30A OR-ing Circuit?

• #25 Reply
  Posted by Magicmushroom666 on 02 Jul, 2015 17:01
• Hi, I'm running an onyx heated plate for my printer too. I'm using 18V at whever current that takes, not actually measured it recently, but 18V was enough to get good heat up times without needing silly sized cables etc.
  
  You could save on the SSR and just use a mechanical relay to save costs, as long as the controller is just off-on. I've used a relay salvaged from a car which happily manages the current.
• #26 Reply
  Posted by PeteD on 06 Jul, 2015 17:46
• Quote from: Magicmushroom666 on 02 Jul, 2015 17:01

  Hi, I'm running an onyx heated plate for my printer too. I'm using 18V at whever current that takes, not actually measured it recently, but 18V was enough to get good heat up times without needing silly sized cables etc.
  
  You could save on the SSR and just use a mechanical relay to save costs, as long as the controller is just off-on. I've used a relay salvaged from a car which happily manages the current.
  

  
  Someone please correct me if I'm wrong, but I believe wire gauge is determined by amperage not voltage.
  
  I did look into using an EMR instead of an SSR, but too may people who have used EMRs in their printers have had issues with false signals messing up the print when the EMR switches. Avoiding that problem is worth the cost of an SSR to me.
• #27 Reply
  Posted by ruffy91 on 06 Jul, 2015 18:57
• Quote from: PeteD on 06 Jul, 2015 17:46

  Someone please correct me if I'm wrong, but I believe wire gauge is determined by amperage not voltage.
  

  The heated bed is fixed resistance so more volts means more ampere.
• #28 Reply
  Posted by PeteD on 06 Jul, 2015 19:32
• Quote from: ruffy91 on 06 Jul, 2015 18:57

  Quote from: PeteD on 06 Jul, 2015 17:46

  Someone please correct me if I'm wrong, but I believe wire gauge is determined by amperage not voltage.
  

  The heated bed is fixed resistance so more volts means more ampere.
  

  
  The heated bed is a fixed resistance, so more volts means more power dissipated, which is the purpose of this exercise. 
  
  However, the amperage flowing through the wires to the heated bed should max out at 30A, and the wire gauge won't change between 12V or 18V or 24V.
• #29 Reply
  Posted by ruffy91 on 06 Jul, 2015 21:40
• No. If the bed is 1.1 Ohm you will get a max. current of 12V/1.1Ohm=11A for 12V or 22A for 24V
• #30 Reply
  Posted by Riotpack on 06 Jul, 2015 22:42
• big rectangle power resistor bolted to the plate and connected to the mains through a thermostat
• #31 Reply
  Posted by LukeW on 07 Jul, 2015 03:20
• Quote from: PeteD on 06 Jul, 2015 19:32

  Quote from: ruffy91 on 06 Jul, 2015 18:57

  Quote from: PeteD on 06 Jul, 2015 17:46

  Someone please correct me if I'm wrong, but I believe wire gauge is determined by amperage not voltage.
  

  The heated bed is fixed resistance so more volts means more ampere.
  

  
  The heated bed is a fixed resistance, so more volts means more power dissipated, which is the purpose of this exercise. 
  
  However, the amperage flowing through the wires to the heated bed should max out at 30A, and the wire gauge won't change between 12V or 18V or 24V.
  

  
  Higher voltage means less current for the same power transfer, meaning thinner wires and less supply current needed, less voltage drop in the wires and less heating, if the heating element is designed for the appropriate voltage.
  
  Let's suppose that your particular 3D printer design needs 100W of bed heater power. This depends on the area, the materials, airflow, thermodynamics, desired temperature at the surface etc.
  But I think 100W is plausible - not 720 watts.
  
  Suppose you've got a 12V power supply available.
  
  So you use a 1.44 ohm heating element, and this gives you 100W of heat power, ignoring loom resistance. And your power supply needs to deliver 8.33A.
  
  But you need to watch the loom resistance - suppose your total loom resistance is 0.5 ohms.
  
  Now your power delivered to the heating element is only 55 watts, so it won't get hot quickly enough. And you're dissipating about 20 watts in the wires.
  
  So suppose you keep the same heating element, and crank the supply voltage up to 24 volts.
  
  Ignoring loom resistance, you're now dissipating 400W in the heated bed, so it will get way too hot. The PSU also needs to be able to deliver 4x the power, twice the current it did before - 16.7 amps at 24V.
  
  If there's 0.5 ohms of loom resistance then the heater power has dropped to 220 watts - still probably too large.
  And you're dissipating 77 watts in the wiring loom.
  
  One solution is to use thicker cable, but a pound of copper costs a bit.
  
  A better solution is to use a higher supply voltage if you can, the highest voltage that is practical, and choose the bed heater resistance appropriately. And use relatively thick connecting wires.
  
  Suppose the supply voltage is 24V and the heating element resistance is 4.7 ohms and the loom resistance is 0.5 ohms. The element resistance should be high relative to the cables.
  
  Then you'll have 100W at the heating element, a manageable 10W in the wiring loom, and a supply current requirement of 4.6A which is pretty manageable.