Boost on 9V battery

Discussion in 'The Projects Forum' started by chimera, Apr 17, 2011.

1. chimera Thread Starter Member

Oct 21, 2010
122
2
here are the numbers for a boost converter running on a 9V battery. The output should be 12V at 1amp. My question is: How much battery life am I looking at with this set up..also..im going to be using the same battery for 555 timer chip. (WISHFUL THINKING!!)

Topology: Boost

Inductance based on the specified minimum load current.

Volts In:
9V
Volts Out: 12 V
Load Current: 1 A
Freq.: 100KHz
Vripple: 0.06 V
Duty Cycle: 30.833333333333 %
Ipp Inductor: 0.1 A
Ipk Inductor: 1.05 A
Irms: 0.950 A
L: 255.917 uH
C: 51.39 uF

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25.2 KB
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2. SgtWookie Expert

Jul 17, 2007
22,182
1,728
Are you talking about a PP3 "transistor" 9v battery?
If so, maybe around 30 seconds. If you have a tailwind.

3. chimera Thread Starter Member

Oct 21, 2010
122
2
yeah..its the PP3 battery.

Well the whole objective is to produce a boost converter to produce a 12V 1A output to charge an iphone, making it portable. The circuit involves generating a 100Khz pulses to power the switching MOSFET.

I can design the circuit, but its going to be useless if its not portable using batteries.

4. Adjuster Well-Known Member

Dec 26, 2010
2,147
298
You want 12V at 1A from a battery supply, using a converter. A small 9V battery won't give that much power - you will need a fairly big battery. Why not go for a 12V battery in the first place, and dispense with the converter altogether?

5. Jeff7 New Member

Apr 17, 2011
11
1
I'd wager that the battery in an iPhone has a good bit more power capacity than a 9V.
And you'll incur efficiency losses going from 9V to 12V, and then further losses when the charger converts that 12V to what the iPhone needs.

This is a portable option, though it's not a cheap one.

6. SgtWookie Expert

Jul 17, 2007
22,182
1,728
I mentioned 30 seconds in my last post; that turned out to be overly optimistic.

A PP3 battery may have an internal resistance of anywhere from 1.7 Ohms to 2.8 Ohms when new. I threw a simulation together using your 100kHz 30.83% duty cycle, 256uH inductor, output cap, added a 100uF input cap, and ran it for 10mS.

The battery was loaded down to 6v, and the output voltage was ~8v.

Besides the low performance, the 9v battery would be very heavily overloaded, and would likely explode due to the excessive heat produced; it would be dissipating around 4.5 Watts internally.

Have a look at this 9v battery short circuit analysis report: http://friedrichengineering.com/web_documents/9volt Battery.pdf

Last edited: Apr 17, 2011
7. chimera Thread Starter Member

Oct 21, 2010
122
2
Okay, then how will i regulate the current at 1amp, assuming that the battery is rechargeable and i have a charging circuit in place

8. Jeff7 New Member

Apr 17, 2011
11
1
I think the point being made here is that this would be exceedingly impractical.

The battery in an iPhone has a power capacity that's more than double that of a good rechargeable 9V NiMH battery.
So at the very least, you'd completely drain 3 9V batteries in order to fully-charge an iPhone once. It's likely going to be even more than that, once you factor in all the efficiency losses along the way.

You're trying to refill a 2L soda bottle with a teaspoon.

9. chimera Thread Starter Member

Oct 21, 2010
122
2

LOL..thts a good way to put it. Okay, let me get some thing cleared up. The amount of time a devices needs to get charged depends upon the current or the voltage?

For instance, I am to charge something at 0.2A at 6V. Will it charge faster at 0.4A at 6V and charge slowly at 0.1A at 6V?

10. SgtWookie Expert

Jul 17, 2007
22,182
1,728
When you're charging a battery, you can't hold BOTH the current and voltage constant.

How you charge depends on battery chemistry, mAh or AH rating, temperature, and a number of other things.

Lead-acid batteries are generally charged at a constant current until the battery reaches a set voltage level for the "bulk charge" phase, then the voltage is held constant at a lower level until the current drops below a threshold value during the "absorption charge" phase, and then "float charging" ensues; charging at a fixed voltage somewhere around 13.6v to 13.8v when the battery core is at room temperature.

Batteries charge at different rates using different methods; and the terminal charge phases can be very different.

Check out: http://BatteryUniversity.com

Last edited: Apr 17, 2011
11. Jeff7 New Member

Apr 17, 2011
11
1
Assuming you could charge a battery that way, feeding it 0.4A would result in a shorter charge time.

But the issue with what you're doing is that you've got a set amount of energy to draw from - a 9V battery.
So if you were to try to charge something at 1A versus charging at 0.1A, you'd still (in a perfect world) only be able to get about 2 watt-hours out of the 9V. Pulling that power out at 1A* or 0.1A would just be the difference between draining the 9V completely dead in 10 seconds* or 100 seconds, but you'd still end up with the same (theoretical) 2 watt-hours being delivered to the iPhone's battery.

(* - these times and current draws are just to illustrate the concept)

But as SgtWookie said, charging a battery properly requires monitoring and control of the voltage and current together. In your case, all that is handled by the charging circuitry - you'd just feed it 12V, and it transparently handles the rest of the work.

12. SgtWookie Expert

Jul 17, 2007
22,182
1,728
Afraid this isn't quite correct.
The more rapidly you discharge a battery, the more power is dissipated across the battery's own internal resistance.

Have a look at some battery datasheets. Many of them have discharge plots for various loads; you'll see that the faster you discharge them, the less total power you can get out of them.

13. Jeff7 New Member

Apr 17, 2011
11
1
True; I was mainly illustrating a more basic concept, and intentionally left out that stuff (though I did not clarify that part).

I've seen that myself too - take a bunch of "dead" AA cells from a high-drain device, and stick them in a clock, and get another 6-12 months of life out of them because of the extremely low current draw.

Last edited: Apr 17, 2011