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Prevent signal when device is turned on

I have one of the output pins of a 16F84A hooked up to a transistor
which in turn drives a relay. Though the software can drive that pin
low, my concern is that during power up there might be enough current
to switch the transistor.

Since this is a kit-type project I can not do major modifications, so
is there a simple way to bypass the base for a brief period of time
and only when the signal is high for a longer time will it actually
switch the transistor ?
 
Hm...have you noticed that that the transistor is turnin on when
during power up ? From what i recall, the pins are in high Z until
confired to be either input or output. There shouldnt be any current.

Seems interesting. Post results or fixes.
 
J

John Fields

Jan 1, 1970
0
I have one of the output pins of a 16F84A hooked up to a transistor
which in turn drives a relay. Though the software can drive that pin
low, my concern is that during power up there might be enough current
to switch the transistor.

Since this is a kit-type project I can not do major modifications, so
is there a simple way to bypass the base for a brief period of time
and only when the signal is high for a longer time will it actually
switch the transistor ?

---
Yes. View in Courier:

+V
|
+------+
|K |
[DIODE] [COIL]
| |
+------+
|
C
PICI0----[R]-----------+------B
| E
C |
Vcc>--[C1]--+--[R1]--B 2N3904 |
|K E Q1 |
[CR1] | |
| | |
GND>--------+----------+--------+
 
J

john jardine

Jan 1, 1970
0
I have one of the output pins of a 16F84A hooked up to a transistor
which in turn drives a relay. Though the software can drive that pin
low, my concern is that during power up there might be enough current
to switch the transistor.

Since this is a kit-type project I can not do major modifications, so
is there a simple way to bypass the base for a brief period of time
and only when the signal is high for a longer time will it actually
switch the transistor ?

Hang a capacitor on the connection between the transistor base and it's base
resistor and the 0V. (something like 1uF).
 
J

John Fields

Jan 1, 1970
0
Hm...have you noticed that that the transistor is turnin on when
during power up ? From what i recall, the pins are in high Z until
confired to be either input or output. There shouldnt be any current.

Seems interesting. Post results or fixes.

---
Well, of course, your highness. Your wish is our command. BTW,
since you're new here and posting from Google groups, it'd be nice
if you'd learn to bottom post like most of the rest of the
non-cretins on USENET do.

Thank you.
 
J

John Fields

Jan 1, 1970
0
I have one of the output pins of a 16F84A hooked up to a transistor
which in turn drives a relay. Though the software can drive that pin
low, my concern is that during power up there might be enough current
to switch the transistor.

Since this is a kit-type project I can not do major modifications, so
is there a simple way to bypass the base for a brief period of time
and only when the signal is high for a longer time will it actually
switch the transistor ?

---
Yes. View in Courier:

+V
|
+------+
|K |
[DIODE] [COIL]
| |
+------+
|
C
PICI0----[R]-----------+------B
| E
C |
Vcc>--[C1]--+--[R1]--B 2N3904 |
|K E Q1 |
[CR1] | |
| | |
GND>--------+----------+--------+

---

Oops... to keep the base from floating:


+V
|
+------+
|K |
[DIODE] [COIL]
| |
+------+
|
C
PICI0----[R]--------------+------B
| E
C |
Vcc>--[C1]--+--[R1]--+--B 2N3904 |
|K | E Q1 |
[CR1] [R2] | |
| | | |
GND>--------+--------+----+--------+
 
Hang a capacitor on the connection between the transistor base and it's base
resistor and the 0V. (something like 1uF).

Okay, I see how that takes care of things while the signal is briefly
high. But what happens if the pin goes low or high impedance? Won't
the capacitor then discharge via the Base-Emitter ?
 
J

john jardine

Jan 1, 1970
0
Okay, I see how that takes care of things while the signal is briefly
high. But what happens if the pin goes low or high impedance? Won't
the capacitor then discharge via the Base-Emitter ?
(assuming of course an NPN transistor)
PIC pin high for a decent period and the cap sits about 0.7V with the
transistor ON. (same action with or without the capacitor)
PIC pin low for a decent period and the capacitor discharges via the
resistor to 0V and the transistor goes OFF. (same action with or without the
capacitor)

If transistor is ON and PIC pin switched to high impedance then capacitor
discharges via BE till transistor switches OFF (about 0.5V on capacitor)

If transistor is OFF and PIC pin switched to high impedance then transistor
stays off. The capacitor even provides some noise filtering.

But ... If the PIC pin is liable to be switched from output to input, then
there should (capacitor or no capacitor) also be an additional pullup or
pulldown resistor fitted somewhere, sufficient to define the transistor
ON-OFF state, under this rather odd operating condition.
 
J

jasen

Jan 1, 1970
0
I have one of the output pins of a 16F84A hooked up to a transistor
which in turn drives a relay. Though the software can drive that pin
low, my concern is that during power up there might be enough current
to switch the transistor.

does startup take long enough for the relay to respond?
IOW: does that really matter?
Since this is a kit-type project I can not do major modifications, so
is there a simple way to bypass the base for a brief period of time
and only when the signal is high for a longer time will it actually
switch the transistor ?

huh?

a capacitor to ground will slow the response a bit.


Bye.
Jasen
 
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