Forward looks odd as well. A deranged damaged diode.
Forward looks odd as well. A deranged damaged diode.
-- Thanks, - Win
From experimentation, some interesting results:
By Mark Jones 1N4001 1N5189 1N4148
1) Vforward @ 5v, z=100M 0.29v 0.01v 0.46v 1) Vforward @ 5v, z=10M 0.32v 0.02v 0.48v 1) Vforward @ 5v, z=1M 0.38v 0.05v 0.55v 1) Vforward @ 5v, z=100k 0.45v 0.10v 0.62v 1) Vforward @ 5v, z=10k 0.56v 0.18v 0.70v 1) Vforward @ 5v, z=1k 0.75v 0.37v 0.90v----------------------------------------------------
2) Ireverse @ 5v, z=0.1R 66fA 1.1uA 2.75mA----------------------------------------------------
3) Vreverse @ 5v, z=100M 6.6mV 4.99v 2.75v 3) Vreverse @ 5v, z=10M 2.1mv *4.10v 2.65v 3) Vreverse @ 5v, z=1M -0.3mV *1.00v *2.24v 3) Vreverse @ 5v, z=100k -0.7mV *0.12v *1.67v 3) Vreverse @ 5v, z=10k -0.8mV 0.01v *1.06v 3) Vreverse @ 5v, z=1k -0.9mV 300uV 0.48v-M
Why? Do you mean 20Kohm/Volt isn't adequate ?:-)
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
-- That looks pretty fishy to me; especially the part about a reverse current of 2.75mA through a 1N4148. How did you make the measurements and what did you use to make them with? In particular, what's the resistance of your voltmeter?
No, actually that's quite interesting. Relabel that column, and measure a proper 1n4148 for a fourth column, then add a fifth red LED column, and you'll have something pretty cool indeed.
-- Thanks, - Win
Oh crap... that 1N4148 ended up actually being a 1N5231, which is a
5.1v zener! It looks damn identical to a 1N4148, no wonder the results were askew... sorry about the misleading data there. :0-M
The IN4148 reverse current looked a bit odd. Just measured one and it's only passing 50na, even at 60V!. regards john
Mark, be aware that you not only measure the DUT, but at the same time the effect of the input current of your voltmeter. Try to find out what comes from what and if with these values the whole thing really makes sense to measure.
-- ciao Ban Bordighera, Italy
Yup! Heisenberg rules! Measuring equipment, when used on a circuit always alters that circuit. So, one must know about that equipment and the circuit, and re-draw the schematic to accomidate. Then use that to determine how useful the readings, and the meaning of the readings. When us magicians use magic, we have to be willing to accept the consequences of that magic...
I ran into that. Tried to measure the voltage drop of a 1N914 which feeds backup 4.5 V to a small CMOS RAM. The current drawn by the RAM is the same order of magnitude as the current drawn by the DVM (10 M input?). Best I could come up with is "somewhere" around 300 mV.
Tam
To get rid of the meter load, use a bridge, like this:
+4.5 | +--------+-------+---->E1 | | [1N914] [R1] | | E3E2 | | [RAM] [R2] | | +--------+-------+ | 0VWhile measuring E1, adjust R1 and R2 until E3-E2 = 0. Then, since:
E1R2 E2 = ------- R1+R2
the drop across the diode will be _precisely_ E1 - E2.
-- John Fields
The characteristic curves for a 1N914 show that 300 mV drop will result from 2 uA of current. A 10 meg input DVM will be passing about
0.03 uA at 300 mV so the meter current is only 1.5% of the RAM current.Jim
If you do measure an LED, make sure it is in the dark.
Regards Ian
-- What offset current? If you're thinking about a d'Arsonval movement, when E2 = E3 there'll be no current flowing through the meter, so it'll vanish as far as the circuit is concerned. Same thing for a digital meter. Think about it... all of the bias and offset currents for the front end are supplied by its own internal supply when it's not connected to anything and it's reading zero, so why should current need to flow across the inputs when they're both connected to the same voltage? And, essentially, shorted together. Perhaps this would have been better: +4.5 | +--------+-------+---->E1 | | [1N914] [R1] | | E3E2 | | [RAM] [R2] | | +--------+-------+ | 0V
But then the decreasing zener voltage with decreasing source resistance would be unexpected, wouldn't it?
John
The offset current of the meter will still influence the result, but when you swap the meter connections and note down the different values you might compensate for that too.
-- ciao Ban Bordighera, Italy
-- OK then, choose the sensitivity you want: +4.5 | +-----------+-----------+---->E1 | | [1N914] [R1] | | E3E2 | | | | [RAM] +--[RHEOSTAT]-->| [R2] | | +-----------+-----------+ | 0V
I might mention that I was reading about .25V across the diode, and about ..35V if I did Vin - Vout. Clearly, in the first case the meter was subtracting from the diode current, in the second case it was adding to it. Since this was for personal use, either number was OK.
BTW, the use was as a standby battery in a Yaesu transceiver. The original lithium battery was not user replaceable, and would have required disassembling the front panel. I just added a 3 x AA battery holder and its own diode. (Lots of room inside for the 3 alkalines). After 4 years, it is still going
Tam
One dirty trick to help, is the "slide-back" method. Put an adjustable supply in series and adjust for zero reading, then read the voltage used for zeroing.
I take it that the batteery was one of those tabbed versions and soldered in? I hate them and always remove them, *then* do what yoou mentioned.
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