ensuring LOW of MCU is 0V.

[allegro]

when my MCU's output is LOW, it can output up to 0.5V.
I need to ensure that this 0.5V wont affect the rest of the circuit, therefore i would like to connect in series to the MCU output a diode.
What should be the VF of the diode?
when the MCU output is HIGH, it sources 8mA,
does it mean that i need to pick a diode whose 8mA @ VF_typ rated?

I have another MCU output which when HIGH, it sources 2mA, So i need to choose 2mA @ VF_typ rated diode?

Thanks in advance.

 

[Nigel Goodwin]

Wrong way to do it - you also need to specify what you're feeding, and if it might cause a problem.

 

[allegro]

what do you mean by "whats your feeding"?
The MCU is provided with 3.3V by a 3.3V regulator.
The MCU output of the 2mA circuit can output up to 4mA.
The MCU output of the 8mA circuit can output up to 20mA.

 

[Nigel Goodwin]

No, what is the output pin feeding to? - that's all that matters.

And a diode would be completely useless.

 

[alphadog]

No, what is the output pin feeding to? - that's all that matters.

And a diode would be completely useless.

Oh, its feeding a BJT's base through resistor.

How come a diode is useless? if 0.5V falls on 0.7v VF diode, then the diode's current will be very low and therefore wouldnt open the BJT.

 

[Nigel Goodwin]

Oh, its feeding a BJT's base through resistor.

How come a diode is useless? if 0.5V falls on 0.7v VF diode, then the diode's current will be very low and therefore wouldnt open the BJT.

A diode under those circumstance won't drop 0.7V, it will drop much less.

Like in the other thread, two resistors is all you need - or perhaps just one series resistor, depending what your source is - certainly PIC's work perfectly well with only a single resistor.

 

[Papabravo]

The ability of an output to drive to 0V or to Vcc depends on the type of output stage that you have AND what kind of load it is driving. Your MPU output will only approach 0.5V if it is sinking several tens of milliamps such as might be the case with driving an IR LED transmitter.

A series resistor into a transistor base is no problem for a CMOS output stage to drive low to nearly 0V. Sourcing enough current into the base to get it to 0.7 V is a problem for some processors like athe 8051 and not a problem for others like the PIC or the AVR.

 

[alphadog]

Papabravo
According the regular diodes (like the 1N400x series), when Vdiode = 0.7V than IF >> mA.
You mean that when input is HIGH, and V_diode is 0.7V, the IF will be more than the MCU can push?
But when input is HIGH, the resistor already limits the base current to be 2mA which the MCU can push, so the V_diode will be around 0.5-0.6V, with IF being 2mA, whats the problem with that?

 

[ccurtis]

Alphadog, the point papa is making is that the output low level voltage depends on the amount of current the output is sinking. The more current the output is sinking the higher the output low level voltage will be.

Likewise, the output high level voltage depends on the amount of current the output is sourcing. The more current being sourced, the lower the output high level voltage will be.

I don't agree that the use of a diode to raise the threshold for turn-on is "completely useless". If the transistor is driving a large load (say a relay), a few milliamps of base current could be necessary to drive the transistor into saturation to energize the load and a base diode does indeed reach its full forward voltage only when a sufficient driving voltage (and thus base current) is reached. It may not necessarily be the best way to do it from a given standpoint, but it can be effective, depending on the load and the effect of small amounts of base current.

BTW, making a statement with the phrase "completely useless" without explaination from someone seeking comment and assistance, is irresponsible and rude -- especially from a "super moderator". It reflects poorly on the forum and its moderators. Maybe the forum's real purpose isn't what it is purported to be?