Best technique for supplying different voltages from one battery source

[markurxn]

I've got an Arduino that I want to send 9 volts to and a electromagnet I want to sent 13 volts to. The battery source is a 14.8v lipo. I have a handful of step-down buck converters. It doesn't seem elegant to use 2 step-downs to deliver the different voltages, but this is the first time I've run across this situation. Would this be an appropriate solution?

 

[zappedagain]

Two buck converters is the simplest. Do you want one supply to enable the other?

Or you could put a resistor in series with your electromagnet from the battery for approx 13V.

Z

 

[itsmoked]

I agree with zapped, use one buck for the 9 Volts and either correct the electromagnet design or just use a dropping resistor or two diodes in series to shave off the volt. The cost and space needed for a 13V buck and the resulting efficiency make little sense.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[markurxn]

Cool, that gives me some options to investigate. Thank you both!

 

[markurxn]

Hey guys, something unanticipated happened and I wanted to verify my results with you. Prior to using the buck converters, I was using a 12v lead acid straight into both the Arduino and magnet with a mosfet in between to turn the electromagnet on. Using that approach, I had measured the amps going into the EM at 1.0 amps. Now with the 14.8 Lipo powering both buck converters, one at 9v to power the Arduino and the other at 13v to power the EM. Measuring the Amps now, i'm getting .04 amps on the EM. It looks like I'm saving nearly a full amp yet getting more voltage to the EM. This is amazing if its true, but I just want to validate the results with you guys.

 

[markurxn]

Never mind, I should know better. Can't beat ohms law.

 

[IRstuff]

There are plenty of multi-output regulators, like:

https://www.maximintegrated.com/en/products/power/switching-regulators/MAX17521.html/tb_tab1

It sells for about $7.20 on Digikey

TTFN
I can do absolutely anything. I'm an expert!
homework forum: //

http://www.engineering.com/AskForum/aff/32.aspx

faq731-376 forum1529

 

[OperaHouse]

And now for something completely different. Turn your electromagnet into a buck converter. I PWM the output to an electromagnet. The final duty cycle will be the percentage of the end to the source. A high speed flyback diode is needed across the coil. Pull in current is always much higher than hold. Relays easily operate at a quarter of their pull in current. you could easily operate your magnet at half current. Pull in at full PWM and ramp down duty cycle over a second to 50% or less. I operate 6V solenoids this way from a 60V source. Super fast and they run cool.

For the 9V source I would run a 7808 linear regulator. The on board 5V regulator needs just that to be stable. A common 7805 can be adjusted up with two resistors to obtain that. Never power anything external with the on board 5V except a couple LED. If more power is needed use a external 5V supply.

 

[LionelHutz]

I'm with OperaHouse, it doesn't make sense to use a regulator for the electromagnet. Just connect the MOSFET circuit right to the battery and then control the MOSFET via pwm to get the required current.

In the case of an electromagnet, the current is what matters.

 

[OperaHouse]

Here is some sample code to decrement down the duty cycle. Normally you would rly=1 to turn on and rly =0 to turn off. If program loops to fast, a loop counter should be added then every hundred loops or so, a decrement is performed. Magnet current could easily be reduced to 1/3 amp.


// PWM RELAY DRIVER SECTION PIN #3
// PWM relay output will reduce power consumption of relay
if (rly == 0) PWM3 = 0;
// set PWM value to zero if relay should be off.

// DECREMENT PWM VALUE EACH LOOP IF RELAY ON
if (PWM3 >125) PWM3 = PWM3 - 1;
// decrement PWM3 till it reaches minimum value
// reliable operation can often be 40%

// TURN PWM RELAY ON BY SETTING INITIAL VALUE
if (PWM3 == 0 && rly == 1) PWM3 = 255;
// when relay first turns on set PWM3 to maximum value
// prior state of relay has to be off in order to set this value
// this could also be used to power a 12V relay from a 24V source

// THIS OUTPUTS THE PWM SIGNAL TO PIN #3
analogWrite(3,PWM3);
// sets status of PWM3 RELAY output pin #3
// PWM values are between 0 and 255


The on board regulator power should be isolated with a diode and large electrolytic capacitor. I had one system that drew 125A off battery. When the battery was a little weak it would sometimes reset the processor from under voltage drop out. The diode would prevent drawing power from capacitor and the processor would br powered by the cap durring this period.

 

[benta]

Do you really think the electromagnet will care if it gets 13.0 or 14.8 V?
I don't.

Keep it simple.

Benta.

 

[ScottyUK]

Depending on distance and conductor size you might find that Ohm's Law takes care of the slight excess of supply voltage without further intervention. I'm inclined to agree with Benta.

 

[OperaHouse]

Just seems a shame to pull 1A out of a Li battery when you don't have to. Computers are supposed to allow us to work smarter. The OP is probably long gone.

 

[jimkirk]

One point I don't see mentioned, if you really do need to generate 13 volts from your LiPO...

Your nominal 14.8 volt battery sounds like it has 4 cells. The typical cell voltage for common lithium batteries is 3.7 volts, times four gives 14.8.

A fully charge cell is usually 4.2 volts (there are chemistry variations that range up to 4.35 volts or so).

A completely discharged cell will have a voltage around 3.0 volts, and the cell impedance will increase, so under load the output voltage will be lower than expected. And discharging below a couple of volts or so will permanently damage the cell, so have some sort of low voltage cut-off.

So your battery voltage will range from something like 12 volts up to 16.8 volts, so to truly generate 13 volts requires a boost-buck converter.

Sure, the electromagnet will work at 12 volts with only a slight degradation, but in general, always design your power supply to accommodate the full range of the expected battery voltage. This is tricky when that voltage goes both above and below your output voltage.