Hi,
I picked up "Getting Started in Electronics" by Forrest Mims and I'm having trouble understanding how a particular circuit works.
It's the "light flasher" circuit on pg 104 of the third edition. I'm hoping someone out there has the book and can help. If not, am I allowed to scan and post the circuit ?
Anyway, I think it's just a basic lack of understanding of circuits in general and capacitors and transistors in particular but I'm not sure I understand how this circuit work. let me take a stab at it:
First I assume the input voltage is DC.
My guess is that Q2 is allowing current flow as soon as power is supplied to the circuit and hence the light is on.
The capacitor begins to charge (rate being controlled by R1) and once it hits .6V Q1 allows current flow.
Does this in effect bypass the light (turning it off) because the resistance is now less through Q1 than through Q2 and the light ? The capacitor would then be grounded and discharge, closing the base at Q1 and starting the cycle over ?
I guess part of what is confusing me if I'm wright is why use Q2 at all if it is always on ?
I'm probably way off the mark but if some kind sole could help me out that would be cool.
Thanks
Marc
I picked up "Getting Started in Electronics" by Forrest Mims and I'm having trouble understanding how a particular circuit works.
It's the "light flasher" circuit on pg 104 of the third edition. I'm hoping someone out there has the book and can help. If not, am I allowed to scan and post the circuit ?
Anyway, I think it's just a basic lack of understanding of circuits in general and capacitors and transistors in particular but I'm not sure I understand how this circuit work. let me take a stab at it:
First I assume the input voltage is DC.
My guess is that Q2 is allowing current flow as soon as power is supplied to the circuit and hence the light is on.
The capacitor begins to charge (rate being controlled by R1) and once it hits .6V Q1 allows current flow.
Does this in effect bypass the light (turning it off) because the resistance is now less through Q1 than through Q2 and the light ? The capacitor would then be grounded and discharge, closing the base at Q1 and starting the cycle over ?
I guess part of what is confusing me if I'm wright is why use Q2 at all if it is always on ?
I'm probably way off the mark but if some kind sole could help me out that would be cool.
Thanks
Marc