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

 

Are you talking about a PP3 "transistor" 9v battery?
If so, maybe around 30 seconds. If you have a tailwind.

 

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.
Any comments???

 

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?

 

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.

 

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

 

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

 

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

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.

 

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

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?

 

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

 

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?

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.

 

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.

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.

 

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.

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.