What Are Saturation and Cutoff Points in Drawing the DC Load Line?

[ranju]

while drawing the dc load line , , we take VCE = 0 and mark it as saturation point but but in saturation state VCE is very small but not zero..! I actually want to make clear that whether its a fact or is it just an assumption to find out saturation point..??
similarly for the cutoff point we take IC=0 !..Please explain this thing!

Also , we define operating point as a point on the dc load line which represents VCE & Ic in ABSENCE OF SIGNAL... this thing I wanted to ask that what does it mean by having no signal..?? Does it mean no input..?? & why we are defininf it for no signal..??

 

[Baluncore]

Saturation, (of an NPN BJT), is defined as V_CE <= V_BE. It is not defined as V_CE = 0

 

[ranju]

yes I knw that..It was in the book ..while drawing the dc load line they taken as VCE=0..

 

[Baluncore]

It is NOT defined as VCE = 0

You misunderstand ?

 

[ranju]

I am not getting what you are trying to say..!

 

[Baluncore]

… , we take VCE = 0 and mark it as saturation point but but in saturation state VCE is very small but not zero..!

Saturation starts when V_C = V_B.
http://en.wikipedia.org/wiki/Bipolar_junction_transistor#Regions_of_operation

 

[ranju]

ohk ..now can you please explain the rest part of my query..>>

 

[dlgoff]

while drawing the dc load line , , we take VCE = 0 and mark it as saturation point but but in saturation state VCE is very small but not zero..! I actually want to make clear that whether its a fact or is it just an assumption to find out saturation point..??
similarly for the cutoff point we take IC=0 !..Please explain this thing!

Setting V_CE=0 gives you the graphical y-axis intercept and setting I_C=0 gives you the graphical x-axis intercept for the load-line equation that follows by assuming I_C≈I_E.

image compliments of http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/loadline.html

All of this comes from an http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_3.htm

 

[ranju]

Also , we define operating point as a point on the dc load line which represents VCE & Ic in ABSENCE OF SIGNAL... this thing I wanted to ask that what does it mean by having no signal..?? Does it mean no input..?? & why we are defining it for no signal..??

Can you please explain this thing too..??

 

[Baluncore]

The operating point is the point on the load line that represents the DC bias of the system without a signal.

A signal causes the V_CE and I_C point to move along the load line, on both sides of the operating point, in proportion to the signal.

 

[NascentOxygen]

while drawing the dc load line , , we take VCE = 0 and mark it as saturation point but but in saturation state VCE is very small but not zero..! I actually want to make clear that whether its a fact or is it just an assumption to find out saturation point..??
similarly for the cutoff point we take IC=0 !..Please explain this thing!

Also , we define operating point as a point on the dc load line which represents VCE & Ic in ABSENCE OF SIGNAL... this thing I wanted to ask that what does it mean by having no signal..?? Does it mean no input..?? & why we are defininf it for no signal..??

Although we construct the load-line so it crosses the entire quadrant on the graph, this does not mean we can cause the transistor to operate at any or every point on that load-line. We can't. As you pointed out, we can't get V_CE right to 0 V. But all of the points we can cause the transistor to operate at do lie on that line.

If you were to apply a tiny AC signal by itself to the base of a transistor nothing much will happen. Unless the base voltage reaches 0.5V or so, no base current can flow. As most signals we'd like to amplify do not reach anywhere near 0.5V, it would seem we are stymied. To overcome this obstacle, we can use DC to hold the transistor in some prearranged condition where base voltage exceeds 0.5V and collector current is already flowing without the AC signal. We then add our tiny signal voltage to this DC and send that combination to the base. The transistor amplifies this combination, then we (usually) use a capacitor at the collector to separate the AC from the DC, and this output AC component represents an amplified version of our input AC signal.

 

[ranju]

but the operating point is being defined particularly for no-signal..! see I want to clear some points..!

firstly , we are using the load line to decide the conditions at which we operate the transistor ..right..??
and preferably , the operating point should lie in the middle of the transistor..
secondly , giving a signal means we are applying voltage.right..?? then , its obvious... we have to give some input to the transistor , then it'll work ..then why a no signal..??

 

[NascentOxygen]

Base current is always the sum of some AC and some DC.

"no signal" conditions mean no AC but plenty of DC current going into the base. This is the condition where you set up the circuit before applying the tiny AC signal.

 

[ranju]

so you mean to say , before applying any ac signal to the signal , we first supply dc , and decide a particular value of current and junction voltage of operation..then at those conditions we then supply the ac signal??

 

[NascentOxygen]

so you mean to say , before applying any ac signal to the signal , we first supply dc , and decide a particular value of current and junction voltage of operation..then at those conditions we then supply the ac signal??

Yes. It's called biasing the transistor. We carefully bias the transistor with DC to operate at a point where amplification will be linear when we add in some AC.