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# transistor as a switch

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Hi
can you explain What R8 (1K) exactly do?
is it a voltage divider??
thanks

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Hi Walid,
You have a very good point. ;D
The circuit can't work if R8 is there!

The supply voltage is 6.8V if the battery is fully charged.
Then the output high of the 555 is about only +5.6V.
R5 and R8 form a divide by 11 voltage divider so the base voltage of T2 is only 0.51V and so T2 never turns on since it needs at least 0.65V.

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hi, just wondering, what does this circuit do?
is it just to change the output voltage of the 555?
if so what does the transistor for?

thanks for the help.

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To guru
thank u very very much you are good and expert man.

hi, just wondering, what does this circuit do?
is it just to change the output voltage of the 555?
if so what does the transistor for?

this is a part of a circuit shown in the following link:
http://www.electronics-lab.com/forum/index.php?topic=10229.0
it is an IC controlled emergency lamp with charger
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hi guru

The supply voltage is 6.8V if the battery is fully charged.
Then the output high of the 555 is about only +5.6V.

5.6/6.8 = 82%
is the output high of the 555 is always about 82% of Vcc?
thanks
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hi guru

5.6/6.8 = 82%
is the output high of the 555 is always about 82% of Vcc?

No. It is about 1.3V, not a percentage of the supply voltage.
The output of a 555 has a darlington emitter follower to pull its output high. It is two base-emitter junctions in series. 0.6V + 0.7V= 1.3V.
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Hi guru

It is two base-emitter junctions in series. 0.6V + 0.7V= 1.3V.

part of the 555 schematic is shown below.
yes, It is ok, there is a darlington emitter follower, but why not two 0.6 or two 0.7 volt. why different VBE?
thanks
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The output transistor passes the load's high current.
Its driver transistor operates at a much lower current, just the base current of the output transistor.
An emitter-follower  transistor with a high current has a higher Vbe than a transistor with a low current.

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