Intro. To Electronics Practice Finals-Help!

[Spitefire11]

Hello, I am currently in Intro. to Electronics (KXE131) at the University of Tasmania. I have 3 years worth of finals that were provided by my teacher. However, I am not sure how to do all of the problems. Could you guys take at long at the finals and provide me with some answers showing the work, it would be greatly appreciated! I will attach the 3 finals and one of my HW. Thank you!

 

[ericgibbs]

hi,
Please post your calculations, so that we can check your work so far.

All the questions are very easy.

 

[colin55]

[FONT=&quot]The electronics teacher is regurgitating the same questions, year after year.
It certainly would not take 2 hours to answer the 5 questions.
Very few transistors have a collector-emitter voltage of 0.6v.
Let me know of a LED that drops 1.4v.
Do you realise meter "2" is connected the wrong way.
Do you realise the 12v battery symbol is incorrectly drawn.
I don't know that the "+" and "-" on the zener diode is supposed to represent.

For Q2: The first thing you do is move the waveform up or down the screen so that the bottom of the waveform is sitting on one of the screen lines (graticules). This makes it easier to read the size of the waveform.
Secondly, I would write: the resistor and capacitor should be placed in a vertical line so that they represent what is commonly known as a "time-delay circuit." - with the "pick-off" point in the middle.

What is the reading of meter 3: If it is a high impedance voltmeter, the reading is 10v.
If is a low-impedance meter, the reading will be less than 10v.
If it is an ammeter, the reading will be 0.6mA
If is it a microammmeter, the needle will hit the stop.
If it is an ohm-meter, the needle will hit the "stop."

As you can see, the answers to these questions can be much more complex than you think. Normally, you assume a "current-reading-meter" has no voltage-drop across it and a voltmeter draws no current from the circuit.

[/FONT][FONT=&quot]How much power is dissipated in the zener diode will depend on the impedance (resistance) of the meters in the circuit. [/FONT][FONT=&quot]If you neglect the impact of the meters, you will have to say the current though the 6v zener is the same as the current flowing through the 1k resistor, which is 6mA. Thus the power [/FONT][FONT=&quot]dissipated in the zener diode will be 0.06mA x 6v = 0.36w or 36mW[/FONT]
[FONT=&quot]


[/FONT]

 

[JimB]

Some of the questions seem badly specified, I remember seeing some unspecified circles which I assume to be meters.

Also a couple of odd little traps, such as meters connected the wrong way round so that they read negative. I am not sure if that was intentional or not.

However, dont sit back and expect anyone here to do the work for you.
Make an effort on your own and you will get lots of help, but not a hand-out.

JimB

 

[Spitefire11]

equations

V1= R1/(R1+R2) X E
V2=R2/(R1+R2) X E
P=VI
I=V/R
R=V/I
V=RI
R=R1XR2/R1+R2
Vr=R2/(R1+R2) X Vu
Vc=Q/C
τ=RC
A1=V1/R1
A2=V2/R2
I1=R2/(R1+R2)X I
P=VI=RI^2=V^2/R

 

[Spitefire11]

V1= r1/(r1+r2) x e
v2=r2/(r1+r2) x e

 

[dougy83]

The electronics teacher is regurgitating...

Looks like the teacher is not the only one regurgitating... Posting every formula you know doesn't constitute 'working'... even if you post it twice. Try applying those equations to the questions.

 

[Spitefire11]

So no one is willing to help me?

 

[colin55]

The whole test paper is decidedly very primitive and "Indian" in appearance. I can understand why you are not learning anything from the lecturer.
I have fully helped you in the first paper, but you have not replied to anything I have done for you.
Learning is a two-way street.
I ran a magazine for 20 years and the only readers who did well were those who built project after project.
You are really wasting your time doing this course.
Go out and buy a 200-in-one electronics kit and start to put things together.
You should not be doing this course until you have constructed at least a few hundred circuits.
That's the cold, hard facts of the situation.

 

[dougy83]

So no one is willing to help me?

Are you not willing to even attempt to try to solve the problems? Or even attempt to work out what approach to take towards each problem?

 

[Spitefire11]

The whole test paper is decidedly very primitive and "Indian" in appearance. I can understand why you are not learning anything from the lecturer.
I have fully helped you in the first paper, but you have not replied to anything I have done for you.
Learning is a two-way street.
I ran a magazine for 20 years and the only readers who did well were those who built project after project.
You are really wasting your time doing this course.
Go out and buy a 200-in-one electronics kit and start to put things together.
You should not be doing this course until you have constructed at least a few hundred circuits.
That's the cold, hard facts of the situation.

Well seeing as my Teacher is Indian its actually quite true to your point of view. I completely agree with you, but to obtain my degree I have to take this course.
I have attempted and found that mos logic gates are in fact very beneficial. (in my dumb down version) That they are beneficial because they use low amounts of energy when not in use and small so more capacitors can be used. I have toyed the other problems but If I had the correct answers I wouldn't be asking for help.

 

[ericgibbs]

hi spitfire,
We are only asking to see your attempt at solving the problems as shown your first post document.

Posting an equation is really not solving the problems numerically, have a try at working out the answers yourself, post your answers and then we can give hints.

 

[dougy83]

I have attempted and found that mos logic gates are in fact very beneficial. (in my dumb down version) That they are beneficial because they use low amounts of energy when not in use and small so more capacitors can be used.

Good attempt. I'm not sure what you mean by "more capacitors can be used"?

CMOS gates have a very small static current (current due to leakage through the transistors when they're not switching). They do, however, have dynamic current that increases with the switching frequency of the logic gates - due to the energy required to charge the gate capacitance, and the shoot-through current while switching.

Each time we change the input state we have to charge (or discharge) the input capacitance. Shootthrough: While the input is charging/changing, between e.g. a valid '0' level to a valid '1' level or vice versa, there can be a time that both of the transistors in the CMOS device conduct, causing current to flow from VDD(5V) to VSS (0V).

I have toyed the other problems but If I had the correct answers I wouldn't be asking for help.

Show your attempts. If you show that you're willing to put some effort in, either myself or someone else will be more than happy to help you.

 

[JimB]

Well seeing as my Teacher is Indian

Well, that is it then!
Based on the test papers you showed and my own practical experience...

... you are screwed!

JimB

 

[Spitefire11]

For the 2008 Paper:

q1.
A) Rt= R1+R2+R3
22+10+12= 44k Ohms
B) 1/Rt=1/r1+ 1/r2
2/200 + 1/ 200
200/3
Rt= 66.7

B) Measuring using Ammeter.

 

[Spitefire11]

Would I use the I1= R2/ (R1+R2) x I
to determine what the Ammeter is reading?

And to determine the power that is supplied by the battery and the power dissipated in the LED you would use the Equation P= V I ???

 

[Spitefire11]

..... thanks guys.....

 

[dougy83]

Here's some answers & working to the first couple of Qs that I did back when you originally started the thread. You didn't make an attempt, so I didn't bother posting it or finishing the paper.

I can't remember if I checked them or not.