Friday, November 20, 2015

Electronics: Germanium Diode- Testing and Analysis

I've never really had the chance to use or test out a germanium diode before, so this was an exciting opportunity for me to try one out! Of course, I first wanted to see if the sayings that there would be voltage drop of around 0.3 volts would be accurate...


I hooked up the diode to the multimeter to test it out. The cathode (the blue-striped end) goes to the negative (black) terminal on the multimeter for the diode test mode.

If the diode is hooked up the other way, then it will not conduct until the reverse breakdown voltage is supplied, which isn't possible for the multimeter. The multimeter would instead return "no reading".

But if we hook up the diode correctly, then we get the right result:


Neat! So this diode drops 0.252 volts.

But what does this mean? Let's look at a sample I-V chart for a diode:


The voltage dropped by the diode in this situation, or 0.252 V, corresponds directly to a certain value of current. In fact, the current supplied to the diode determines exactly how much voltage is dropped.

After consulting Fluke about my Fluke 76 multimeter, I learned that it supplies 800 uA, or 0.8 mA, in the diode test function. That means that the point (0.8 mA, 0.252V) is a part of this germanium diode's I-V curve.


By identifying this diode's part number (1N695), which can be found from its stripes, I was able to scrounge up a datasheet from IHSDatasheets360. The datasheet has this little table:


Is this chart useful in confirming the behaviors of this diode? Not really. The only values that are relevant are from the "Forward Voltage" line, which mentions that the diode should drop a maximum of 1.0V at 100mA. There's no information about what happens at 0.8 mA, which is a different point on the graph.

The 1.0V 100mA behavior shouldn't occur in a steady state, though, because the "Average Power Dissipation" is 80 mW. How do we know this? Through P=IV, we can find that the "Forward Voltage" power dissipation is 1.0V*100mA, or 100 mW. To avoid damage to the diode, the 1.0V, 100mA state could only be at certain point of an input that alternates between a 100mW value and a lower power.

Basically, all that we know about the I-V curve is this (zoomed in part of graph above, not to scale):


We know what the multimeter test value is (0.252 V at 0.8 mA) and what the datasheet says is a maximum value for forward voltage (1.0 V at 100 mA). In total, not very helpful, but interesting nonetheless. It is unlikely that any datasheet value will match the voltage drop that the multimeter finds, simply because the multimeter value is not a maximum or minimum but somewhere in the middle. However, you may be able to find I-V curves for diodes online, which can be useful for comparing multimeter values.

If the multimeter value was lower in voltage but higher in current than the datasheet value, we could expect something wrong (maybe a short). If the multimeter value was higher in voltage but lower in current than the datasheet value, something could be wrong as well.

I mentioned in the last blog post that germanium diodes are highly sensitive to temperature. This appears to be shown right here- just take a look at the datasheet sample above. The "Inverse Current" entry is repeated twice just for different temperatures because the leakage (the value measured) increases by a factor of 10 over 45 degrees Celsius. I also mentioned that the "Peak Inverse Voltage" (see above) for a germanium diode would be pretty low- this is true here, with a value of only 20 volts for the 1N695. (The 1N914, a common silicon diode, has a peak inverse voltage of 75 V.)


Interesting information! That's all for now about the 1N695 germanium diode. Coming up next week, there will be some graphics of germanium diodes that I have created... see you then!