I am building two separate circuit boards, one for driving 12-24 vdc high current and the second to provide 5 volts for the logic circuit and atmel controller (low current). Essentially I have 12 - 24 volts dc available but have to reduce it to 5 volts for the control board. It appears to me voltage regulators would waste a lot of current to heat dropping to voltage that much. Therefore I suspect it would be better to use a step down buck converter. I think I could use a better understanding of the efficient use of the buck converter. Any suggestions which one to use would be helpful. Need to keep it simple as this is driving a pump that works off a solar panel and has to conserve as much energy as possible.

Thanks,
LaurenceR

 

A 5 V regulator might waste about 200 mW & buck reg not too far different at low current?? Do you know 5 V supply current?

 

I believe the ICs draw on the order of 20-30 milliamps. putting a signal on an opto isolator, another 20ma. Total in the neighborhood of 100ma.
Atmel attiny26 IC

 

The attiny26L consumes 15 ma at 5 volts. Just looked it up.
Thanks

 

Probably not worth the trouble of a switcher. The linear would waste about 300 mw, the switcher 60mw.

 

Thank you for your help. I didn't realize the little difference at lower current draws, not to mention how much easier it is to set up.

Many thanks.

 

Have to ask this question.
If I am dropping 24 vdc down to 5 vdc at approx. 25ma, would it work to first use a voltage divider ahead of the linear voltage regulator.
example:
voltage divider R1 600 ohms, R2 500 ohms, Vin=24 volts, Vout=9.6 volts with no load. This would draw approximately .0045 amps or (5.0 volt * .0045a=.0255 watts.
Now if you subtract the 5 volts from the 9.6 volts for the regulator you have 4.6 volts * .050 amp power consumption of the circuit, this would give you another 0.23 watts.
Adding the .23 watts and the .0255 watts gives a total of .2555 watts which is much less than if you dropped 24 volts down to 5 volts with the regulator alone.
19 volts * .050 current draw of circuit = .95 watts.
I am probably missing something here and if so I would appreciate it is someone could explain what.
Thanks again.

 

The voltage divider would waste exactly the same amount of power as it would save in the regulator. Your calculation is wrong in that the power in the voltage divider would be 24V times the current, not 5V times the current. Also, the current through the regulator would also be passing through the upper resistor, using more power still.

Bob

 

Thank you. I can see what you mean about the voltage divider and the regulator. Six of one and half a dozen of the other. Appreciate the explanation.

Larry

 

Do you think an LM7805 would handle this with minimal wasting of current, or should I take another approach.
Thanks,
Larry

 

Using a Step Down Switching voltage Regulator LM2576-5 5V/3A output, eff=75% when Vin=12V.

 

Thank you, I'll try it.