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Bc557 pnp transistor problem...

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yusuf

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Hello guys i am trying to use bc557 pnp transistor as switch...

I am using 5v power supply..

I have connected the collector pin to led via resistor and emitter pin to positive terminals of power supply..

And i have added 4.7k resistor to base..

The problem is i am getting 4v on base..
Without giving the signal to base...

That is as i power this circuit i am getting 5v on emitter and 4v on base without given base signal...
THE LED TURNS ON IF I TOUCH THE BASE...



I have replaced several bc557 pnp transistor.. But the problem is same..

Is there any fault in basing..
 
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A BC557 is an PNP transistor, it requires a Negative Base and Collector voltage, not Positive
 
a bc557 is an pnp transistor, it requires a negative base and collector voltage, not positive

actually i want to take pin no.1 high of cd4093 with the help of bc557 pnp transistor..

Can you provide simple circuit...
 
actually i want to take pin no.1 high of cd4093 with the help of bc557 pnp transistor..

Can you provide simple circuit...

This is one way to use a BC557 to control the 4093 pin #1
 
this is one way to use a bc557 to control the 4093 pin #1
what is mean by " when test 1 is low pin 1 is high "
dear moderator , i want if test 1 is high pin 1 is high .. And if test 1 is low pin 1 is low..

ACTUALLY I WANT IF THE ELECTRICITY IS AVAILABLE THE OPTO-COUPLER CONDUCTS AND IT WILL GIVE THE SIGNAL TO THE BASE OF PNP TRANSISTOR ..
SO THE PNP WILL WILL CONDUCT AND TAKE PIN 1 HIGH..

I HAVE ATTACH THE ELECTRICITY DETECTOR CIRCUIT PLEASE HAVE A LOOK....
Thanks again...
 
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You don't need a PNP transistor. Just connect the collector of the photo-transistor directly to pin 1 (assuming +V is in the range 3-15V).
 
you don't need a pnp transistor. Just connect the collector of the photo-transistor directly to pin 1 (assuming +v is in the range 3-15v).

i am using 5v dc power supply
but i am getting 2.11 v at pin 3 of opto-coupler... Which is not taking pins of 4063 high...

IS PIN 4 OF OPTO-COUPLER PC817 IS A COLLECTOR....
 
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I have attached the current working of my circuit........... In my attachment..

The output voltage is 2.11v from pin 3 of pc817 opto-coupler...
This voltage is not taking pins high of cd4093...

So to take pin high what should i do friends...........
 
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Remove R4 and replace the top 4k7 resistor with a 470 Ohm one.
 
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remove r4 and replace the top 4k7 resistor with a 470 ohm one.
if i remove r4 then the led flickers..

Why we will have to replace 4.7k with 470 ohms...

And why i am getting 2.11v at output of outo-coupler...
 
The opto-coupler turns on and off at the mains frequency so the average DC output voltage is less than half of 5V.
 
The opto-coupler turns on and off at the mains frequency so the average DC output voltage is less than half of 5V.
but I want to take pin high of 4093 logic ic... from the output of opto-coupler...
how should i do...
 
You don't need a PNP transistor. Just connect the collector of the photo-transistor directly to pin 1 (assuming +V is in the range 3-15V).
and what is this ... I didn't understood...
 
if i remove r4 then the led flickers..
It turns on and off every other half cycle, i.e. 50 times per sec.
Why we will have to replace 4.7k with 470 ohms..
Because 4k7 won't give enough current to light the LED noticeably.
 
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It turns on and off every other half cycle, i.e. 25 times per sec.
Because 4k7 won't give enough current to light the LED noticeably.
what about taking pins high of cd4093........

as the phase is detected it should also take pin 1 of 4093 high...
 
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Just connect the collector of the photo-transistor directly to pin 1 (assuming +V is in the range 3-15V).
Which bit of that don't you understand?
 
but I want to take pin high of 4093 logic ic... from the output of opto-coupler....
The output of the opto-coupler does go high, but at the mains frequency. Your DC meter shows the average DC voltage that is about +2.1V.

The circuit rectifies the mains sine-wave.
 
If you add a filter capacitor on the output, you can get 5V out. See attachments.
The half wave circuit requires a larger capacitor, and several cycles to get to a full logic level.
The full wave circuit uses a smaller cap, and can get to a full logic level within a half cycle.
 
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if you add a filter capacitor on the output, you can get 5v out. See attachments.
The half wave circuit requires a larger capacitor, and several cycles to get to a full logic level.
The full wave circuit uses a smaller cap, and can get to a full logic level within a half cycle.
thanks roff... But from where in this circuit i remove the connection to take pins of cd4093 high...
 
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