Polyphase rectifier: Calculating the diodes needed.

  • Thread starter Thread starter Simple10
  • Start date Start date
  • Tags Tags
    Diodes Rectifier
Click For Summary
SUMMARY

The discussion centers on calculating the necessary diodes for a three-phase rectifier system using a theoretical coil with specific parameters. The user derived a voltage output of 111.78V DC at 46A after rectification, based on a coil with 170 turns and a magnetic field of 4871 Gauss. The user inquired about the voltage and amperage ratings for diodes, as well as the impact of additional phases on diode requirements. Recommendations for diode specifications include a reverse voltage rating of at least 37V and a continuous current rating of at least 15A, with considerations for voltage drop across the diodes.

PREREQUISITES
  • Understanding of Faraday's Law of Electromagnetic Induction
  • Knowledge of three-phase rectifier circuits
  • Familiarity with diode specifications and ratings
  • Basic principles of AC to DC conversion
NEXT STEPS
  • Research diode specifications for three-phase rectifiers
  • Learn about the effects of adding phases on rectifier performance
  • Investigate the use of simulation tools like Wolfram Alpha for electrical calculations
  • Explore the relationship between magnetic field strength (Tesla) and current output
USEFUL FOR

Electrical engineers, hobbyists designing rectifier circuits, and anyone involved in power electronics or renewable energy systems.

Simple10
Messages
15
Reaction score
0
Hello,

I have created a theoretical coil of
170 turns*.4871 Tesla's*.0045m^2/.06 seconds=6.21V (Faraday of course.)

ohm's of the coil=.405
6.21V/.405ohms=15.33A

The equation I used
-N * curl*B*A/curl*t
N=number of turns; B=Tesla's; A=area of the coil; t=Time; curls cancel out.
(To calculate Time I used 1000 rpm/60=16.666 rps;1 second/16.666 rps=.06 seconds)
(To calculate Tesla's I chose a magnets surface gauss, 4871 and converted to Tesla's; 10,000 gauss=1 Tesla)

Lets say I have 9 coils with 3 phases:
http://www.allaboutcircuits.com/vol_3/chpt_3/4.html#03267.png
How do I calculate the Diodes I will need for the 3 Phase Rectifier? (see above link for schematic.)
If I had more phases will that change what diodes are needed? (see above link for 6 phase schematic.)
What voltage and amperage ratings for diodes?
What reverse voltage ratings for diodes?
Will the voltage drop across the diodes depend on the diodes I choose for the system?
Is a dump load nescessary if I were charging batteries and the batteries were full?


After rectification 111.78Vdc @ 46A. (This does not account for any voltage drop through the rectifier.)

Google seems to return non pertinent info about calculating for diodes.
Would Wolfram Alpha be better in this area?

P.S.

My goal is to understand the generating of Alternating Current and it's conversion to Direct Current.
 
Last edited:
Engineering news on Phys.org
My voltage and amps out is incorrect.
For the 3 phases it's 37.26 Volts and 46 Amps

I checked mouser form the first sticky for diodes.
I know that diodes allow voltage and amps to travel in one direction and prevent it going in the opposite direction.
My Volts on one side is 18V the other side 37V. Reverse Voltage I think would be 37V.
Second, Continuous current through the diode. I have 15 Amps.
I think I need a diode that would handle <37V and at least 15A.
I found Power Diodes from 200V reverse and 16A continuous up to 1400V reverse and 16A.
Voltage drop for all of these is 1.2V per Diode.

Would the 200V, 16A be sufficient?

If I add 3 more phases for 6 phases total that would be 37V @ 92A. Is this correct?

I also noticed the higher the Tesla's the more current produced.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
13K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 11 ·
Replies
11
Views
8K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
7K