[HomeBrew] Need some help!
pete wokoun, sr.
[email protected]
Fri, 13 Feb 2004 11:25:20 -1000
The surest way if you have about a 20 volt DC power source is to put the
zener in series with about a 470 to 1000 ohm 2 watt resistor and put this
combination across the 20 volts. Put your voltmeter across the diode.
You'll measure either about 0.6 volts if you have the diode in the forward
current direction or the zener voltage (12 v) when it's in the zener
configuration. When you're reading a positive zener voltage on the
voltmeter, the meter negative probe is the anode and the positive probe is
the cathode.
>From: [email protected]
>To: [email protected]
>Subject: [HomeBrew] Need some help!
>Date: Fri, 13 Feb 2004 10:08:53 -0500
>
>** Please do NOT cross-post messages to multiple mailing lists on the "To:"
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>
>I have a SK550 (918) zener diode that I believe to be rated at 12VDC @
>50W. There appears to be ( barely visible) an arrow pointing away from
>the metal threaded housing towards the cathode end. When tested with the
>VOM on the diode scale, it conducts (value of 770) with the negative
>(black) lead on the metal base and the positive lead on the insulated
>end. No meter indication when the leads are reversed. So far so good!
>What concerns me is that I checked it with another Western Electric
>zener diode clearly identified with numbers and symbols and the
>positions of the two VOM leads must be reversed in order to get it to
>show a value. In this test, the red lead (+) is on the base and the
>negative lead is on the output end ( direction of arrow ) . This
>particular 50w zener was working fine on a two hole 8877 I built some
>time ago;however, I pulled the zener to do some extensive bias rewiring
>and cannot recall how it was originally wired. The question is basically
>which orientation should I use. Follow the arrow or reverse the diode and
>follow the wiring scheme dictated by the VOM . Is there anyone familiar
>with the SK550 in order to:
> A. verify its identification?
>
> b. help get the orientation correct?
>
> For obvious reasons, I'd rather not use the cut and try approach. Any
>help would be appreciated. Dealing with lots of power so I want to ensure
>I get it right the first time. Ron W2CQM/3
>
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