NCEES 124

I understand the numerical derivations of the problem, up to the point in the solution where it says,

"A positive swing of 5.4V will result in saturation of the BJT."

"A negative swing of 4.6V will result in clipping of the output signal."

Since V_c = -5.4V (it's of negative value), does this mean in general that any transistor having a negative collector voltage is being driven into saturation?

What do they mean by "positive" and "negative" swings.

And I think the 4.6V comes from 10V-5.4V = 4.6V, correct?

chicago said:

I understand the numerical derivations of the problem, up to the point in the solution where it says,
"A positive swing of 5.4V will result in saturation of the BJT."

"A negative swing of 4.6V will result in clipping of the output signal."

Since V_c = -5.4V (it's of negative value), does this mean in general that any transistor having a negative collector voltage is being driven into saturation?

What do they mean by "positive" and "negative" swings.

And I think the 4.6V comes from 10V-5.4V = 4.6V, correct?

Click to expand...

The way this is biased, vo is sitting at around -5.4 VDC. "Voltage swings" just means it swings up and down from there because of the sinusoid. If it swings negative 4.6 from there, it will go to -10 which is the power supply voltage (so you are basically right). THe output cannot exceed the supply voltage, or that is hitting the rail. So it clips. THE MOST IMPORTANT LESSON ON THIS PROBLEM - YOU ARE DONE AT THIS POINT! You do not care what happens when the output goes positive because 4.6 is the smallest answer given, so unless there is such a thing as a lopsided sinusoid, it has to be the right one. So why even worry about the positive swing. Why waste time.

I am not good on trans operating regions, but htis is what happens on the positive side, just in case you do need it for some problem -

When it swings the other direction - positive - THe base is grounded, and that puts Ve around .7 V. On the positive side, if the voltage ouput goes up +5.4 that will put the Vc at 0. THerefore, since Vc is approximately equals to Ve the transistor is in saturation (this happens when Vce is very low).

I may have made a little mistake here, but that is how I understand it

benbo said:

The way this is biased, vo is sitting at around -5.4 VDC. "Voltage swings" just means it swings up and down from there because of the sinusoid. If it swings negative 4.6 from there, it will go to -10 which is the power supply voltage (so you are basically right). THe output cannot exceed the supply voltage, or that is hitting the rail. So it clips. THE MOST IMPORTANT LESSON ON THIS PROBLEM - YOU ARE DONE AT THIS POINT! You do not care what happens when the output goes positive because 4.6 is the smallest answer given, so unless there is such a thing as a lopsided sinusoid, it has to be the right one. So why even worry about the positive swing. Why waste time.

Click to expand...

Thanks for the vote of confidence in the other post Benbo.

After your explanation, I was able to visualize the max voltage swing of 4.6V before it clips > 10V.

Q1: Going back to the term "biasing" now. You mentioned that understanding biasing circuits for BJT's is a must. They way EERM explains it is that biasing is done to establish the Q or operating point with no input signal. Is this sort of like determining the residual voltage or current in a BJT?

Q2: Does NCEES 121 qualify as a biasing circuit because it has no input signal, except for Vcc battery voltage of 20 VDC?

Q3: In contrast, NCEES 124 does have an input signal Vi at the emitter terminal, whereas a 10V battery source is connected to the E-C terminals. So, why are we considering Vo as the "biasing" voltage sitting at -5.4 VDC?

chicago said:

Thanks for the vote of confidence in the other post Benbo.
After your explanation, I was able to visualize the max voltage swing of 4.6V before it clips > 10V.

Q1: Going back to the term "biasing" now. You mentioned that understanding biasing circuits for BJT's is a must. They way EERM explains it is that biasing is done to establish the Q or operating point with no input signal. Is this sort of like determining the residual voltage or current in a BJT?

Q2: Does NCEES 121 qualify as a biasing circuit because it has no input signal, except for Vcc battery voltage of 20 VDC?

Q3: In contrast, NCEES 124 does have an input signal Vi at the emitter terminal, whereas a 10V battery source is connected to the E-C terminals. So, why are we considering Vo as the "biasing" voltage sitting at -5.4 VDC?

Click to expand...

Don't get hung up when I use the word biasing. I'm probably not using it correctly. You are right, technically, when they say "biasing" they are referring to setting the DC - Operating or quiescent point - or finding Vce, Ic. Generally when we think of biasing an amplifier we think of that voltage divider network on the base of an NPN transistor that we studied ad nauseum in school

I don't have the book, so I can't really answer all your specific questions, but when I use the term "biasing" I am using it in a broad (and perhaps imprecise) sense to mean doing a DC analysis of the supply voltages, transistor currents, and resistor network in the amplifier to determine what the DC voltages and currents are at various places in the circuit. Obviously, Vo is an output, it doesn't bias anything. For the circuit in 124 (if I remember correctly) I mean that we can get the emitter current by following the Emitter Base path to ground, then use that to figure out the collector current, the Vce, and at what DC voltage the AC output sits.

I can try to answer your question about 121 when I get home if you like, or if you explain the problem a little more, but I sense you understand my meaning. I suspect you are right in what you say here. If you can do the DC analysis for these NCEES problems, that is the type of thing you are going to see in the exam.

I am stumped on where the .7 comes from in the first line of the solution IE= 10-0.7/20k?

John

jdd18vm said:

I am stumped on where the .7 comes from in the first line of the solution IE= 10-0.7/20k?
John

Click to expand...

John,

The 0.7 volts is the typical voltage drop across the Base-Emitter junction of a silicon transistor. If it's something other than silicon, the voltage drop will be a different value. I think the EERM will list values for a couple of possible materials.

Jim

IFR_Pilot said:

John,
The 0.7 volts is the typical voltage drop across the Base-Emitter junction of a silicon transistor. If it's something other than silicon, the voltage drop will be a different value. I think the EERM will list values for a couple of possible materials.

Jim

Click to expand...

This was also confusing me, but I found the 0.7V value on p. 43-11 in the EERM. Of course, I needed to read the solution and the added explanation from some other people on here before I even got to that point!

IFR_Pilot said:

John,
The 0.7 volts is the typical voltage drop across the Base-Emitter junction of a silicon transistor. If it's something other than silicon, the voltage drop will be a different value. I think the EERM will list values for a couple of possible materials.

Jim

Click to expand...

you know i thought that but when i looked up Ge and Si it was something different like 6 and 3 or vice versa. But thats the answer I'm good.

Thanks