Yes
and no.
?
I
didn't quite catch how your circuit is build. A phototransistor does have
"variabel resistance". The more light you apply, the more Ic you get (light here
is used as the base current).
?
However, if you have the phototransistor turning on/off
a relay. And you have a piezo on the other side of the realy, the this resitance
shoulder matter. Cause the relay can only be on or off, nothing in
between.
?
Did I
misunderstand you here?
?
Mounir
toggle quoted message
Show quoted text
Hello
Everyone,
I am new to the group and know just enough about electronics
to be
dangerous. The circuit that I am trying to complete is 9V DC. I have
placed a phototransistor just before a relay. The relay is 5V. A
piezo
alarm is connected to the "Normally Open" lead on the relay.
For some
reason, I cannot get enough power through the
phototransistor to activated
the relay switch. When I test the
circuit, the alarm quietly sounds and
becomes slightly louder when
more light is applied.
It is as though
there is resistance with regards to the
phototransistor.
Please
forgive me is my terminology is incorrect. I hope that someone
understands
this. Any help would be greatly appreciated.
Sincerely,
Kenyon
Jones
To
unsubscribe from this group, send an email
to:
Electronics_101-unsubscribe@...
Your
use of Yahoo! Groups is subject to the .
