This is a schematic from the ladybird book called 'learnabout Simple Electronics' circa 1979......I am working my way through electronic theory and practice from the ground up, I built this circuit on plugboard and it does exactly the opposite of what is tells me in the text, it tells me that when I attach the flying lead to ground, the bulb will slowely light, it doesn't, the bulb starts at full brightness and slowely goes out, now looking at the circuit I agree with my actual results since DC current will be blocked progressively as the capacitor charges up via the transistor and the base resistor.....can someone elighten me to what is going on here please?

Neil.

 

Hey, what can I say? You circuit work as expected. Maybe they print the wrong diagram.
Also notice that there is no path for capacitor to discharge. So this circuit will work only once.
So we need manually discharge the capacitor.

 

Yes Jony, indeed I have to discharge the li'l bugger each time, I just wanted conformation of what I was actually experiencing was correct instead of what I was being told.

thank you

Neil.

 

You're right, the book is wrong.

 

For your next assignment;

Fix the circuit so the bulb will slowly light up. No extra parts are allowed.

 

I havn't tried it but looking at the circuit again I would remove the bulb from the collector lead and insert it in the emitter lead?

Neil.

 

I would remove the bulb from the collector lead and insert it in the emitter lead?

Can you explain why you think that this change will work?

 

Because as the transistor starts to shut down, the emitter voltage will rise and so lighting the bulb?

Neil.

 

There's at least one other (imho, slightly better) way to do it, too. hint: think about what you said in your original post!

Also, just to join the chorus, you're definitely right. They're definitely wrong.

 

From my perspective for a book entitled 'learnabout Simple Electronics' it does not appear to be that usefull a circuit?
I would have thought the emphasis would have been on something that might be of a more practically use or method?
Max.

 

Ooh ooh I know..... I could take the flying lead to the positive rail instead right?

Neil

 

Because as the transistor starts to shut down, the emitter voltage will rise and so lighting the bulb?

Neil.

Are you sure that you know how PNP transistor work??
How can any emitter or collector current flow if base current is 0A?
In which of this two circuits the bulb will light?

 

The bulb lights in the first circuit since taking the base positive will turn the transistor off.

Neil.

 

I need to set my breadboard up again, I'm really confused.... Pnp needs 0v to conduct and non needs a positive voltage above. 6v to conduct?

 

PNP need a path for a base current, from "+" battery---->emitter-base junction --->some resistance---> "-" battery
http://forum.allaboutcircuits.com/showthread.php?p=165135#post165135

The emitter-base junction act just like a diode.
But Ie is beta + 1 larger then Ib current

 

Oh wait..... I see what you are saying, yes there is no current flow so nothing will happen...... The plot thickens!

 

Thanks for that link Jon, I need to find time to build that circuit and take measurements, I have all the parts.

Neil.

 

Jony, I still don't understand why the bulb BL2 is off in that picture.
ib2 = Vbe2/R2
ic2 = beta*ib2

 

So the only difference between pnp and non is the direction of current flow as directed by the arrow in the circuit, I get it.... Thankyou

Neil.

 

Jony, I still don't understand why the bulb BL2 is off in that picture.
ib2 = Vbe2/R2
ic2 = beta*ib2

But notice that Q2 is a PNP transistor so the base need a path to ground for his base current. Do you see any direct path from Q2 base to GND ?

So the only difference between pnp and non is the direction of current flow as directed by the arrow in the circuit, I get it.... Thankyou
Neil.

Yes that's the main difference.