# Best way to get 2.5 volts from somewhere? (Vcc = 5 volts)

Discussion in 'Embedded' started by Tomás Ó hÉilidhe, May 13, 2008.

1. ### Tomás Ó hÉilidheGuest

I've got a bi-colour LED. It has two pins. Internally it consist of
two LED's in parallel except they face in different directions.

I'm looking into ways of using one micrcontroller pin to control the
LED as follows:
Pin High = Light up Red
Pin Low = Light up Green
Pin as Input = Nothing lights up

A friend of mine suggested to me today to connect one of the LED pins
to the microcontroller, and the other to 2.5 V. That way, if the uC
pin is high, it will source current from 5 volts to 2.5 volts. If it's
low, it will source current from 0 volts to 2.5 volts. (Of course I'd
have a resistor somewhere).

So the only question is how I'd put one of the pins at a constant 2.5
volts. My first thought was to use a zener diode, i.e. take a pin from
the LED, put into one side of the zener, and tie the other side of the
zener to ground. I'm not entirely sure if this will work though.
Another complication would be that I'd need two zeners in parallel
facing the opposite direction in order to let current flow in both
directions.

Do you think the whole 2.5 volts idea is good? What's the best way of
getting one of the LED pins to sit at 2.5 volts?

Tomás Ó hÉilidhe, May 13, 2008

2. ### linnixGuest

You can use another uC to PWM switch a voltage source to an op-amp
integrator, since cost/components are not an issue for you.

linnix, May 13, 2008

3. ### johnspethGuest

I've got a bi-colour LED. It has two pins. Internally it consist of
You don't need a microcontroller pin to generate 2.5V. You just need
a
constant voltage source. You can use a simple voltage divider off 5V
which
uses two equal valued resistors that serve as a voltage divider AND
current
limiters.
2.5V
from 5V. An implied requirement is it works. If you try it and it
works,
it's a good idea. You're the judge of "good".
There is no way. Your datasheet will tell you the pin will either be
0V or
5V if it's an output. It will be at high impedance if it's a an input
and
your external circuit will determine what voltage it will be.

JJS

johnspeth, May 13, 2008
4. ### Spehro PefhanyGuest

Predict the current through each LED when it is on, over unit-to-unit
variations and temperature. Then see if you think it's a good idea.

Best regards,
Spehro Pefhany

Spehro Pefhany, May 13, 2008
5. ### NilsGuest

Maybe I'm just dump, but:

Why don't you simply use a 4066 bilateral-switch, a current-limiting
resistor plus some kind of inverter (can be as simple as a NPN-tranny in

It's cheap and just works...

Nils

Nils, May 13, 2008
6. ### Tomás Ó hÉilidheGuest

Tomás Ó hÉilidhe, May 13, 2008
7. ### linnixGuest

linnix, May 13, 2008
8. ### Tomás Ó hÉilidheGuest

I have a question...

If I have an LED that has about 2 volts across it, then is it OK to
put a 2 volt power supply across it without a current limiting
resistor?

My overall power supply would be 9 V coming from a square battery, but
I'll be putting it thru a voltage regulator to give me out 2 V.

I'll then be putting the 2 V across the LED.

Can I leave out the LED's current-limiting resistor, or is there still
a chance of there being too much current that would fry components?

Tomás Ó hÉilidhe, May 13, 2008
9. ### linnixGuest

My overall power supply would be 9 V coming from a square battery, but
Please ignore my last post, since you are changing spec on me. You
said you were getting 2.5V out of 5V with the LM317.

linnix, May 14, 2008

LEDs are not compliant with C99.

DSP and Mixed Signal Design Consultant
http://www.abvolt.com

11. ### linnixGuest

So, we need 2 resistors at the uC end, and 2 resistors at the voltage
regulator end. Oh, great, we eliminated 2 LED resistors with a
regulator and 4 other resistors.

linnix, May 14, 2008
12. ### Andrew SmallshawGuest

Potential problem here. What voltage is present on the MCU pin?
5V strongly suggests TTL compatible inputs/outputs to me which are
_not_ 0V and 5V. From memory low is up to 0.8V and high is at
least 2.0V. There is a possibility that your pin could be 'high'
and delivering 2.0V which is still _less_ than the 2.5V on the
other end of the LED.

This is of course the worst case scenario, but the LED's barrier
voltage is also conspiring against you. The exact value varies
depending on the device but typically around 1.7V is needed for
the LED to conduct. 2.5V+1.7V=4.2V which is a big ask from a 5V
device.

Andrew Smallshaw, May 14, 2008
13. ### Andrew SmallshawGuest

I thought that I replied to this yesterday but I can't see it here
so I'll post again...

There's a potential problem here. Consider what voltage you will
be getting from the MCU pin. It will _not_ be 0V, 5V, or Hi-Z.
Most 5V devices aim for TTL compatibility. From memory that allows
a low to be up to 0.8V and a high to be as low as 2.0V. Therefore
it is possible for your 'high' voltage to be below the 2.5V centre
voltage.

This is of course the worst case scenario, but the LED barrier
voltage is also conspiring against you. This varies from device
to device but is typically around 1.7V. 2.5 + 1.7 = 4.2V needed
from the pin in its high state, which is asking a lot from a 5V
device, particularly when you are drawing current from it.

Do yourself a favour. Put an H-bridge in there.

Andrew Smallshaw, May 15, 2008
14. ### Andrew SmallshawGuest

I thought that I replied to this yesterday but I can't see it here
so I'll post again. Having news server issues at the moment.

There's a potential problem here. Consider what voltage you will
be getting from the MCU pin. It will _not_ be 0V, 5V, or Hi-Z.
Most 5V devices aim for TTL compatibility. From memory that allows
a low to be up to 0.8V and a high to be as low as 2.0V. Therefore
it is possible for your 'high' voltage to be below the 2.5V centre
voltage.

This is of course the worst case scenario, but the LED barrier
voltage is also conspiring against you. This varies from device
to device but is typically around 1.7V. 2.5 + 1.7 = 4.2V needed
from the pin in its high state, which is asking a lot from a 5V
device, particularly when you are drawing current from it.

Do yourself a favour. Put an H-bridge in there.

Andrew Smallshaw, May 15, 2008
15. ### linnixGuest

And for the rest of us dumb old timer (in the OP's mind), we just
drive LEDs low with separate pins. Sometimes, we need to KISS up.

linnix, May 15, 2008
16. ### TomGuest

From p.135 of the PIC16F684 datasheet:
Voh min = Vdd - 0.7V at 3.0 mA and Vdd=4.5V
Vol max = 0.6V at 8.5 mA and Vdd=4.5V
So no problem getting 4.3V with a 5V supply. Actual values at room temperature
are even better than this.

A bigger concern is how he plans to multiplex the LEDs. The maximum current
into/out of Vdd or Vss is 95mA. So if he's using 25mA per LED and has more
than three LEDs connected in this manner, all it takes is one small glitch
somewhere to turn them all on at once and there goes the CPU. This might be an
acceptable risk for his hobby project but not something I would ever do in a
commercial product.

Tom, May 17, 2008
17. ### Andrew SmallshawGuest

At 3.0mA. I would expect the voltage to drop further as the LED
draws, say, 25mA. You may end up with LEDS that are lit but unusably
dim.

Andrew Smallshaw, May 17, 2008