Diode i-v characteristic via oscilloscope

[sandy.bridge]

Hello all,
We utilized a differential attenuator to display the i-v characteristic of a 1n4005 diode. However, I was a bit confused as to how I could determine what the y-axis was calibrated as. I know that the vertical axis is actually current, but is there some sort of tactic for determining whether it is mA, A, etc? I have a printout of the oscilloscope's screen and it says:
$1-500\stackrel{m}{v}$ and $2-20.0\stackrel{m}{v}$

 

[vk6kro]

To measure current, you must be measuring the voltage across a small resistor.

So, you find out how much resistance that resistor has and then use Ohm's Law to calculate the current needed to produce, say, 1 volt across that resistor.

Then you read the scale on the oscilloscope to see what voltage is actually being produced.

Then you can work out what the current is.


OR, you can put a 1000 ohm resistor in the circuit instead of the diode, and calculate the current through the resistor and see what deflection that gives on the oscilloscope

 

[sandy.bridge]

Sweet, good idea! Thanks

 

[sandy.bridge]

Sorry to bump this thread, but I have one question:

When I display the i-v characteristic of a zener diode (3.6V, 0.77V drop forward biased), the graph is only clear at low frequencies such as 50Hz. When I increase the frequency more and more, the contribution along the x-axis splits into two lines, illuminating an elongated elipse. Is there some sort of explanation for this? I have tried numerous other zener diodes, and they all show the same result.

 

[f95toli]

My guess is that the hysterisis is simply due to the capaciatance of the diod, the time-constants can become quite long; especially if the load is an oscilloscope with a high inpute impedance.

 

[vk6kro]

There is always a current limiting resistor in series with a Zener diode and, if this is large, the power source cannot charge or discharge the capacitance of a Zener (when it is reverse biased) fast enough.

So this capacitor voltage lags behind the supply voltage and this produces the ellipse.

There is no need to do anything about it. Just test at the lower frequency.