while I need ony 10mA)
PIC out =0V - full brigtness
PIC put =3.3V - it will be only 0.7V drop and LCD will be off.

Thanks

Any other ideas?

 

P

Pommie

Well-Known Member

Most Helpful Member

• Mar 20, 2008

• #4

If you set the pin to input before the sleep command then you will use even less power. If you use a mosfet then make sure to include a tiedown resistor on the gate.

Mike.

 

A

Andy_123

Member

• Mar 20, 2008

• #5

I don't really think that I need a mosfet at all,
- setting pin as output with "0" state will turn LCD on,
- setting pin as input will turn it off, just need to select pin that can handle 5V.

 

P

Pommie

Well-Known Member

Most Helpful Member

• Mar 20, 2008

• #6

Andy_123 said:

I don't really think that I need a mosfet at all,
I don't really think I suggested that you did.
- setting pin as output with "0" state will turn LCD on,
- setting pin as input will turn it off, just need to select pin that can handle 5V.

Click to expand...

No, you need a pin that can handle 0V.

Mike.

 

A

Andy_123

Member

• Mar 20, 2008

• #7

If pin set as input it will have 5V potential on it through the LED connected to +5V.

 

K

kchriste

New Member

Forum Supporter

• Mar 20, 2008

• #8

Well, not really. Remember that the IO pin on your PIC has built in diodes to VDD and VSS so the voltage at the pin will be clamped at 4V ( 3.3V + 0.7V ). It may still be OK due to the voltage drop of the LED in the backlight but you'll have to experiment to be sure:

 

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A

Andy_123

Member

• Mar 20, 2008

• #9

According to the datasheet pins that have no analog inputs can tolerate up to 5.5V

 

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K

kchriste

New Member

Forum Supporter

• Mar 20, 2008

• #10

Ah, I see from your other posts that you are using the 18F66J60 which operates from a 3.3V supply and has some 5V tolerant IO pins. If so, then your idea looks like it will work.