0
\$\begingroup\$

enter image description here

As you can see, both the right terminal of diode and top terminal of the resistor is negative.

This has something to do with shorting the negative terminal of an oscilloscope, but I don't under stand why. Why do these terminals both need to be negative?

From what I remember, the negative terminal of an oscilloscope is ground, so if the top resistance is positive, there will be a short circuit. I don't have any idea what that is or if that really was what I heard.

The experiment was to find the I-V characteristics of a diode.

\$\endgroup\$
  • \$\begingroup\$ As far as I can tell the circuit shown is incorrect. In what context it was drawn I have no idea. \$\endgroup\$ – Andy aka Sep 21 '15 at 13:59
  • \$\begingroup\$ The experiment was to find the I-V characteristics of a diode. \$\endgroup\$ – studious Sep 21 '15 at 14:05
1
\$\begingroup\$

See the below image:

enter image description here

As PlasmaHH says, the (-) sides of each channel of a typical inexpensive oscilloscope are common with each other so you cannot simply connect the (-) side to the lower pin of the 1K resistor without shorting out the 1K resistor (which could destroy the diode, damage the scope or do other bad things).

As shown the X axis shows voltage across the diode and the Y axis shows the negative of the current through the diode. By inverting the Y channel (usually there is a switch to do this) you can get your classic I-V plot, or simply flip the curve if no such switch exists.

\$\endgroup\$
0
\$\begingroup\$

You might want to read a bit more about how not to blow up your oscilloscope (google that term). I am not sure what exactly the "-Ch+" should indicate in your image, so let me give a quick superficial hint.

The ground clip on a scope is usually earth referenced, so when you put it at two different positions on your circuit, you create a low impedance path between those two (aka. blowing up your scope, maybe).

Additionally unless your circuit is floating (which we have no indication of in your example) it is ground referenced somewhere too, and you will create a low impedance path between that point and the point you put your ground clip (which may also result in a spectacular light show).

For that reason people that know what they do sometimes put the DUT behind an isolation transformer to remove any ground referencing.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.