1. The transistor in circuit (Fig 31F), resistor R has a resistance of 150k(ohms), RL has a resistance of 750k(ohms), and the direct current gain is 80.
Assume that there is negligible potential difference between the base and emitter, calculate :
a) the base current Ib
b) the potential difference between the collector and emitter.
Ib=V/R=9/150000=6*10^-5A
then
Ic=6*10^-5*80=4.8*10-3A
Vce=9-4.8*10-3A*750000=-3591V. this is where my problem is

Attached Images

Fig 31F.bmp (907.1 KB, 6 views)

__________________
To each his own.

I am a bit rusty on this topic, but what happens i guess is that the transistor goes into saturation . In this condition the CE vltage is very low even lower than the BE voltage of ~ 0.6V.
I may be wrong, but if the gain is such that the collector current can achieve a value greater than 9 / 750 k, then the device goes into saturation.
When we have a resistor and an active device in series with it , The active device often decides what happens. E.g. if we connect a variable voltage source to a 1k resistor in series with a diode (with its cathode connected to the ground and anode to 1K) and slowly increase the voltage, then, initially the diode is off thus representing a hi resistance and most of the voltage will be across it. Once it conducts howver, the situation gets reversed.

Ic cannot be greater then
Ic <= 9V / 750 kOm = 1,2 10^-5 A.

That's what i thought, but my calculations don't say it. Well, I've done something wrong somewhere.

__________________
To each his own.

In the fixed biased transistor amplifier circuit, Rb should be always greater than R(L)

Probably Rb = 150kohm and RL = 7.5 kohm.

ok i changed the values of R and RL around, but still get a -ve answer for the voltage, -135V.

__________________
To each his own.

What I mean is that the given values of resistances are not correct.

Suppose Ic is 1 mA. V(CE ) = 9V - 7.5V = 1.5V

IB = (1/80) mA So Rb = 9/(1/80) = 720 kohm.

The above example gives the range of Rc and Rb.

I have just gone through and done the calculations assuming you were correct and were wrong and swapped the values of the resistors. You definitely have a wrong resistance value somewhere. When you get a drop across RL that is larger than 9 volts, then something is wrong, because the 9 volts applied is the only place a voltage comes from. So some of it must be across RL and the rest across the transistor from the collector with respect to the emitter.

Are you sure RL is 720 kOhms and not simply 720 Ohms. If it were 720 Ohms would get a Vce of 5.544 V which makes far more sense. Your base resistance has to be much larger or the collector resistor must be much smaller for the circuit to be analyzed to yield reasonable results.

Craig

oops! my bad, its R=150kohms, and RL=750ohms. So my problem is solved! thanks!

__________________
To each his own.

This question didn't even involve a capacitor nor the RC time constant such that the charging time for a capacitor is not even involved. I'm quite sure you are just grabbing stuff on line and posting it as you reply only to get your profile shown hoping more people may click on your links so you can sell that crap.

People are here with legitimate questions for which they are asking for help with, not for a sale from you. Try a different means of getting your product our to the public, because you are starting to annoy the sh*t out of me with your worthless posts.

Transistor problem due to voltage drop? What in the hell are you talking about. People no longer deal with electrical problems? What in the hell is an electric server and what does it have to do with properly biasing a BJT?

Yeah I want a customized term paper from an idiot, if I want a failing grade maybe. Are you just going copy and paste whatever you find into my custom term paper, because I can train a monkey to do that. Maybe that's what you are, a trained chimp, in which case I apologize as for a chimp you type better than some people. You do make about as much sense as a chimp, so that follows. Idiot!