Driving LEDs from PIC weak pullup, saving resistors?

What voltage does that weak pull-up current source develop across the LED and what internal CMOS is looking at it?

ent.

Your driving the LED with an input pin? That must be a mistype.

This should be prohibited under the law since it contributes to the global warming. You should drive the ROHS LED by the green PWM at maximum environmentally friendly current, so the energy waste would be minimal.

VLV

On a sunny day (Sun, 6 Jan 2008 09:08:03 -0800 (PST)) it happened "

No it is no typo, the inputs can have a weak pull up MOSFET, and that current is enough to light a modern high brightnes LED. The pull ups can be enabled and disabled via software too.

urr=3D

rt.=3D

xt -

Many IC pins have programmable input resistors, there for line termination and to control floating lines and so on.

Hmm, it's called an "input" for a reason, I'm unsure why you would do that. Do you have the PIC number so I can take a quick look at the data sheet.

On a sunny day (Sun, 6 Jan 2008 09:43:38 -0800 (PST)) it happened "

It goes for most, if not all, PICs I know, I am using a 16F690 now. The reason is, as I stated, that moderne high brightness LEDs are way too bright at say even 1mA, and people like to drive these at levels as low as 400uA these days, to reduce eye damage :-) This requires some resistors, but using the weak pullups has enough current for normal LED signalling use, and saves resistors. This became only possible now the LEDs shine bright already at 200uA.

te=3D

Ok, most of those pins appear to be I/O pins. What pin are you using?

On a sunny day (Sun, 6 Jan 2008 09:59:13 -0800 (PST)) it happened "

Yes, they are almost all IO pins. The TRIS (TRISB for port B) bits indicate if the pin is an input or output.

So you can do: bsf TRISB, 1 and that pin (RB1) will be configured as input. You need to globally enable weak pull ups with bit NOT_RABPU in OPTION_REG, then have individual control for each pin in the WPUB register. See page 72 of the 294 page document on the PIC 16F690.

You could certainly configure the port to be an output and drive the led pwm to control brightness, but I think the energy used would be the same.

Unless you add a series inductor, in which case the whole thing becomes a synchronous switcher.

John

ted=3D

Hm, so your drawing 200uA, what's the voltage drop across the diode?

It is better then shorting a pullup to the ground; as for the PWM control, it is not too obvious because of the nonlinear conductance and capacitance of the LED.

Did you account for the energy required to build the inductor and to put it on the board? :)

VLV

On a sunny day (Sun, 6 Jan 2008 10:56:06 -0800 (PST)) it happened "

Click for pizza

Jan mentioned that he got dark by reconfiguring the port, so no power is wasted then.

I assumed that the copper was recycled from the recent remodel of Al Gore's mansion.

John

Ok, that's to high to be a valid low voltage, is that a valid high voltage? Data sheet says a valid high is 2v or [.25xVdd +.8] or are you operating in an invalid voltage range?

I think your input is oscillating and burning up that pin, but that's just me......

Interesting point. Even if it's not oscillating, the input cmos transistors may be biased in their linear range and wasting a lot more current than the chip is dumping into the led.

We recently had some tiny-logic cmos gates powered from +5 but logically driven from +3.3. They work, but glow nicely on a thermal imager.

John

On a sunny day (Sun, 6 Jan 2008 11:31:12 -0800 (PST)) it happened "

I did not notice any current increase, you cannot burn a PIC with 2.5 mA Idd. In fact I think any voltage on the input is valid, but some thresholds are set for logic 0 and logic 1. It is irrelevant if it is a valid high if you drive a LED, and no, it is not oscillating.

On a sunny day (Sun, 06 Jan 2008 11:39:57 -0800) it happened John Larkin wrote in :

I think you should take into account that many of the PORT inputs also can be configured as AD channels, and the diagram shows some sort of analog mux at the input, then followed by a D flop, and also branched to a schmitt trigger for the other logs. Page 75 etc of the pdf.

Perhaps using PIC inputs that include Schmitt triggers would keep the supply current down. To be sure, I would have to dive into the data sheet for the particular PIC in question.

Well that's up to you. This is digital Logic right, so you can only have a high or low. Any other voltage level is undefined, and the IC is no longer guaranteed to function, that's why they put those numbers in the data sheet.

What I don't know is what other logic circuit inside the device is looking at that level ~ wondering what to do with it ~ I already closed the data sheet. Your not really driving the LED, your sourcing current out of an input pin ~ could be 50 other CMOS FETs inside with invalid input voltages.

Leroy ElectricaL Engineer; 1984