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[DPRG] Over dioded?

Subject: [DPRG] Over dioded?
From: Bob Jordan rljordan at airmail.net
Date: Tue Dec 28 13:39:29 CST 2010

Dale,

 

Can you use one of the LEDs AS the protection diode and get any light out of
it??

 

Putting the LED on the driven side of the transistors with a jumper or a
land to cut if it needed to have a jumper installed seems a better idea.

LEDs and jumpers optional even.

 

Just random thoughts!

 

Bob Jordan

 

 

From: dprglist-bounces at dprg.org [mailto:dprglist-bounces at dprg.org] On Behalf
Of Dale Wheat
Sent: Tuesday, December 28, 2010 11:20 AM
To: dprglist at dprg.org; jdrumm9015 at aol.com
Subject: Re: [DPRG] Over dioded?

 

John,

 

Thanks for looking at this for me.  I'm leaning towards keeping the
protection diodes, but moving the LEDs a little farther to the right on the
schematic, so that they, too, enjoy the protection afforded by the other
diodes.

 

Dale

 

 

From: jdrumm9015 at aol.com 

Sent: Tuesday, December 28, 2010 9:39 AM

To: dale at dalewheat.com ; dprglist at dprg.org 

Subject: Re: [DPRG] Over dioded?

 

 

 

 

Dale,  

  The inherent "diode-ness" will only work if you knew what the other end
was referenced to.  The protection diodes work well since one end is
connected to ground, you know what the reference is.  When the other end
drops one forward biased diode drop (0.6-0.7 volts Si) then the diode
conducts and protects the line from negative transients.

 

  The diodes in series with the data are referenced to the base of the
transistors (Q1, Q2, Q3).  Is that diode forward biased or not?  What would
a negative transient see on that line?  Maybe it depends on the logical
value on the other end of the series diode, or maybe it depends on if the
transistor base-emitter is forward biased.  I believe a negative transient
is almost guaranteed to blow out the base-emitter junction of the
transistors, Q1, Q2, or Q3 since the series diode on state is somewhat
nebulous given the unknown state of the base-emitter on state.

 

  Just my two cents.

 

John Drummond

 

PS:  The series resistor (high tech battery charger circuit) might give you
all the protection you need anyway since it will limit any transient current
flow, you still might be able to eliminate the protection diodes, but not
because of the series LEDs.


 

-----Original Message-----
From: Dale Wheat <dale at dalewheat.com>
To: dprglist <dprglist at dprg.org>
Sent: Mon, Dec 27, 2010 6:31 pm
Subject: [DPRG] Over dioded?

I'm designing a new RS-232 to TTL adapter for the NXP LPC Cortex-M0 and 
Cortex-M3 processors.  It uses some of the modem control lines to reset the 
chip and put it into bootloader mode.  I'm using very simple 
resistor-transistor level shifting to constrain the incoming signals to 
typically a 0V-3V range (or whatever is supplied on the Vcc pin).
 
My original design works well, even with "barely compliant" USB to RS-232 
adapters.  I went and got all fancy by adding some LEDs inline with the 
incoming signals, instead of adding an additional driver transistor for each

signal.  Using high-brightness LEDs, I can just see the LEDs twinkling when 
bits are flowing to and from the chip.
 
My question is this:  With the LEDs in series with the incoming signals, do 
I still need the protection diodes (D1, D2, D3) to shunt potential negative 
levels to ground?  Or will the inherent diode-ness of the LEDs prevent this 
from ever happening?
 
Here is a PDF of the schematic-in-progress:
 
http://dalewheat.com/info-content/RS232%20to%20TTL%20bootloader%20schematic.
pdf
 
Any suggestions are appreciated!
 
 
Thanks,
 
Dale Wheat
http://dalewheat.com <http://dalewheat.com/> 
(877) DALE WHEAT
(972) 486-1317 
 
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