The output of U1B will never go to 5 volts. Probably over a volt or more
from 5 volts. The diode D2 has a forward drop of .7 volts. The total is >1.4
volts. This gives you about 3 volts charging the cap. Replace U1B with a
BS170 and put a pullup on the drain.
Rick

Chris Emerson wrote:

> I'm confused about a circuit I've been experimenting with.  The circuit
> is at http://www.nosreme.org/tmp/timer1b.png, using a 74HC14 hex Schmitt
> trigger inverter.
>
> U1A (buffered by U1B) produces a low duty cycle square wave, which
> slowly charges C2 through D2 and R3.  So far so good.  The voltage on C2
> starts to rise with the expected exponential curve, but then stops at
> about 2.6V (Vcc is 5V from a 7805), and I don't understand why.  I would
> expect it to rise close to 5V minus a diode drop.
>
> I've tried the following:
>
> 1. Disconnecting U1C, in case input leakage was causing problems - no
> effect on C2.
>
> 2. Changing C2 does change the time constant (I've tried 0.1uF and
> 100uF) as expected, but doesn't change the final voltage.
>
> 3. Reducing R3 does increase the final voltage on C2, but not by very
> much.  With R3 around 20-30k C2 gets up to about 3V.
>
> I'm stumped.  Can anyone enlighten me?
>
> The overall intention is a 4-5 minute timer (turn on, and a few minutes
> later it beeps), using the first IC I happened to pick up.  I already
> know it's a terrible circuit.  :-)
>
> Chris
>
> P.S. Thanks to everyone who helped with my previous question about DC
> motors; the thread is still in my mailbox waiting for me to have time to
> do something about it!
>
> --

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