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I am trying to build a simple circuit that takes a 9v battery through a voltage regulator that outputs 1.5v across a load resistor . The load will be just a piece of metal that gets heated up.

My initial understanding of voltage regulators was wrong. I thought I wouldn't be wasting a lot of power using one, but from further reading I have discovered that they are just used for rectifying an ac source and when (vi-vout) is a large number a lot of power is dissipated/wasted as heat in the regulator. I want as much of the power from the battery as possible to be dissipated across the load resistor @ 1.5v.

So, my questions is how can I build this circuit with a 9v input and a 1.5v across load dissipated as heat with the smallest amount of wasted power through a regulator or other components?

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  • \$\begingroup\$ You need a switchmode converter. But how much heat are you expecting to get? 1 Watt would be a lot of power for a 9V battery. \$\endgroup\$
    – Dave Tweed
    Commented Mar 3, 2013 at 18:41
  • \$\begingroup\$ Somewhere from 175 degrees C to 200 degrees C. I am going to use a load with a high resistance. So this switchmode converter won't dissipate a significant amount of power? I was hoping to make this as efficient as possible. Thanks. \$\endgroup\$
    – Sam
    Commented Mar 3, 2013 at 18:55
  • \$\begingroup\$ I read up on switch mode converters and can see that they are significantly more efficient than using a voltage regulator. Thanks. \$\endgroup\$
    – Sam
    Commented Mar 3, 2013 at 19:11
  • \$\begingroup\$ Do you think the LM25576 would be a good choice for my application? : ti.com/product/lm25576#buy Thanks. \$\endgroup\$
    – Sam
    Commented Mar 3, 2013 at 19:25
  • \$\begingroup\$ I asked "how much heat", not "what temperature". What is it you're heating up, and how fast does it lose heat to the environment? \$\endgroup\$
    – Dave Tweed
    Commented Mar 3, 2013 at 20:30

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Couple of $0.02 thoughts.

  1. Can you use a heater with higher resistance so that it can run directly off 9V? Then you wouldn't need a regulator.
  2. Pulse-width modulation (PWM). Connect the load through a MOSFET (or a BJT) to the battery. PWM signal is applied to the MOSFET gate. In this approach, PWM duty cycle controls the RMS voltage across the heater.
  3. Buck converter is a type of a switch-mode power supply. It's more complex that, say, a linear regulator, but is can have 90% efficiency. May be, you can find an integrated power supply module like this one, which would suit your input voltage and output voltage & current. If you can't find a module, you can roll your own.
    edit: LM25576 mentioned above is an IC around which a buck converter can be built. "Buck converter" and "step down switching converter" can be used pretty much interchangeably.
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  • \$\begingroup\$ 1. I could, but would the 9v last as long without some kind of regulation? Do you think I would be better off just using a higher resistance? I am trying to make this device work for as long as possible on one battery. So do you think it would lose less power with a higher resistance and no switch mode converter. Or a lower resistance with the switch mode converter? 2. Would a buck converter work better than the LM25576 mentioned above? Thanks. \$\endgroup\$
    – Sam
    Commented Mar 3, 2013 at 22:15

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