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transistor base voltage

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1610286555388.png
Hi,

I’m trying to calculate the voltage needed at the transistor base to “open” the transistor. I know it’s got to be ≥ 0.7v.

Saw the schematic above. I calculate the resistance in parallel as R = 470x1000 / 470+1000 = 320Ω. Therefore current is I = 7v (9v – current drop of LED) 7/320 Ω = 0.02ma. Therefore voltage going through transistor base is V = 0.02x320 = 6.4V.

Is that calculation correct and is 6.4V too high a voltage for the transistor to handle?

Thank you!
 

KlausST

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Hi,

Your descriptiin is confusing, I can't follow it.
R1 and R2 is already given....so what "R" do yo try to calculate?
R = 470x1000 / 470+1000 =
= 2000
How do you come to 320 Ohm?
I = 7v (9v – current drop of LED) 7/320 Ω = 0.02ma.
What is this? It seems you are mixing something. Please go step by step so we can follow your ideas.

Klaus
 

barry

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your calculations make no sense. First of all, there’s no “parallel restance”. And I don’t think you want to “open” the transistor, you want turn it on, i.e, drive it into saturation.. When using a transistor as a switch, the rule of thumb is that Your base current should be one-tenth the collector current.

For your case, assume 2volt drop across the LED. So, Ic=(9-2)/470= 14mA. Ib should then be 1.4 mA.
(9-0.6)/1.4mA=Rb=8.4K.
 

00kam

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Hello
you have two closed loop current path when you close the switch one from B-->E and from C-->E so you have two closed circuits the first path is (+ 9 battery-->R1-->LED-->C-E transistor junction-->-9 battery which is the end loop) the second path is (+9 battery -->R2-->B-E transistor junction-->-9 battery which is the end loop) the two circuits's components are connected in series if you want to make current,volt, resistance calculations

kamal
 

Audioguru

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Why is the base resistor value as low as 1k ohms? Then the base current is (9V - 0.7V)/1k= 8.3mA which is almost the same as the collector current of (9V - 2V)/470 ohms= 14.9mA.

EDIT: why do you want to calculate the base voltage? Look in the datasheet.
The base-emitter is a diode with a voltage of 0.6V when the base current is low and is 0.8V when the base current is very high.
 

danadakk

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A general rule of thumb is to operate the transistor as a switch, to turn it on and put it
into saturation, is to set Ib = Ic / 10. Most datasheets specific the Vcesat of the
transistor at this level of base current.

So nominally you would compute R1 for a design Ic, then compute R2.

R 1 = (Vbatt - Vled - Vcesat ) / Ic and R2 = (Vbatt - Vbe ) / (Ic / 10 )

Vcesat is a transistor spec, saturation V collector to emitter, and graphs
of it versus Ic in datasheet. Also base current levels. Here is a curve for 2N3904
where Ib = Ic / 10 (term is "forced beta") -

1610368179004.png


Regards, Dana.
 
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Many thanks for your responses.

I see I have confused a number of issues and used a bad schematic example.

I have made a number of simple circuits where I have overheated the transistor which I thought was due to excess voltage going through the base. I see now I have been looking at this the wrong way.

Very useful insights – thank you!
 

Audioguru

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The base of a transistor is usually the input that uses a low current. Then the base-emitter is delicate and is easily damaged by too much current.
A little 2N3904 transistor has a maximum allowed collector current of 200mA. Then as a switch its base current is never more than 20mA.

If the base current is too low then the transistor will not be a good switch and will have voltage across it and current through it which causes too much heat in it.
 

barry

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At the risk of sounding pedantic, voltage doesn’t “go through” the base, or through anything, for that matter. CURRENT goes through, voltage is across.

Also, transistors are current, not voltage, devices. It’s not the voltage AROSS the base that controls the collector current; it’s the current into to the base that matters.
 

crutschow

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Also, transistors are current, not voltage, devices
You do realize that may generate a response from other pedantic types who will say that the solid-state physics equations of a BJT shows that it is a voltage controlled device, not current controlled.

Of course, for large-signal and switching applications it's much easier to use a current-controlled black-box model,.
The voltage-controlled model is really only useful for the design of small-signal BJT circuits.
 

danadakk

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V vesus I control, like wave particle duality in physics, the coin has two sides. V is
the one, but don't throw away your current controlled tools and understanding,
both very useful.

.

Regards, Dana.
 

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