NCEES 122

For those not having it handy:

An Si diode used as a 1/2 wave rectifier and is drawn in series with a 220VAC source. The 220 VAC Source in parallel with a 50 microfarad Cap The cap is drawn after the Diode. The minimum PIV (V) rating necessary to avoid reverse breakdown is most nearly A.)220, B)310, C)440 or D) 620

Answer is 620 with the soultion explained as " Since the Cap charges to 220 x1.4 V and holds that charge while the supply goes to its negative max -220x1.4 the PIV must be 440x1.4=616.

This seems relatively harmless, can someone walk me through the solution, offer a better explanation? Where does 1.4 Volts come from?

John

jdd18vm said:

For those not having it handy:
An Si diode used as a 1/2 wave rectifier and is drawn in series with a 220VAC source. The 220 VAC Source in parallel with a 50 microfarad Cap The cap is drawn after the Diode. The minimum PIV (V) rating necessary to avoid reverse breakdown is most nearly A.)220, B)310, C)440 or D) 620

Answer is 620 with the soultion explained as " Since the Cap charges to 220 x1.4 V and holds that charge while the supply goes to its negative max -220x1.4 the PIV must be 440x1.4=616.

This seems relatively harmless, can someone walk me through the solution, offer a better explanation? Where does 1.4 Volts come from?

John

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I think the 1.4 is really 1.414 or rather sqrt(2). I am not sure, because I haven't got the book right here, but I think the 220 is RMS and you need the peak AC. So if the cap charges up to 220*1.414 and then the source cycles to -220*1.414 (the negative peak) the maximum voltage across the diode is 440*1.414. Like I said, I can't see the problem and what they say about the source voltage, so I hope someone will correct me if I am wrong.

John,

I don't have my book handy but I remember the problem.

The 220 VAC is an RMS value. What you are interested in for the capacitor is the peak voltage (Vp).

Vp = sqrt 2 x Vrms = 1.414*220 = 311 Volts.

The peak-to-peak voltage is 2xVp = 622 volts.

As the voltage wave form goes positive, the capacitor will charge to a DC voltage value equal to the peak voltage minus the voltage drop value for the diode. As the wave goes negative, the diode will stop conducting and the capacitor will remain charged. The voltage drop across the diode at the negative voltage peak will be nearly the peak-to-peak voltage of 622 volts.

Hopefully, I haven't just repeated the NCEES explanation.

Jim

benbo said:

I think the 1.4 is really 1.414 or rather sqrt(2). I am not sure, because I haven't got the book right here, but I think the 220 is RMS and you need the peak AC. So if the cap charges up to 220*1.414 and then the source cycles to -220*1.414 (the negative peak) the maximum voltage across the diode is 440*1.414. Like I said, I can't see the problem and what they say about the source voltage, so I hope someone will correct me if I am wrong.

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it clearly says 1.4V wonder if there is a relationship to Silcone Diode forward Voltage is normally .7 volts times 2 is 1.4?

IFR_Pilot said:

John,
I don't have my book handy but I remember the problem.

The 220 VAC is an RMS value. What you are interested in for the capacitor is the peak voltage (Vp).

Vp = sqrt 2 x Vrms = 1.414*220 = 311 Volts.

The peak-to-peak voltage is 2xVp = 622 volts.

As the voltage wave form goes positive, the capacitor will charge to a DC voltage value equal to the peak voltage minus the voltage drop value for the diode. As the wave goes negative, the diode will stop conducting and the capacitor will remain charged. The voltage drop across the diode at the negative voltage peak will be nearly the peak-to-peak voltage of 622 volts.

Hopefully, I haven't just repeated the NCEES explanation.

Jim

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OH :brickwall: okay yes thanks Jim that makes sense, and Benbo was also on the right track. Ignore my comment above, the way it was written in the solution through me. I should have known that!. Im struggling with these electronics questions. Hell im struggling with everything.

John

jdd18vm said:

OH :brickwall: okay yes thanks Jim that makes sense, and Benbo was also on the right track. Ignore my comment above, the way it was written in the solution through me. I should have known that!. Im struggling with these electronics questions. Hell im struggling with everything.
John

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As I recall they are not real clear in the explanations. But, the actual problems on the exam are sometimes written in a funny looking way, so that is one of the main reasons the NCEES Sample Questions are a must have, to get used to their funky way of writing things. I swear, on my exam it looked like they drew out the problems using the drawing toolbar from MS Word!

benbo said:

As I recall they are not real clear in the explanations. But, the actual problems on the exam are sometimes written in a funny looking way, so that is one of the main reasons the NCEES Sample Questions are a must have, to get used to their funky way of writing things. I swear, on my exam it looked like they drew out the problems using the drawing toolbar from MS Word!

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you know they do seem to take short cuts and over simplify. I'll keep at it. Thanks Benbo

John

jdd18vm said:

you know they do seem to take short cuts and over simplify. I'll keep at it. Thanks Benbo
John

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I hated this electronics problems!!!!!!!!!!!

BringItOn said:

I hated this electronics problems!!!!!!!!!!!

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My favorite bit with these problems is the application of an AC signal to a little amplifier circuit with capacitors. For the typical problem, the capacitors end up being short circuits for the AC. The problem usually ends up being dirt simple but the capacitors are **** useful at making the problem LOOK like a total bite.

I spent a good bit of study time getting proficient at determining bias/operating points for simple transistor circuits. My strategy was to get good enough in all areas to smoke the morning session in hopes of giving myself a cushion for the afternoon. It felt like it worked when I left the exam and I ended up passing. YMMV

Jim

IFR_Pilot said:

My favorite bit with these problems is the application of an AC signal to a little amplifier circuit with capacitors. For the typical problem, the capacitors end up being short circuits for the AC. The problem usually ends up being dirt simple but the capacitors are **** useful at making the problem LOOK like a total bite.
I spent a good bit of study time getting proficient at determining bias/operating points for simple transistor circuits. My strategy was to get good enough in all areas to smoke the morning session in hopes of giving myself a cushion for the afternoon. It felt like it worked when I left the exam and I ended up passing. YMMV

Jim

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Thats the sort of feedback that is helpful. As I read through Text and try to grasp concepts, it seem much more in depth than actually required required on the Sample questions (or so it seems). I keep wanting to just skip ahead on this electronics stuff, but I too think I need to have a much broader range.

So any time you can offer advice like that above please do.

John

IFR_Pilot said:

John,
I don't have my book handy but I remember the problem.

The 220 VAC is an RMS value. What you are interested in for the capacitor is the peak voltage (Vp).

Vp = sqrt 2 x Vrms = 1.414*220 = 311 Volts.

The peak-to-peak voltage is 2xVp = 622 volts.

As the voltage wave form goes positive, the capacitor will charge to a DC voltage value equal to the peak voltage minus the voltage drop value for the diode. As the wave goes negative, the diode will stop conducting and the capacitor will remain charged. The voltage drop across the diode at the negative voltage peak will be nearly the peak-to-peak voltage of 622 volts.

Hopefully, I haven't just repeated the NCEES explanation.

Jim

Click to expand...

I don't mean to resurrect this resolved topic back from the dead, but how can I relate Jim's explanation in his paragraph with the reverse breakdown concept for a diode's performance characteristic?

EERM states on p. 43-11 "The peak inverse (reverse) voltage, PIV or PRV, is the maximum reverse bias the diode can withstand without damage."

Does this mean that after this circuit experiences more than 622V across the diode, the diode will enter breakdown region? Pardon my little experience in electronics, but that region is of no use to a diode and undesirable, correct?

The problem has you calculating the minimum acceptable PIV for the diode in this particular circuit. This is determined by finding the maximum voltage expected to be applied across the diode. Since the maximum voltage across the diode will be just less than 620 volts, the diode must be rated at 620 or greater. If this diode were placed in a higher voltage circuit, it would fail.

Jim