Substituting a 74HCT74 for a 4013

Can someone tell me how I could use a 74HCT74 in place of the 4013 in
this circuit...
http://www.uoguelph.ca/~antoon/gadgets/rc-sw.htm

I tried adding a pull up resistor to both sets and the second reset.
Then I ran the first Q output into one of the inverters on a 7404, with
the capacitor to ground and then the output of the inverter into the
first reset but it still doesn't seem to be working. Any ideas? I
really don't know what I am doing here. Oh and I just need the high or
low output, I don't need the TMOS FET.

--
Chris W

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--- ___ _____
The HC74 needs a logic low asserted on the SET and RESET inputs in
order to make them work, while the 4013 needs a logic high to do the
same thing. Therefore, if you want to substitute an HC74 for a 4013,
you'll need to invert the signals driving SET and RESET on the 4013.

 

I guess I wasn't clear, but that is what I am trying to do. After more
checking, I think the inverter I have isn't working right. With no
connections the inputs all read 1.6v and it takes a resistor no larger
than 100 ohms to take pull it down below .1v. Having a resistor in
there is obviously going to interferer with my capacitor charging.
Shouldn't the capacitor pull an input down to 0 volts or is this thing
defective? The markings on the chip say DM7404N.

--
Chris W

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You don't really need an inverter. The chip has both Q and /Q logic
outputs for both flipflops. Thus, just connect the timeout circuit to /Q
on the first flipflop and tie the unused sets and reset to 5V instead of
ground. Everything else should be the same.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

---
I missed the part about the 7404 before, but that's what's keeping it
from working, since you can't pull its input down with R1 and the pot
in there. Sinse the sense of SET and RESET is backwards for the HC74
than it it for the 4013, if you connect the unused SET and RESET lines
to Vcc instead of ground and drive the RC from Q- instead of from Q,
as shown below, it ought to work.


VCC>-----+------+-----------------+
| | |
+-------------------+ |
| | | | |
| | +-------+ | +-------+
| +--|D S- Q| +----|D S- Q|---->OUT
| | _| | _|
IN>--+------|> R- Q|----+----|> R- Q|
+-------+ | +-------+
| [47K] |
| | VCC
+---->[100K]
| |
+--------+
|
[22nF]
|
GND>---------------------+

 

That worked, thanks John. I don't know why I didn't think of that, as
an experienced computer programmer I don't know why it is taking so much
to get my head wrapped around these logic circuits. I think I am just
trying to make it more complex than it really is. Thanks again.

--
Chris W

Gift Giving Made Easy
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give the gifts they want
http://thewishzone.com

 

Being a programmer you're not accustomed to have the value of a variable and
its inverted value available at the same time. So you implemented an
"invert" operation. Always wondered why some people tend to split up the
world in hardware and software with hardly any appreciation for each other.
From the first time I touched a computer I've always been aware one needs
both hard- and software to run that machine. You apparently experienced the
same

petrus bitbyter