J
John Fields
Guest
On Wed, 31 Aug 2005 04:17:33 GMT, Robert <hedrock@insightbb.com>
wrote:
Originally, the plan was for the cap to act like a differentiator
when the supply was turned on and send a positive going pulse into
the base of Q1 to turn it off and then to let it charge gradually,
pulling the base of Q1 down to ground gardually, thus turning Q1 on
gradually. Unfortunately, to get a time constant of, say, one
second, the cap would have to have a capacitance of:
T 1s
C = --- = ------ = 1.96E-6F ~ 2000ľF
R 510R
Not bad, but the hitch is that the base only has to go about 0.7V
more negative than the emitter in order to turn the transistor on,
and as soon as it even _starts_ turning on, current will be flowing
into the cap from the collector-emitter diode, further shortening
the turn-on time. In short, unless you really use a huge cap, it
won't make much difference what you put in there. Here's a solution
that works:
+18>--> | Q1
|S1 PNP
O--+------E C------+----> >---+
|R1 B | |
[100k] | | |
| C Q2 |+C2 |Rl
+------B 2N4401 [4700ľF] [50R]
| E | |
|+C1 |R2 | |
[47ľF] [510R] | |
| | | |
GND>-------+--------+--------+----> >---+
You can vary the length of the turn-on by changing either R1 or C1,
and the length of the turn-off with C2.
As shown, your turn-on ramp will be about half a second long.
--
John Fields
Professional Circuit Designer
wrote:
---On Tue, 30 Aug 2005 18:34:40 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:
He said the 470uf cap wasn't needed.
Originally, the plan was for the cap to act like a differentiator
when the supply was turned on and send a positive going pulse into
the base of Q1 to turn it off and then to let it charge gradually,
pulling the base of Q1 down to ground gardually, thus turning Q1 on
gradually. Unfortunately, to get a time constant of, say, one
second, the cap would have to have a capacitance of:
T 1s
C = --- = ------ = 1.96E-6F ~ 2000ľF
R 510R
Not bad, but the hitch is that the base only has to go about 0.7V
more negative than the emitter in order to turn the transistor on,
and as soon as it even _starts_ turning on, current will be flowing
into the cap from the collector-emitter diode, further shortening
the turn-on time. In short, unless you really use a huge cap, it
won't make much difference what you put in there. Here's a solution
that works:
+18>--> | Q1
|S1 PNP
O--+------E C------+----> >---+
|R1 B | |
[100k] | | |
| C Q2 |+C2 |Rl
+------B 2N4401 [4700ľF] [50R]
| E | |
|+C1 |R2 | |
[47ľF] [510R] | |
| | | |
GND>-------+--------+--------+----> >---+
You can vary the length of the turn-on by changing either R1 or C1,
and the length of the turn-off with C2.
As shown, your turn-on ramp will be about half a second long.
--
John Fields
Professional Circuit Designer