Jump to content
Electronics-Lab.com Community

Search the Community

Showing results for tags 'rheostat'.

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


Forums

  • Electronics Forums
    • Projects Q/A
    • Datasheet/Parts requests
    • Electronic Projects Design/Ideas
    • Power Electronics
    • Service Manuals
    • Theory articles
    • Electronics chit chat
    • Microelectronics
    • Electronic Resources
  • Related to Electronics
    • Spice Simulation - PCB design
    • Inventive/New Ideas
    • Mechanical constructions/Hardware
    • Sell/Buy electronics - Job offer/requests
    • Components trade
    • High Voltage Stuff
    • Electronic Gadgets
  • General
    • Announcements
    • Feedback/Comments
    • General
  • Salvage Area

Find results in...

Find results that contain...


Date Created

  • Start

    End


Last Updated

  • Start

    End


Filter by number of...

Joined

  • Start

    End


Group


Website URL


Yahoo


Skype


Location


Interests

Found 1 result

  1. I have a circuit (see pic) that includes the primary winding of a 1:20 self wound transformer. Without some form of additional resistance this circuit will draw too much current from the power supply and damage the FET I’m using so I plan to put a rheostat inline with the primary winding to control the maximum current flowing in it. Given the basic equation for power expended in a resistor of P=I^2R, I’m trying to work out the maximum power rating for the rheostat. If I set the input voltage to 50V and set a 100R rheostat to 50Ohms then my current should be 1.0A. In that case the power rating of the pot will need to be 50W x2 (for de-rating 50% of the track)= 100W i.e. a very beefy rheostat! My query then is this - is it normal practice in circuit design to insert a fixed value resistor, in this case say 100R (as shown), so that the maximum current with the pot at zero is still 0.5A and the max power dissipation 12.5W? On the other hand, if the rheostat is set to zero then will it have no power dissipation capacity at all and be damaged - or are they built differently to potentiometers? Any other thoughts on how I can safely reduce the current flow welcomed. Thank you
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.