Simple question about this "trip wire" circuit.

[PizzaCombo]

Hi all, first post here and still getting the hang of electronics .

So I am looking at making the circuit found on this page http://www.instructables.com/id/Extremely-Easy-and-Safe-Trip-Wire/ and I think I understand it all, except for one part.

So when the "trip wire" has been broken, the piezo buzzer makes noise as the current is forced to go through the base of the transistor, and lets the circuit flow.

What I'm not understanding is when the trip wire is still connected, why is there not any current being applied to the base of the transistor? Why does the current just go right past the base and straight to ground? I have a feeling it's something to do with "Electricity wants to take the path with least resistance" but I'm not entirely sure what's going on.

I hope this is clear enough, thanks.

 

[davenn]

hi there
welcome to the forums

I have a feeling it's something to do with "Electricity wants to take the path with least resistance" but I'm not entirely sure what's going on.

yes that's basically the answer. to take it further, the transistor needs ~ 0.7 of a volt on the base to turn on to make a conduction path between the collector and emitter to operate the buzzer

With the loop intact the base is held at 0V ( the negative of the battery) so the transistor cannot turn on
When the loop breaks, the base voltage will rise to ~ 0.7V or so and the transistor will turn on and allow the buzzer to operate

cheers
Dave

 

[PizzaCombo]

hi there
welcome to the forums



yes that's basically the answer. to take it further, the transistor needs ~ 0.7 of a volt on the base to turn on to make a conduction path between the collector and emitter to operate the buzzer

With the loop intact the base is held at 0V ( the negative of the battery) so the transistor cannot turn on
When the loop breaks, the base voltage will rise to ~ 0.7V or so and the transistor will turn on and allow the buzzer to operate

cheers
Dave

Thanks Dave, good explanation. So essentially by being connected straight to the negative terminal of the battery, it makes the base of the transistor grounded?

 

[davenn]

Thanks Dave, good explanation. So essentially by being connected straight to the negative terminal of the battery, it makes the base of the transistor grounded?

yup, so a voltage cannot develop on the base pin in that condition

when the loop is broken, a voltage divider is produced by the 2 resistances
1) the 10k resistor from base to +V and
2) the base - emitter junction resistance of the transistor

knowing that it takes ~ 0.7 - 1V drop across the base - emitter junction to turn the transistor on
we then know about 5V will be dropped across the 10 k resistor
from that we can work out the current flowing through the 10k resistor and base - emitter jnct
if we wanted to..... it will be very low !

Dave

 

[PizzaCombo]

yup, so a voltage cannot develop on the base pin in that condition

when the loop is broken, a voltage divider is produced by the 2 resistances
1) the 10k resistor from base to +V and
2) the base - emitter junction resistance of the transistor

knowing that it takes ~ 0.7 - 1V drop across the base - emitter junction to turn the transistor on
we then know about 5V will be dropped across the 10 k resistor
from that we can work out the current flowing through the 10k resistor and base - emitter jnct
if we wanted to..... it will be very low !

Dave

Cheers Dave, helped me understand it.