Transistor Hard Saturation - pg224 - Malvino

[veek]

'A designer who wants a transistor to operate in the saturation region under
all conditions often selects a base resistance that produces a current gain
of 10. '

http://pix.toile-libre.org/?img=1493526531.png
http://pix.toile-libre.org/upload/original/1493526562.png

I don't understand what he's doing with the math (it's not complicated math
but he's trying to illustrate an idea that I don't get).

1. betaDC=Ic/Ib but we only have control over Ib so.. how has he computed
Ic=2mA?
2. Then he does Ib = 2mA/50 okay but then again the normal way would be to
work out Ib first and multiply with the given BetaDC to work out Ic..

If he wants IcSat directly then that's at Vce=0 so you do Vcc/Rc = IcSat
What's he doing in Link-2 (Link-1 is his reference ckt)

 

[rickman]

On 4/30/2017 12:35 AM, veek wrote:

'A designer who wants a transistor to operate in the saturation region under
all conditions often selects a base resistance that produces a current gain
of 10. '

http://pix.toile-libre.org/?img=1493526531.png
http://pix.toile-libre.org/upload/original/1493526562.png

I don't understand what he's doing with the math (it's not complicated math
but he's trying to illustrate an idea that I don't get).

He isn't deriving anything, he's just calculating some values. It is
not a mathematical proof.

1. betaDC=Ic/Ib but we only have control over Ib so.. how has he computed
Ic=2mA?

He isn't calculating it, he picked it as an example by setting the Vcc
and Rc. 20 V / 10 kohms = 2 mA.

2. Then he does Ib = 2mA/50 okay but then again the normal way would be to
work out Ib first and multiply with the given BetaDC to work out Ic..

Again, he has a max Ic already. Now he wants to pick an Ib that will
saturate the transistor. Since the beta is 50, use a gain of 10 which
will use a lot more base current than needed and put the transistor into
saturation.

If he wants IcSat directly then that's at Vce=0 so you do Vcc/Rc = IcSat
What's he doing in Link-2 (Link-1 is his reference ckt)

I don't see any links.

He picks a value for Ic and sets Rc and Vcc to get that. Now to set the
value of Rb he knows the beta is 50, so using a value of 10 is assured
to provide much more base current than needed and puts the transistor
into saturation. So Ib = 2 mA / 10 = 0.2 mA which with Vbb = 10 V
makes Rb = 100 kohms. Er, um, wait, that's not right! Ib of 0.2 mA
with Vbb of 10 volts would be 50 kohms (ignoring the 0.7 volt Vbe drop).

Don't know why he is showing 100 kohms on the schematic, but do you see
the logic here?

--

Rick C

 

[veek]

rickman wrote:

On 4/30/2017 12:35 AM, veek wrote:
'A designer who wants a transistor to operate in the saturation region
under all conditions often selects a base resistance that produces a
current gain of 10. '

http://pix.toile-libre.org/?img=1493526531.png
http://pix.toile-libre.org/upload/original/1493526562.png

I don't understand what he's doing with the math (it's not complicated
math but he's trying to illustrate an idea that I don't get).

He isn't deriving anything, he's just calculating some values. It is
not a mathematical proof.


1. betaDC=Ic/Ib but we only have control over Ib so.. how has he computed
Ic=2mA?

He isn't calculating it, he picked it as an example by setting the Vcc
and Rc. 20 V / 10 kohms = 2 mA.

2. Then he does Ib = 2mA/50 okay but then again the normal way would be
to work out Ib first and multiply with the given BetaDC to work out Ic..

Again, he has a max Ic already. Now he wants to pick an Ib that will
saturate the transistor. Since the beta is 50, use a gain of 10 which
will use a lot more base current than needed and put the transistor into
saturation.


If he wants IcSat directly then that's at Vce=0 so you do Vcc/Rc = IcSat
What's he doing in Link-2 (Link-1 is his reference ckt)

I don't see any links.

He picks a value for Ic and sets Rc and Vcc to get that. Now to set the
value of Rb he knows the beta is 50, so using a value of 10 is assured
to provide much more base current than needed and puts the transistor
into saturation. So Ib = 2 mA / 10 = 0.2 mA which with Vbb = 10 V
makes Rb = 100 kohms. Er, um, wait, that's not right! Ib of 0.2 mA
with Vbb of 10 volts would be 50 kohms (ignoring the 0.7 volt Vbe drop).

Don't know why he is showing 100 kohms on the schematic, but do you see
the logic here?

Thanks yeah I got it - I did it the hard way as well - calculated Ib for
beta=10 and beta=50.

(He could have just said it out - instead of showing me the math - he's done
these calculations before so the math is totally unimportant in that
context)