# Quescent point and ac and dc load lines

1. Nov 26, 2015

### NIKHEL RAINA

Can any one explain me in simple words what is quescent point and what is shows
2- what is ac and dc load lines what they represents actually
and what is difference between ac and dc load lines

2. Nov 27, 2015

### meBigGuy

There are many explanations and tutorials on line to explain load lines. You need to fight through a few examples to get it.

The quiescent point is the steady state DC bias point, that is, the operating point with no signal.

3. Nov 27, 2015

### tech99

The quiescent point shows the operating point with no signal, the resting condition, for instance, the collector voltage and current with no signal.
The DC load line applies if the collector load is a resistor. It shows how the collector voltage will swing up and down as the current varies, going from zero to the supply voltage. Its slope is equal to the load resistance.
The AC load line is when the load resistance has a DC path, so the DC resistance is zero. For instance, if the load consists of a transformer with a resistor across the secondary. In this case the load line has the same slope as for the dc case, but the collector voltage can swing from zero to twice the supply voltage.

4. Nov 27, 2015

### NIKHEL RAINA

Please elobrate what is operating point with no signal (is this point a operating point which depends upon the manufacturing of semiconductor)
being new and ignorant still in confusion
Try to elobrate more

5. Nov 27, 2015

### tech99

A transistor is an amplifying device, so that a small current at the input give a big change at the output. However, the output cannot normally swing more than from zero to the supply voltage, so to make best use of this range, a small bias current is injected into the input to set the operating point half way. When you add a resistive load, the characteristic of the device is altered (basically it is spoiled a bit), so that is why you need to draw the load line. For some cases, you can guess the result without drawing it.

6. Nov 27, 2015

### NIKHEL RAINA

If there is two quescent points in the load line then it means the nonlinear part of circuit works between these points am i right
Thanks again to continue discussion and elobration

7. Nov 27, 2015

### tech99

Only one quiescent point, where the no-signal current and voltage are plotted. The load line passes through this point. The signal causes the voltage and current to swing between two points on the load line.

8. Nov 27, 2015

### LvW

The transistor is a strongly non-linear voltage-controlled (!) device because the output current Ic depends on the input voltage Vbe according to the famous Shockley eqtation: Ic=Ic,o[exp(Vbe/Vt)-1)].
For amplification purposes we desire to have a linear input-output relationship.
For this reason, we must bias the transistor with suitable DC values to get an operating point in the quasi-linear part of this exponential characteristic.
Now - when an ac voltage (to be amplified) is connected to the input the output current will move around this bias point - and if the amplitudes are not too large, the linearity will be acceptable.

9. Nov 27, 2015

### meBigGuy

10. Nov 28, 2015

### NIKHEL RAINA

I read this topic from page you assign to me in the given fig Q Point is not described

According to the theory of this page the points where characteristic curve and load line intersects is possible operating point(Q points of the circuit) at these points the current and voltage parameters of both parts of circuit match
But in the earlier post some one told that the quiscent point(Q) is the operating point with out input signsl
Again if there is not input then how we find this point as it is necessary to draw the graph between current and voltage both parameter.
How to assign definite values without input

Last edited by a moderator: Nov 29, 2015
11. Nov 28, 2015

### meBigGuy

Think of an NPN transistor with DC bias resistors. It is operating at a DC bias point (collector voltage and current) with no input signal. That is the Q point. Now, connect a capacitor to the base so you can inject signal. An AC input signal will cause the currents and voltage to change along the load line. If you inject enough signal to saturate, that will define the upper operating point. If you cutoff, that will define the lower operating point. Draw a line between the two points and you get the different operating points as the input signal varies. 0 volts input takes you back to the Q point.

http://collections.infocollections.org/ukedu/en/d/Jgtz040e/7.1.1.html explains a way to determine the axis intersections for the load line

12. Nov 28, 2015

### LvW

meBigGuy - of course, it is clear that the "DC bias point" is the same as the "Q point".
However, I am not sure if the terms "lower and upper operating point" are well defined - and I am afraid, these terms could cause some confusion for newcomers.
This is, because I know that sometimes the Q point also is called "operating point".
I rather think, that we should use the terms "momentary" (upper and lower) signal values instead.

Last edited: Nov 28, 2015
13. Nov 28, 2015

### NIKHEL RAINA

Thanks to explain
No doubt these notes are pretty good for understanding i understand much more but these are also insuffecient to explain the quescent point accurately
U talk about npn transistor with dc biased resistor. It is opeating at dc biased with no input signal
please explain what is dc bias with no input signal and how it works as there is no input
again if there is zero input at dc bias voltage from where it gets current for generating signal
Again thank ful for link given and discussion

14. Nov 28, 2015

### CWatters

A transistor only works as a transistor when the voltages on the base, collector an emitter are within certain ranges. For example the base voltage must be around 0.7V wrt the emitter. So bias resistors are used to set the base to this voltage with no input signal. This is the q point for the base. The input signal only changes the base voltage very slightly away form the q point. Similarly the collector voltage must also be biased to a q point or the transistor wont work as intended.

Edit: I've added typical DC bias voltages or Q-point voltages to the drawing assuming a 9V supply voltage.

Last edited: Nov 28, 2015
15. Nov 28, 2015

### meBigGuy

You seem to be having some difficulty understanding how a transistor amplifier actually works. You should probably understand that in a basic way before trying to understand the load line and Q point.

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npnce.html

No signal means (in this circuit) no AC input through the capacitor. Even when there is no AC signal, the transistor is conducting current due to the bias currents into the base through the base resistors. The transistor needs to biased such that there is a "midpoint" output voltage when there is no signal input. The input signal can then cause the output voltage to go above and below the "no-signal" bias point as shown in the picture.

16. Nov 28, 2015

### NIKHEL RAINA

इन
In this figure the input signal comes to base through capacitor C1
Actually when input is given the base must be dc biased and this point when ac of input signal becomes zero and dc base becomes dc biased is Q point.
If im true then how alone this capacitor can convert ac in to dc

17. Nov 28, 2015

### meBigGuy

You need to read some basic electronics texts that describe the operation of RC circuits.

If you remove the transistor what would you see at the junction of the resistors and capacitor when you apply AC to the capacitor.

18. Nov 29, 2015

### CWatters

The capacitor does not "convert" AC to DC,

The AC input is centred around 0V. The base voltage is centred around 1.7V. The capacitor blocks the DC difference. The capacitor allows the AC component of the input to be superimposed on the base.

19. Nov 29, 2015

### NIKHEL RAINA

Being new to this subject i read all the topics the basics about capacitor, resistors, inductors as well as their working but im fail to understand under what conditions the three terminals of transistor must work how to bias and what amount of voltage is needed for each terminal to work please elobrate
Im thankful if you add some basics also
Again i put stress on theoretical portion im not aware of practicals

20. Nov 29, 2015

### Staff: Mentor

The thread topic here relates to Q point and loadlines. That material has been amply covered by a number of volunteer helpers.

Transistor characteristics and biasing, while related, are a separate topic outside this thread. Best advice is that the wealth of online material should be the first recourse of anyone self-studying the subject. These forums cannot substitute for coursework, that is not their function.

Thanks to all contributors.