PFET Reverse Polarity Protection

[hgg]

Hi,

I am trying a simple but very effective little circuit for reverse polarity protection which
you may have seen.



Same circuit with a small modification:




I have just added a little diode and an LED so that when the polarity is reversed the LED
will light up and give you a warning.  The diode is to protect the LED from reverse voltage
under normal operation and its the BAT45S.  It works great.

My question is that when the polarity is correct I can measure 456mV across the LED.
(2.5mV across the diode).     Do you know why is that?

Thank you.

[T3sl4co1l]

Hmm, bet the voltage drop across the LED is even higher.  You could test with a JFET source follower, or JFET opamp follower, to see about what voltage it's at.

You would expect with, say, 8.2V in as shown, that "no" current flows, so R1 has 0V across it and the voltage between the diodes is undefined (anywhere between 0 and 8.2V).  In reality, BAT45 is a very leaky diode, and will drop very little voltage relative to the LED, which might be a very good diode (i.e., low reverse leakage).  Depending on ambient light level, that is.

Worth asking... did you measure this with real instruments (and if so, which), or in the simulation (that looks like... Multisim?)?

Tim

[hgg]

Depending on ambient light level, that is.

You found it!  I forgot that LEDs become voltage sources under a light source.
When I cover the LED, the voltage drop goes from 456mV to 10mV and when
I shine a very strong light on it, the voltage drop goes up to 1.3V !!!  It acted
like a solar panel.   That is a nice way to burn any sensitive components.   

Mystery solved.  Thanks!   

p.s. No simulations.
With an even stronger flashlight, 2.1V...

[hgg]

Correction for those who want to try it.

If I leave the resistor as is, then when polarity is reversed I get a -3.0V across the load...
It should be connected from the anode of the warning LED to ground and the MOSFET gate
tied directly to ground.  Not sure why it works like that.   I guess Vgs becomes lower than it
should be, because of the voltage drop caused from the diode and the LED.


Corrected