There is a lot of miss understanding on diodes.

The very high voltage diodes are lower quality in many respects
than lower ones. The Peak Inverse volage of 1000 means a 1000v
reverse voltage can be applied without a lot of current flowing.

The leakage in both directions is higher than say a 100v model.
The 100v one would vaporize on high voltage so each have their job.


If you are running motors in 240 or 375v range, the 1000v would be ok.
Magnetic loads can create double voltage spikes.

Martin

On 1/15/2011 11:58 PM, Ignoramus25553 wrote:
On 2011-01-16, Bill wrote:
On 1/14/2011 10:24 PM, Cydrome Leader wrote:
Bill wrote:
On 1/14/2011 1:35 PM, Terry wrote:
My 13" SB metal lathe has been retrofitted with a PM DC motor. It's
not a treadmill motor, it's continuous duty according to the tag and
can handle up to 180 V. I assembled a speed control from a
treadmill-type board. Unfortunately it doesn't appear that the
controller provides enough oomph.

What I'd like to try instead is a 120v 20amp variable transformer,
output to a bridge rectifier, then filter the DC and send it to the
motor. By happy coincidence I have all the parts, total cost will be
simply a bit of time.

The rectifier I have is a GBPC1210W; data sheet says 12 amp, 1000 v.
It is rectangular with a hole through the center. It looks like it
ought to be mounted to a heat sink, but the data sheet I saw doesn't
say anything about doing so.

The question is: should a bridge rectifier be heat-sunk (sinked?) in
such an application? If so, how big (roughly) should the heat sink
be? Would it be sufficient to bolt the rectifier to the 4" square
metal enclosure? Thanks for your input.

it's amazing how much just plain wrong information you are getting.

Vfd on a silicon diode is .7v. Fullwave rectifiers have two in the
forward path conducting on each half wave. 1.4V drop. If you pull 12
amps, power is therefore 16 watts. But, it is AC, so you won't pull the
12 amps continually.

maybe you should check a datasheet for a bridge rectifier some day, or
takes some actual measurements.

Vishay is even kind enough to tell you the dissipation in watts for such
items.


ok, have it your way - here is a reference showing that what I said is
right http://www.electronics-tutorials.ws/diode/diode_6.html, there are
at least 10,000 other references - perhaps you have found some new kind
of silicon, with new semiconductor properties. Go ahead and give more
bad information to the OP, this is not my problem. In fact, maybe YOU
should take some measurements and check your facts.

and no, I will not respond, I have no need to get into a screaming
contest for no money with people who have no information. I will let the
OP do his own research and determine what he wishes to do. I've offered
my recommendation, YOU choose to tell me I am wrong.


Guys, can I ask an ignorant question. I thought that voltage drop
depends on the maximum reverse voltage that the diode would withstand?
Is that true or not?

In other words, is the voltage drop on a 1,000V rated diode, the same
as on a 30v rated diode?

i