I'm sure this has been answered plenty before but I cant seem to get the search terms. I'm wanting to use a 3.7v battery to power a small 9v device and a 12v device at the same time using a boost converter. Would it be better practice to have each converter be powered from the battery, or would it be better to attach a converter to the 9v converter and boost the 9v to 12v which would also let me use just one switch to turn off both devices. Or is this idea madness to begin with?
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\$\begingroup\$ Welcome! Can both your 9 V and 12 V device share ground? Are they connected in any way? \$\endgroup\$– winnyCommented Jul 4 at 11:32
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\$\begingroup\$ @winny yup, same device so they'd share a ground (I'm making something be powered off battery(9v) and also have a backlight(12V) \$\endgroup\$– luminousfoxCommented Jul 4 at 11:36
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2 Answers
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I would probably boost to 12 volts and, use a buck converter from 12 volts to 9 volts. However, if the 12 volt device current was significantly smaller than the current taken by the 9 volt device, I would consider two boost circuit; one to 9 volts and, one to 12 volts.
If there are isolation requirements then, you would likely use isolating DC-to-DC converters.
Boosting 9 to 12 volts can cause overshoot of the 12 volts if you are hand designing the boost converters.
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\$\begingroup\$ Ooh I didn't even think of stepping the 12v down to 9v, I probably should have included what I'm trying to do, Making a (9v)toy with a screen have a (12v)backlight as well. so the 12v is only for a very small strip of LEDs and the 9v is the main show? \$\endgroup\$ Commented Jul 4 at 11:43
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\$\begingroup\$ My answer covers that scenario @luminousfox \$\endgroup\$– Andy akaCommented Jul 4 at 11:45
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\$\begingroup\$ ahh yes it does, thanks for the help and ideas! \$\endgroup\$ Commented Jul 4 at 11:47
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\$\begingroup\$ @luminousfox maybe you should wait for an hour or two for some other options (more answers) before taking a decision on accepting your preferred answer (What should I do when someone answers my question?) \$\endgroup\$– Andy akaCommented Jul 4 at 11:52
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If you are making generating 9V from 3.7 and generating the 12V from 9V or vice versa, You are wasting some power at first regulator as the power for second regulator also will have to come through 1st regulator which in turns increase the power requirement at 1st and thus increasing the loss at first. Assuming both regulators have 80% efficiency, and both 9V and 12V consume 1W (just for calculation),
Two regulators with 3.7V input, each regulator will have a loss of 0.25W. Total loss across will be 0.5W and the combined efficiency will be 80%. If you daisy chain the regulator, the loss at 1st regulator will be 0.56W an at 2nd 0.25W. which totals to 0.81W. and it will reduce the combined efficiency to 71.11%
If your second regulator is taking very less power the loss will be less but still using separate boost regulator from 3.7V will be better idea if you don't have any other constraints or issues.
