Diode Peculiarities

Forward looks odd as well. A deranged damaged diode.

--
 Thanks,
    - Win
Reply to
Winfield Hill
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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
  • 1N5189 showed extreme tempco in the volt range
  • 1N4148 showed measurable tempco in the mV range
  • Vr measured across dropping resistor to ground.

-M

Reply to
Mark Jones

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  |
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I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

--
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?
Reply to
John Fields

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
Reply to
Winfield Hill

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

Reply to
Mark Jones

The IN4148 reverse current looked a bit odd. Just measured one and it's only passing 50na, even at 60V!. regards john

Reply to
john jardine

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
Reply to
Ban

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...

Reply to
Robert Baer

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

Reply to
Tam/WB2TT

To get rid of the meter load, use a bridge, like this:

+4.5 | +--------+-------+---->E1 | | [1N914] [R1] | | E3E2 | | [RAM] [R2] | | +--------+-------+ | 0V

While 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
Reply to
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

Reply to
Jim Meyer

If you do measure an LED, make sure it is in the dark.

Regards Ian

Reply to
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
Reply to
John Fields

But then the decreasing zener voltage with decreasing source resistance would be unexpected, wouldn't it?

John

Reply to
John Smith

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
Reply to
Ban

--
OK then, choose the sensitivity you want:

                      +4.5
                        |
            +-----------+-----------+---->E1
            |                       |
         [1N914]                   [R1]
            |                       |
     E3E2
            |   |               |   |
          [RAM] +--[RHEOSTAT]-->|  [R2]
            |                       |
            +-----------+-----------+
                        |
                       0V
Reply to
John Fields

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

Reply to
Tam/WB2TT

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.

Reply to
Robert Baer

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.

Reply to
Robert Baer

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