BJT question

[Bo]

I'm quite rusty on BJTs and am taking a class to refresh my memory. I have
the following problem and am uncertain if I'm doing it right or not.

Given 2n3904, desired use amplification at ambient. Desired Ic=50mA.

a) what is Ib? I calculated as Ib=50/60=0.83mA using hfe=60 from data sheet

b) what is Vce? This I am totally unsure about-- I said Vce = Pd/Ic =
625mW/50mA = 12.5V

c) what will Vce be if we want to drive the 2n3904 to the brink of
saturation?

I found a datasheet curve for CE sat vs. Ic and read the value of Vce=~0.08.

Am I on the right track? I'm confused that Vce of 12.5 is much, much more
that Vsat of 0.08 from the curve. Is this kosher? Like I said am am RUSTY on
discretes....

Thanks,

Bo

 

[Jim Thompson]

I'm quite rusty on BJTs and am taking a class to refresh my memory. I have
the following problem and am uncertain if I'm doing it right or not.

Given 2n3904, desired use amplification at ambient. Desired Ic=50mA.

a) what is Ib? I calculated as Ib=50/60=0.83mA using hfe=60 from data sheet

b) what is Vce? This I am totally unsure about-- I said Vce = Pd/Ic =
625mW/50mA = 12.5V

c) what will Vce be if we want to drive the 2n3904 to the brink of
saturation?

I found a datasheet curve for CE sat vs. Ic and read the value of Vce=~0.08.

Am I on the right track? I'm confused that Vce of 12.5 is much, much more
that Vsat of 0.08 from the curve. Is this kosher? Like I said am am RUSTY on
discretes....

Thanks,

Bo

Rusty? Indeed! There is too much to explain here. Go back to your
textbook and _read_ carefully!

...Jim Thompson

 

[Jon Slaughter]

I'm quite rusty on BJTs and am taking a class to refresh my memory. I have
the following problem and am uncertain if I'm doing it right or not.

Given 2n3904, desired use amplification at ambient. Desired Ic=50mA.

a) what is Ib? I calculated as Ib=50/60=0.83mA using hfe=60 from data
sheet

b) what is Vce? This I am totally unsure about-- I said Vce = Pd/Ic =
625mW/50mA = 12.5V

c) what will Vce be if we want to drive the 2n3904 to the brink of
saturation?

I found a datasheet curve for CE sat vs. Ic and read the value of
Vce=~0.08.

Am I on the right track? I'm confused that Vce of 12.5 is much, much more
that Vsat of 0.08 from the curve. Is this kosher? Like I said am am RUSTY
on discretes....

You need to know the following equations:

Ie = Ic + Ib
Ic = beta*Ib
Vb = Ve + 0.7V


Now if you do not use any biasing for the transistor(a bad idea) and want
Ic=50mA with beta = 50 then Ib = 1mA and Ie = 51mA.

So just sending in 1mA will get you about 50mA "out"(or in at the
collector).

To do that you can ground Ve so that Vb = 0.7V(you can approximate this as
0V if you want but is essentially a diode drop).

Vcc ---- R ---- Vb (Base of transistor) "diode" ---- Emitter --- (ground)

R can then be calculated to get 1mA to the base, i.e.

(Vcc - Vb)/R = 1mA

If, say Vcc is 5.7V then

R = 5.7/1mA = 5.7kOhm

So using a 5.7k resistor will get you about 1mA base current which will get
you about 50mA collector current.

This is not a good design though because its very unstable(beta will change
with temperator and other factors).

You should read a book or look on the net for various "biasing" methods that
help stabilize the transistor so that it is "independent" of beta and
depends only on the components around it.

In the above case obviously Vce depends on the load we attach to the
collector or emitter. (we have 51mA going through the emitter)

 

[Bo]

You need to know the following equations:

Ie = Ic + Ib
Ic = beta*Ib
Vb = Ve + 0.7V


Now if you do not use any biasing for the transistor(a bad idea) and want
Ic=50mA with beta = 50 then Ib = 1mA and Ie = 51mA.

So just sending in 1mA will get you about 50mA "out"(or in at the
collector).

To do that you can ground Ve so that Vb = 0.7V(you can approximate this as
0V if you want but is essentially a diode drop).

Vcc ---- R ---- Vb (Base of transistor) "diode" ---- Emitter --- (ground)

R can then be calculated to get 1mA to the base, i.e.

(Vcc - Vb)/R = 1mA

If, say Vcc is 5.7V then

R = 5.7/1mA = 5.7kOhm

So using a 5.7k resistor will get you about 1mA base current which will
get you about 50mA collector current.

This is not a good design though because its very unstable(beta will
change with temperator and other factors).

You should read a book or look on the net for various "biasing" methods
that help stabilize the transistor so that it is "independent" of beta and
depends only on the components around it.

In the above case obviously Vce depends on the load we attach to the
collector or emitter. (we have 51mA going through the emitter)

Jon,

Thanks for the help. I think I got it now...

Bo

 

[whit3rd]

Given 2n3904, desired use amplification at ambient. Desired Ic=50mA.

a) what is Ib? I calculated as Ib=50/60=0.83mA using hfe=60 from data sheet

Isn't the data sheet hfe value a minimum? It's not '=0.83 mA',
really,
but rather '<0.83 mA'. It's probably a LOT less, and you should
refrain from quoting two decimal place numbers here...

 

[JosephKK]

Rusty? Indeed! There is too much to explain here. Go back to your
textbook and _read_ carefully!

...Jim Thompson

Jim; you should have at least directed him to work on operation point
/ DC solution / quiescent point (Q-point), and the definition of edge
of saturation.

 

[Jim Thompson]

I'm quite rusty on BJTs and am taking a class to refresh my memory. I have
the following problem and am uncertain if I'm doing it right or not.
[snip]

Thanks,

Bo

Rusty? Indeed! There is too much to explain here. Go back to your
textbook and _read_ carefully!

...Jim Thompson

Jim; you should have at least directed him to work on operation point
/ DC solution / quiescent point (Q-point), and the definition of edge
of saturation.

I would have, but the recitation of his problem sounded to me like he
was reading from a homework sheet rather than expressing his true
confusion.

If he'd expressed _any_ comprehension at all I would have jumped in to
fill the gaps.

We already have enough bad engineers and technicians ;-)

...Jim Thompson

 

[Fred Bloggs]

I'm quite rusty on BJTs and am taking a class to refresh my memory. I
have the following problem and am uncertain if I'm doing it right or
not.

Given 2n3904, desired use amplification at ambient. Desired Ic=50mA.

a) what is Ib? I calculated as Ib=50/60=0.83mA using hfe=60 from data
sheet

b) what is Vce? This I am totally unsure about-- I said Vce = Pd/Ic =
625mW/50mA = 12.5V

c) what will Vce be if we want to drive the 2n3904 to the brink of
saturation?

I found a datasheet curve for CE sat vs. Ic and read the value of
Vce=~0.08.

Am I on the right track? I'm confused that Vce of 12.5 is much, much
more that Vsat of 0.08 from the curve. Is this kosher? Like I said am
am RUSTY on discretes....

In your case, you just forget about power until you compute all the
voltages and currents. Just stick to Ohm's law and the current loops in
the transistor circuit. In the case of Vce you simply have
Vcc=IcRc+Vce+IcRe+Vee, assuming an emitter bias resistor and power
supply (Vee may be 0V), making Vce=(Vcc-Vee)/(Rc+Re). That formula
neglects the Ib component of Ie but in most cases it is so small it is
justified, less than 2% in your case. As for transistor saturation,
recall that this is the condition of having both junctions forward
biased. "On the brink of saturation" is taken to mean the BC junction is
neither forward or reverse biased but has 0V across it. Then using the
fact that Vce=Vcb+Vbe and Vcb=0, makes Vce=Vbe which will be around
0.75V for most small signal stuff at Ic=50mA. The transistor saturation
curves show Vce vs Ic in deep saturation where the base drive is much
much greater than the base drive in the active region, usually at
Ib=Ic/10, so you don't want to use those curves.

 

[Fred Bloggs]

making Vce=(Vcc-Vee)/(Rc+Re).

whoops..that should be Vce=(Vcc-Vee)-Ic(Rc+Re).

 

[David Harmon]

On Thu, 13 Mar 2008 08:57:31 -0700 in sci.electronics.design, Jim

Rusty? Indeed! There is too much to explain here. Go back to your
textbook and _read_ carefully!

I read the original post and figured "here is exactly the kind of
question Jim Thompson can answer if he wants to" so I marked the
thread for retrieval, hoping to learn some electronics. Silly me.

 

[Jim Thompson]

On Thu, 13 Mar 2008 08:57:31 -0700 in sci.electronics.design, Jim


I read the original post and figured "here is exactly the kind of
question Jim Thompson can answer if he wants to" so I marked the
thread for retrieval, hoping to learn some electronics. Silly me.

If I posted one of my past tutorials, would you read it? Or ask
disconnected, unreferenced questions like the OP?

I have a ton of stuff, some of which is already lurking there on my
website, if you're interested.

...Jim Thompson

 

[JosephKK]

I'm quite rusty on BJTs and am taking a class to refresh my memory. I have
the following problem and am uncertain if I'm doing it right or not.
[snip]

Thanks,

Bo


Rusty? Indeed! There is too much to explain here. Go back to your
textbook and _read_ carefully!

...Jim Thompson

Jim; you should have at least directed him to work on operation point
/ DC solution / quiescent point (Q-point), and the definition of edge
of saturation.

I would have, but the recitation of his problem sounded to me like he
was reading from a homework sheet rather than expressing his true
confusion.

If he'd expressed _any_ comprehension at all I would have jumped in to
fill the gaps.

We already have enough bad engineers and technicians ;-)

...Jim Thompson

Legitimate viewpoint. I had perceived it as being so far down in the
mud as to no longer be able to state a coherent question.
Nor did i give OP anymore information than what i had stated.

 

[David Harmon]

On Sun, 16 Mar 2008 14:54:41 -0700 in sci.electronics.design, Jim

If I posted one of my past tutorials, would you read it? Or ask
disconnected, unreferenced questions like the OP?

I always have plenty to read, so don't put yourself out on my
account. I also read more than I ask questions. If you post an
answer to any of my questions, I will certainly read it.