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Electrical Quizzes - QUIZ 8 POSTED

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WaveRider

Fri Feb 24 2006, 02:17PM

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

The amount of charge moved from the capacitor by the 5A "load" must be replaced by the current flowing through the diode.

Steady state operation would imply that this charge would need to be replenished on each capacitor once every cycle. Hence, charge/time for each half cycle would average out to 5A. Hence, the meter reads 5A. RMS current will depend on the form-factor of the current waveform..

EVR can have his cookie, tho'!!

EDIT: On more thought..I neglected to consider that the diodes are off most of the time, hence more discharge of the cap is going to occur.. Hence, EVR's 15A could be reasonable....but it will depend on the conduction angle of the diodes, impedances, etc, as Steve said... /EDIT

I've built a log amp like this..(using a little gain compensation in the feedback loop) and it worked great! When small signals are applied, lots of noise appears because gain increases with reduced signal..

HV Enthusiast

Fri Feb 24 2006, 02:20PM

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

wrote ...

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.

Yes, that is correct. In fact any Vin > 25mV the amplifier is potentially unstable.

However, for this question, lets assume the op-amp is stable.

WaveRider

Fri Feb 24 2006, 02:33PM

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

Shouldn't that be: Vin < 25mV? The output usually goes wild for small input voltages..

Here's another quiz (it's an easy one). What is this and how and why is it used?

Steve Conner

Fri Feb 24 2006, 03:25PM

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

Easy, it's a Smith Chart Of Doom and it's used for scaring off people who express an interest in RF design

If WaveRider is wrong in his analysis, I must be wrong too. The way I see it is: If the circuit is running in the steady state, then all the charge removed from the capacitors by the load must be replaced through the diodes.

IamSmooth

Fri Feb 24 2006, 03:30PM

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

I was hoping with regard to quiz 6 someone could straighten out my lack of knowledge.
First, EV, where are you getting 15A as the RMS? RMS is the Voltage/squareRt(2).
Steve, if the capacitor passes the ripple and allows the DC to pass (5A), wouldn't meter 1 show a higher current than 2 since it must be the sum of the ripple AC and the average DC current?
Finally, what is the physics behind the coil meter showing average current and what type of meter is needed to display RMS current?

Thanks.

Steve Conner

Fri Feb 24 2006, 03:53PM

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

imsmooth: Yes, meter 1 shows the sum of the average DC current and the average ripple current. But what I'm saying is that the ripple current through a capacitor must always average to zero. If not, that would imply that capacitors pass DC. So meter 1 should read just the average DC current.

The physics behind it: The torque on the movement of a moving-coil meter is proportional to the current. The mechanical inertia of the movement averages it so the needle ends up indicating the average current.

In a moving iron (aka iron vane) ammeter, the torque is proportional to the square of the current. When this is averaged by inertia the result is a mean-square reading. These meters have a funny non-linear dial which effectively square-roots the result, giving an RMS reading.

HV Enthusiast

Fri Feb 24 2006, 04:03PM

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

WaveRider wrote ...

Here's another quiz (it's an easy one). What is this and how and why is it used?

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.

Thanks

Dr. Dark Current

Fri Feb 24 2006, 05:22PM

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

Steve Conner wrote ...

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)

I always thought RMS and average are the same

Can someone explain the difference to me ?

IamSmooth

Fri Feb 24 2006, 05:31PM

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

Steve,

I'm going to do this later, but if I wanted to search on the rms current meter would good key words be iron vane ammeter or rms current meter? You wouldn't by change have a good link to one?

Also, why does the ripple average to zero? The diode only allows positive current to pass so shouldn't the ripple always be above zero and hence the ripple average > 0? Or is the ripple above and below zero because the load is connected to both capacitors (which connects both diodes)?

Steve Conner

Fri Feb 24 2006, 05:39PM

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

If you want to discuss this, better start a new thread and leave the other quiz to run.

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