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Please help about transistor as switch

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This circuit came out of a book? This is not a very good example of how to bias transistor as a switch...how do they explain R2?
 

iimagine said:
What seems to be the problem? Have you breadboarded the circuit? What do you mean by drawing...?

PS: You can get rid of R1, it is not needed, just connect switch to ground
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i mean by drawing the circuit

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Switch at position A
lamp on
lamp on
lamp on

switch at position b
lamp on
lamp on
lamp off

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i rely dont know i simulate circuit it work like that, i am a beginner
 

FvM said:
I think the problem is that you didn't fully understand the operation of the circuit. (I must confess that I also didn't realize the built-in problem at first sight).

By design, you get 9.9V across the load resistor when the transistor is "on" and still 9.3 V for Q1 and Q2 when they are "off". Only the voltage at the Q3 load resistor drops to zero in off-state. This happens because the base-emitter voltage of the succeeding stage can't rise above 0.7 V.

To get a clear off-state for Q1 and Q2, you need to add base series resistors of e.g. 10 or 20 K.
Click to expand...

FvM has been right all along. For the circuit to work, you will need some base bias resistors, the value of each base resistor must be atleast 10 times the load resistor
 

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The problem is that people have different ideas what should be considered as on-state or, referring to hardware, which kind of indicator for on-state should be used.

In the original schematic, if you observe voltage drop across the load resistors, you run into confusion. godfreyl is of course right, that you still get the expected state indication if you observe the collector current. But that's only reasonable if you don't want to connect a real hardware indicator, e.g. a filament lamp, LED or relay.

A base resistor gives full collector voltage swing and allows to connect a voltage driven state indicator.
 

iimagine said:
fvm has been right all along. For the circuit to work, you will need some base bias resistors, the value of each base resistor must be atleast 10 times the load resistor
Click to expand...

yes like this is working if i add resistor in base thnx? But in parallel no

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SO I HAVE TO ADD BASE RESISTOR? BUT WHY IN BOOK HE SAY WITHOUT RESISTOR IN PARALLEL? I DONT KNOW HOW IT WORKS CIRCUIT?
 

Circuit in book only has ONE lamp.
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Yes, and it's expected to work as is. The curious thing is that michael 1978 is seeing three "bubbles" (lamps? meters?) somewhere, but we don't know in which circuit. Neither we know which circuit produced the strange ampere readings in post #20. The "book" circuit in post #22 is given without any resistor values.
 

michael 1978 said:
yes but i read one book, he put three bubles...
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This is the circuit from the book.
I don't see three bubles.
What do you mean by "bubles"?


@FvM: I think this is hopeless, the OP's command of English is too poor. Maybe if he told us what his home language is, somebody could explain to him in his own language.
 
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godfreyl said:
circuit in book only has one lamp.
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yes but i read how it works in book? And i put the lamp in place of resistors

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godfreyl said:
This is the circuit from the book.
I don't see three bubles.
What do you mean by "bubles"?


@FvM: I think this is hopeless, the OP's command of English is too poor. Maybe if he told us what his home language is, somebody could explain to him in his own language.
Click to expand...

yes you right
 

Hello dude,May i know reason for how power loss is more
 

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