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Electrical Quizzes - QUIZ 8 POSTED

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LAN_403

WaveRider

Fri Feb 24 2006, 09:15PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Waverider,

Please remember the rules of this thread. Only one quiz may be posted at a time, and the answer or solution has not been provided for the previous quiz yet.

Oops! Sorry...

BTW- Steve... You got it!!!

HV Enthusiast

Sat Feb 25 2006, 02:38PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

[quote]
**************************************
QUIZ 7 ANSWER
**************************************

Okay, time for something tougher . . .

Assume the following:
a. Vin(initial) = 0.010V
b. No offset voltage
c. operational amplifier is stable at unity gain closed loop
d. assume BJT beta is very large

QUESTION: If the input voltage is increase to 0.010V (factor of 10), how much does Vout change? (need an actual number, not an equation) You must provide the reason too or you get no credit.

1140785976 15 FT1067 Quiz07

[/quote1140878029]

Here is the answer:

It will move more negative by 83mV.
The Ebers-Moll equation relates the collector current (IC) of a bipolar transistor to its emitter-base voltage (VBE):

Ic=Io*e(qv*B*E/kT)
where:
Io=collector leakage current
q=charge on electron
k=Boltzmann's constant
T=absolute temperature

Rearranging you get:
Vbe=(kT/q)*ln(Ic/Io)

At room temperature, kT/q is about 25mV. At the same time, increasing Ic tenfold changes the logarithm term by ln10 which is 2.3. So the increase in Vout is 2.3 *25mV = 83mV.

Of course as was noted, for Vin > 25mV, then the feedback gain is greater than 1 so the op-amp may be unstable. But for this example, it was assumed the op-amp was stable. (This was actually the second part of the problem, but Steve already answererd it)

c4r0

Sat Feb 25 2006, 09:27PM

Registered Member #151 Joined: Sun Feb 12 2006, 02:53PM
Location: Poland
Posts: 153

[POST DELETED - This is for electrical circuit quizzes only - see rules and thread title]

What is this powder?

[REMOVED DOUBLE POST BY C4R0]

wrote ...

It WAS electrical quiz. It was a pile of SMD resistors on the photo ... I know the rules and the thread title and there is nothing about circuit ...

[EVR RESPONDS]
Sorry, it wasn't. The last 8 quizzes on this thread were electrical circuit related so you definitely had a good idea of the theme of this thread. Also, even if your post was electrical related as you said and was a bowl of SMD resistors, your question, "What is this powder?" surely didn't imply it was electrical components.

For the future, please keep quizzes to:
Electrical circuits and Electrical theory related problems

Thank you

...

Sat Feb 25 2006, 11:06PM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

Oh well, how about this one...
You have a 10v peak-peak 10hz sign wave (50% duty cycle) going through a 10ohm resistor; how much power is the resistor dissipating?

HV Enthusiast

Sun Feb 26 2006, 12:16AM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

... wrote ...

Oh well, how about this one...
You have a 10v peak-peak 10hz sign wave (50% duty cycle) going through a 10ohm resistor; how much power is the resistor dissipating?

Assuming you mean that voltage is being "dropped" across the resistor, the resistor would be dissipating approximately 1.25 watts (rms)

...

Sun Feb 26 2006, 02:06AM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

Wow, that was fast...

Here is another one...

You are in an ideal world (no losses), and you have 2 plates that are 10"square, are 10" apart, with 10v across them. You move the plates 20" apart. What is the voltage now?

Steve Conner

Sun Feb 26 2006, 11:10AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

There's not enough information to answer the question. Did you charge the plates up to 10v and disconnect them from everything? or leave them connected to a 10v source (in which case the answer would be 10v)

...

Sun Feb 26 2006, 03:58PM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

the plates are no longer connected to anything, they were charged up then the source removed.
Bonus points if you can give the exact voltage (not a decimal approximation)

HV Enthusiast

Sun Feb 26 2006, 07:36PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

... wrote ...

Here is another one...

You are in an ideal world (no losses), and you have 2 plates that are 10"square, are 10" apart, with 10v across them. You move the plates 20" apart. What is the voltage now?

Was going to let someone else take a stab at it since i quickly answered the last one, but no one has jumped.

Assuming the charge, q, is constant (no losses)

q = C*V

C=(k*Eo*Area)/d

Increase d by a factor of 2, capacitance reduces by a factor of 2. Since q remains constant, voltage must double.

So the answer is: 20 volts

...

Sun Feb 26 2006, 08:40PM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

good!

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