[Test-Equipment] Need to shift pulse up 5V

[Barry]

I see it now.  I was thinking of it incorrectly.  When the NPN is conducting, the two equal-value resistors form a voltage divider between +5V and -5V and when it is not conducting, that point is seeing +5V so - non-inverting.

One small point: doesn't the transistor's two diode-drops come into the equation effectively shifting that zero point when it is conducting?

I have a handful of 2N3904s so I will be trying this soon and, I agree - definitely a more elegant solution than a comparator.

BTW, these are strobe signals that set latches so not sure if an inverted clock would be okay but it might.  With a non-inverted pulse, though, I won't worry about it.

Thanks again,
Barry - N4BUQ

----- Original Message -----
> From: "Mike Manes" <mrmanes at gmail.com>
> To: "Discussion of Electronic Test Equipment" <test-equipment at mailman.qth.net>
> Sent: Friday, May 1, 2015 12:10:44 PM
> Subject: Re: [Test-Equipment] Need to shift pulse up 5V
> 
> No, it doesn't invert the clock.  When the emitter is pulled low by the
> -5V clock phase, the NPN turns on, and the collector voltage drops to
> zero.  When the input clock goes high to 0V, then the NPN turns off and
> the collector voltage goes to +5 (or whatever +V bus it's fed from).
> 
> If the clock sig is just a freq reference, then inverting it wouldn't
> be no never-mind.
> 
> 73 de Mike W5VSI
> 
> On 5/1/15 10:23, David DiGiacomo wrote:
> > On Fri, May 1, 2015 at 10:13 AM, Barry <n4buq at knology.net> wrote:
> >> If I understand this correctly, when the clock is at -5V, the collector is
> >> at +5V and when the clock it at 0V, then the collector is at 0V.  Is that
> >> correct?  If so, then doesn't that invert my clock signal?  If I'm
> >> thinking about it correctly (and I may be wrong), the comparator can be
> >> set to do either - depending on which way the inputs are connected,
> >> correct?
> >>
> >> I like the single transistor idea but if it inverts the signal, I'm not
> >> sure that's what I'm needing.
> >
> >
> > The common base circuit is non-inverting.
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