From a 1969 ad for RCA Institutes, a mail-order electronics training course. There is no trick or secret, the resistor value doesn't matter, it's just a straight-up beginners question. What is your answer?



ak

 

21.2........

 

To a first order, 21.4 Vpp.

That's making some pretty obvious assumptions about things like the resistor being sufficient to keep keep the reverse-biased Zener in breakdown but below it's max current and also assuming nice, nearly vertical characteristics in both forward conduction and reverse breakdown.

 

If the resistor is ~0 ohms or 1 G-ohm the answers, the solveability changes.

What if the diodes were silicon carbide. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.1011.4098&rep=rep1&type=pdf

Temperature?

 

If the resistor is ~0 ohms or 1 G-ohm the answers, the solveability changes.

What if the diodes were silicon carbide. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.1011.4098&rep=rep1&type=pdf

Temperature?

"beginner's question"

 

There is no trick or secret,

 

0V, because the upper output is connected to the shadow, not the circuit. (What’s that about?)

Okay, I’ll play along. Looks like 21.3 to me.

Bob

 

What if the diodes were silicon carbide.

It states the diodes are 10V zeners.

 

Answer portion of the same ad, upside down in the original.

Again, ***RCA*** Institutes.

Globus carnis sugat.

ak

 

It states the diodes are 10V zeners.

But it doesn't state that they aren't 10 V silicon carbide zener diodes.

However, it's a beginner's question and thus perfectly reasonable to assume the standard beginner's model for zener diodes.

 

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ak

They are using an even simpler model, namely that a zener diode has no forward voltage drop. I don't know how reasonable that is, since by the time students are introduced to zener diodes, they should be quite comfortable with the notion that physical diodes exhibit a more-or-less constant voltage drop when forward biased.

I assume that they gave full credit to people that included a forward drop in their answer.

 

They are using an even simpler model, namely that a zener diode has no forward voltage drop. I don't know how reasonable that is, since by the time students are introduced to zener diodes, they should be quite comfortable with the notion that physical diodes exhibit a more-or-less constant voltage drop when forward biased.

I assume that they gave full credit to people that included a forward drop in their answer.

Although not directly related to this question, during my early years of circuit design, I would consider by default that a diode had a forward voltage drop of around 0.7V.

It soon became apparent to me that this was not necessarily the case – especially where the diode was within a low current circuit and therefore the resultant diode volt drop might be closer to 0V.

 

Hello there,

Ha! The output is 0vpp because the input is not connected to anything

 

Although not directly related to this question, during my early years of circuit design, I would consider by default that a diode had a forward voltage drop of around 0.7V.

It soon became apparent to me that this was not necessarily the case – especially where the diode was within a low current circuit and therefore the resultant diode volt drop might be closer to 0V.

It would have to be VERY low current to get close to 0 V for a typical silicon diode. Even with an ideality factor of 2, it would be around 120 mV/decade of current.

I generally use 0.7 V unless it's a power device in which case I'll use 0.6 V or even 0.5 V. For higher than normal currents I might use 1 V.

But I can't recall ever having to discover that it wasn't some fixed voltage. I think my first exposure to diodes was in 10th grade when I started taking a ham radio class. My memories are a bit vague, but I seem to recall him drawing an exponential curve and then talking about how the voltage increases rapidly up to the knee voltage at a very low current and then stays pretty much constant over the typical range of current, two or three orders of magnitude, and can be approximated well by 0.7 V.

 

The output is 50 volts because the crappy Chinese Zeners failed the second the power was applied.

 

Y'all are overthinking don't you have anything better to do .

 

Y'all are overthinking don't you have anything better to do .

You need multiple choice and "Chose the best answer".

In a college class I challenged some question on a an exam. I told him, my answer was correct. he finally said OK, "Your not supposed to know that yet"

 

Y'all are overthinking don't you have anything better to do .

No.

 

Y'all are overthinking don't you have anything better to do .

Hi,

Yes, but we did that stuff already and had some time left

I played a chess match at the old RCA building near here long time ago. So long ago i almost forgot.