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How a Transistor connected to 12v and out 4.7v !!!!!!? Answered

Hello All,

I have a problem with a circuit i am building, I use a 2N2222 transistor to drive a Common Cathode big 7 Segment  with a 595 shift register..
I connected a 12v to Emitter, the collector out a 4.7 volts !!
How is this possible ? could some one explain please ?

Thanks all

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steveastrouk
steveastrouk

Best Answer 9 years ago

You need to put a "real" load on the circuit, your meter is effectively an open circuit, so there is no voltage developed across it.

And PLEASE draw your circuits correctly. ALWAYS put the ground rail at the bottom, and make the circuit flow inputs from the left, outputs to the right !!!!

Steve

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EZELab
EZELab

Answer 9 years ago

Steve,

This circuit is real beside me now.. i connected the Transistors to 7 segments already, and when I measured the transistor output, it gives 5v, Thats why i tried to test it on a simulation.. Sorry for my bad drawing, Its just a test .

Thanks
Kim

0
steveastrouk
steveastrouk

Answer 9 years ago

Probably, reduce your base resistor, you aren't saturating the base of the transistor, so it isn't turning on properly. I assume you have got a limiting resistor in series with the segments ???

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EZELab
EZELab

Answer 9 years ago

I reduced the base resistor to 1 ohm !! and i am getting same value !!
And yes I use a limit resistor with segments ..

0
steveastrouk
steveastrouk

Answer 9 years ago

What LED current do you want ? You may have a damaged transistor.

Steve

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EZELab
EZELab

Answer 9 years ago

The 7 Segments i use is a 4" common cathodes. The seller didn't provide a datasheet for it.
I tested it with 12v & 470 ohm resistor and worked good..
Then tested it with 5v & 150 ohm resistor, But it didn't work. this is why i am trying to use 12v.

Is there any other way to control them ?
I control them with 74595 shift register & 2N2222 transistors.
And coz they are comm cathodes, I connect +12 to transistor..
I should get comm anodes, But I must find a way to control them..


0
steveastrouk
steveastrouk

Answer 9 years ago

What you have made is an emitter follower: you won't GET more than base voltage - Vbe across the load !!!!

To drive these things you need a PNP transistor and an NPN transistor.

Steve

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EZELab
EZELab

Answer 9 years ago

Steve,

Is there any 74xxx or 45xx or 40xx IC can do this job ? i have couple pins need this circuit, So if there's a IC do this job it will be much better

Thanks

0
EZELab
EZELab

Answer 9 years ago

Worked. Thank you

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EZELab
EZELab

Answer 9 years ago

I just figured this from little ago (http://img36.imageshack.us/img36/3779/arduinoclock.png)

I'll use BC328 & BC237 for this.. i'll give a try.

Thanks

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steveastrouk
steveastrouk

9 years ago

If you just want a BCD display use a 4511 !

Steve

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EZELab
EZELab

Answer 9 years ago

Yes I am using 4511, With 12v , Arduino work with 5 volts.. How i can control 12v chip with 5 volt Arduino, That's why I am building this circuit, instated of using your circuit 4 times, I am searching for a IC do this job..

Can 4016 do this ? 12 volt as supply & 5v input signal ?

Thanks

0
lemonie
lemonie

9 years ago

Why have you got that arrangement (diagram)?
One of the meters appears to wired in series with the transistor, you appear to have the transistor connected the wrong way too.

L

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EZELab
EZELab

Answer 9 years ago

I just draw this circuit to test why my circuit dose not work in real world.. so I am sorry for bad drawing.. It's just a test anyway..

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lemonie
lemonie

Answer 9 years ago

It doesn't look right - check what you're doing with that first.

L

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EZELab
EZELab

Answer 9 years ago

Here's a better drawing, With a relay load. It doesn't work

Capture.PNG
0
verence
verence

Answer 9 years ago

No, it _does_ work. It works exactly as designed. Only the design is wrong. :-)

You apply 5V across Base/Emitter of Q1. Enough to open it! That's what you want to do. But as current flows through the relay a voltage is created there through to the resistance of the relay. That voltage (that you do measure) 'lifts' the emitter of Q1 from 0V to 4.2V. So from the point of Q1, it 'sees' only about 0.8V (5V-4.2V) across the Base/Emitter junction. Q1 doesn't know that there is a common ground, it only follows the voltages/currents on/through its three pins.

What you can do is:

a) if the load doesn't have to be connected to GND, put the relay between 12V and the collector of Q1. That way, the emitter of Q1 will see no voltage drop and be always connected to GND, so the base will always see 5 (increase the base resistor!).

b) if the load has to be connected to GND, use steveastrouk circuit with two transistors.

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EZELab
EZELab

Answer 9 years ago

Thank you, I used Steve circuit and works great.
Thank you all

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lemonie
lemonie

Answer 9 years ago

Yes, Steve knows this stuff.

L