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From: firstname.lastname@example.org (Jumbaliah)
Subject: Re: blocking signal to unpowered CMOS
Date: 20 Nov 2002 12:08:51 -0800
NNTP-Posting-Date: 20 Nov 2002 20:08:51 GMT
Winfield Hill wrote in message news:...
> Bob Wilson wrote...
> > Jumbaliah email@example.com says...
> >> Is it good practice to use an NPN (3904) transistor to block a signal
> >> to an unpowered CMOS IC? I have a circuit that has a 5V signal on the
> >> collector all the time. However when I switch off the circuit, the
> >> base is left connected to other unpowered circuitry and the emitter of
> >> the 3904 is connected to the input of the unpowered CMOS IC via a 10k
> >> resistor. The ground for the circuit is always connected. When the
> >> circuit is powered up the IC gets power and the base of the 3904 also
> >> receives power thus allowing the signal to pass. (I know they're may
> >> be timing issues as to which powers up first)
> >> I've read about clamping diodes, but they only limit the signal's
> >> voltage range and not block it from an unpowered IC.
> >> Is this good? Bad?
> Jumbaliah, I cannot follow your description of the 3904 connection,
> can you please be more explicit?
pwr---\/\/\-base--|| 2n3904 pwr
(~12V) |\ |
| 10kohm |
I tested it on a bread board and it seemed fine. I was just unsure of
how it would react when the pwr is switched off (at both the CMOS IC
and the base of the 3904). I tried measureing (emitter to gnd) with
my DMM but the voltage reading fluctuated from mV to up around 20V. I
figured it was because the emitter is 'floating'. But the p-type
region of the transistor has no voltage source and therefore no
current to allow the signal at the collector to pass through into the
I came across previous posts about floating bases on transistors to
make a white noise generator. Not what i had in mind. So, I figured
I'd see what you guys thought.
I looked more into this type of circuit and I believe it's like a 'bus
switch'. Only I'm using a BJT and not a FET.
See figure 5
> > An unpowered CMOS IC will do funny things when voltage is applied to
> > its input. What happens is that the voltage goes through the upper
> > of the protection diodes, and onto the interal Vcc line, and the
> > IC springs to life.
> > There are a variety of ways to ensure the inputs are left "unpowered"
> > choose one that works for you.
> I agree, some method is necessary. At least a series resistor.
> One that seems nice, although I haven't used it, employs JFETs.
> Its purpose is to reduce the value of the series input protection
> resistor to retain a fast logic connection, without increasing
> the fault current into the unpowered device's protection diode.
> . ,--------,
> . | |
> . ____V_ |
> . | | |
> . ----' '--/\/\--+---
> Use a small low-capacitance low-threshold-voltage JFET, such as a
> 2n3687 - pn3687. Hmm, I'm sure there's a more available choice.
> Another method employs small MOSFETs, with their gates connected to
> the destination +5 supply. When that supply is off the FET is off.
> Low-threshold-voltage MOSFETs are necessary, because the destination
> logic is robbed of a that voltage, say 0.7 to 1.5V for a 2n7000.
> This is not a problem for HCT logic, etc., with 5V input swings and
> a 1.2V logic threshold.
> - Win
Nice, I'll keep that in mind.
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