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Electrical Quizzes - QUIZ 8 POSTED

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IamSmooth

Thu Feb 23 2006, 06:53PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

We only consider positive cycle. Capacitive reactance is about 1/.314 ohms
Current peak is 69A
Current RMS is about 49A if this was a full sinusoidal; however, half of cycle is zero
Therefore current is average of 49 and zero which is about 24A
Now, we add the load of 5A on parallel limb to get about 29A current read on meter.

Dr. Dark Current

Thu Feb 23 2006, 09:30PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

IamSmooth wrote ...

Now, we add the load of 5A on parallel limb to get about 29A current read on meter.

So without the 5A load the meter wil read 24Arms?

It should read zero.

My theory is that input power=power to load: The power on the load is 5*Uout watts. The peak mains input voltage is Uout/2. Because charging of the capacitors only occurs at the peaks, the RMS input current is 10A (at half volts=same power). This divides into two half-waves, but the meter reads only one, so it will read 5A RMS.

IamSmooth

Fri Feb 24 2006, 12:41AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

+2.5A top (read by meter)

Steve Conner

Fri Feb 24 2006, 11:03AM

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

jmartis has the right answer, but for the wrong reason, so he only gets half a cookie

Hint: Moving coil meters do NOT read RMS (you must know this if you solved quiz 1)

WaveRider

Fri Feb 24 2006, 11:56AM

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

Moving coil meters measure average current...

Do I get the other half of the cookie?

Steve Conner

Fri Feb 24 2006, 12:47PM

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

Yes, you can have the other half

The argument I used to figure this out was:
*Moving coil meters read average current
*Capacitors block DC and pass AC. DC is the same thing as average. Therefore, C1 passes no current on average, just an alternating ripple current.
*Therefore the average current in the two meters must be the same: 5A

HV Enthusiast

Fri Feb 24 2006, 12:59PM

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

wrote ...

jmartis has the right answer, but for the wrong reason, so he only gets half a co

Actually, I get the other half of cookie.

Jmartis stated the meter would read 5A RMS. He was half-right in that the meter would read 5A, but wrong that it would read RMS. Its really 5A AVG.

I stated the meter would read 15A RMS. Which is actually the CORRECT answer should the meter have read in RMS, however the meter reads in AVG.

**************************************
QUIZ 7
**************************************

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.

WaveRider

Fri Feb 24 2006, 01:28PM

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

It is a logarithmic amplifier. increasing the input voltage 10 times increases the output voltage ln(10) times. This is because the emitter current is exponentially dependent on Vbe...and since the transistor beta is very high, collector current very nearly equals emitter current... Assuming "infinite" amplifier gain and op-amp input resistance, the output voltage will be

Vo = -Vt * ln(Vin / (R * Io))..

or thereabouts.. Vt is the thermal equivalent voltage (about 0.026V) and Io is the reverse saturation current.(around 1e-14A).

All this assumes that the transistor is in the forward operating region..

Steve Conner

Fri Feb 24 2006, 01:49PM

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

Sorry EV, no half cookie for you! The relationship between average and RMS is a complex one that depends on a bunch of things, like conduction angle of the diodes, supply inductance, etc. If it works out as "3" then it's by sheer luck :P

Your log amp would be unstable even if the op-amp was unity-gain stable, due to the extra loop gain added by the transistor. So I'll guess Vout will be about 20V of high frequency crud.

HV Enthusiast

Fri Feb 24 2006, 01:57PM

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

Steve Conner wrote ...

Sorry EV, no half cookie for you! The relationship between average and RMS is a complex one that depends on a bunch of things, like conduction angle of the diodes, supply inductance, etc. If it works out as "3" then it's by sheer luck :P

Sorry Steve. Do the math, or even do the simulation. The RMS current is indeed 15A. I did this first by calculation, and then verified it with PSPICE. I'll even build the circuit up in the lab to prove this. So give me my damn cookie! I want a whole one now!

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