NPN brain teaser puzzle

[echo47]

Can you solve this little puzzle?
If you already know the (surprising) answer, please don't reply too quickly and spoil the fun for everyone else.

 

[_i386sx25_]

10V ?

 

[wwfieee]

0V? or floating.

the transistor acting as diode, and collector is open.

 

[echo47]

No, try it on your bench. Many engineers find the results astonishing.

You can use almost any NPN transistor.
Be sure to use a DVM or other high-impedance voltmeter.

 

[flatulent]

The BE junction will be broken down with a zener effect to make about 5 V. There will be leakage current to the collector. You will probably measure 5 V or so.

 

[truebs]

10.7 V , As the base emitter junction is reverse biased it would at up to Vbe .

But if leakage is considered (say for eg Ico = 1 ma ) then volatage recieved would be 9.7 V !

 

[aryajur]

It goes to a negative value !?

 

[movingbait]

ok i did a sim using circuitmaker

So how is the result explained, i assume the sim is right(a damn long shot) and varied the resistance of the voltmeter appropriatly?

movingbait Ü

 

[chinito]

VBE is negative. NPN in cut-off --> zero volts reading

 

[echo47]

Good job Aryajur, you actually tried it! Simulators and textbooks won't give the correct answer.

Now who can explain it?

 

[aryajur]

The explanation I can think of is:
The excessive reverse bias across the base emitter junction, causes the depletion region to expand in the base region. Some of the charge that comes to expand the depletion of the base emitter region comes from the depletion of the base collector region. And so the base collector region depletion width decreases, causing a sort of forward bias, and thus we get a negative voltage in the voltmeter.
Please comment.

 

[Jack// ani]

Now its your turn echo47!!!!!! Please reveal it.....

 

[echo47]

Yes this circuit does output a negative voltage. I measure about -0.35V. Although the available current is small, only a fraction of a microamp, the transistor really does convert from positive to negative without using any capacitor, inductor, or charge pump.

Semiconductor equations usually don't give negative values. The circuit is not oscillating. So what's happening? A new form of energy? An April fool's joke?

No, it's an old form of energy.

I'll give the explanation soon.

You'll see!

Literally.

 

[pixel]

It seems like C and E are inversed.

It should be off (at the edge of inverse active region), but then voltatge should be -0.6V?

 

[echo47]

Here's the unexpected answer:

The reversed-biased base-emitter junction breaks down and emits yellow light. The light illuminates the base-collector junction, where the photoelectric effect generates the negative voltage.

Just for fun, you can cut the top off a metal-case transistor and examine the die under a microscope. A large transistor such as 2N3055 looks quite pretty, like an aerial view of a small town at night.

Note: Driving a transistor's B-E junction into reverse breakdown my injure the device, so don't put it back into your stock bin.

 

[Karthikeya]

it was indeed an unexpected answer........
can u plz explain this photoelectric effect in CB jn in little more detail.....

regards

 

[wwfieee]

how in the world did you find this out? This is like a DC-DC converter.

 

[echo47]

A few years ago, Bob Pease, a famous analog engineer at National Semiconductor, briefly mentioned this effect in his "Pease Porridge" monthly magazine column. I wish I could give you more technical info, but he did not go into detail, and I can't find any literature on this little-known effect.

Bob is an infamous practical jokester. He sent this puzzle to the engineering team at a rival company. The joke apparently worked, because those guys started tearing their hair out trying to figure out why their cherished semiconductor theories were suddenly flying out the window.

Heeeeeerrrr's Bob!
**broken link removed**

 

[flatulent]

Is this a thermal effect? Can someone who has measured it adjust the transistor temperature with spray and see?

 

[echo47]

I freeze-sprayed my 2N3904. The output voltage increased slightly, from -0.32V to -0.40V.

Right now I can't find my decapitated 2N3055. I recall that increasing the input current caused more and more tiny pinpoints of yellow light to appear, like turning on streetlights. Fascinating!