Need to increase current of voltage booster

[quaternion]

ASA
I wanted to make a voltage booster in order to power up some projects that need more than 9V supply with only 9V battery.
I searched in local market for chips that make voltage boost but I didn't find any.So ,I decided to build it my self.I did make a fly-back sps but the output current was very low.I decided to make a boost sps in order to decrease loss of more components.I searched the web a little and found some threads among them i found this
12V To 24V DC-DC Converter Circuit
I start to simulate and notice unstable behavior and not easy analyzed.So I modified it in order to make it in a form that I can understand.
And at last ,I implemented the attached circuit.
The problem is that I can't get from it more than 400mWatt(20V,20mA using 1Kohm load).
I did tuned the loop gain varing the feedback resistors but it seems more or less open loop gain decrease the output power in addition to that i can't use small resistors in order not to dissipate more power nor i can use large resistors in 1M range in order not to have currents comparable with op amp biasing ones.

I want to get max 1Watt from a regulated output

I appreciate any help, thanks in advance.
-I hope that the topic is in place ,I wanted to put it in Hobby Circuits and Small Projects Problems forum -

 

[FvM]

The circuit seems to be intended to supply ampere range voltage supplied from a car battery, assuming a suitable inductor is used. The dimensioning doesn't fit a small 9V battery.

 

[Raza]

I do not think you will be able to use this battery. Even the Industrial type of Energizer (Alkaline) here is the usage graph:

 

[quaternion]

for FvM
what exactly in the attatched circuit that doesn,t fit the 9V battery??

for Raza
Thanks alot for those usage curves but i can't relate them well to my case, for example if i get 15V output can't i reach a current of 50mA?(say 1hour per day?)

 

[FvM]

what exactly in the attatched circuit that doesn,t fit the 9V battery??

The low inductance and respective high peak current.

 

[quaternion]

What about increasing the frequency by decreasing C3 or it will be comparable with parasitics? :|

 

[FvM]

LM358 is a rather poor switch mode driver due to it's slow slew rate of 0.5 V/us. The best way to overcome it's limitations (if you can't afford a real switch mode controller or faster OP like TL084) is to reduce the switching frequency and increase the inductance respectively. Simply design it for continuous conduction mode with moderate (30 to 50%) current ripple. To avoid excessive capacitances, transistor and diode shouldn't be oversized.

 

[Raza]

On your original question:

Need to increase current of voltage booster

You are asking to get 1W approximately for your experiments. So I=W/E² = 1/9x9=.0123 = 13mA (approx). And I do not know what will be your circuit equivalent resistance. Hence the battery volts will drop and instead of boosting the current circuit may not be able to cope up the demand (booster circuit losses not included). This was my concern to say that battery may not be able to support your demand.

 

[quaternion]

Can you make it more clear for me.

I said 1W i.e. 1Watt for 9Volt ---> I = P/V = 1/9 = 0.111 = 111mA(approx), what is this I=W/E² ??? is this used to calculate voltage battery drop at certain current drained , I think the circuit won't stop to work if input voltage dropped a little (say till 7Volt) ???????



Dear FvM
I think according to C3(1nF) & the -ve fb resistor the oscillating frequency won't exceed 10KHz of periodic time 100usec so isn't 0.5V/usec slew rate is enough?

 

[FvM]

so isn't 0.5V/usec slew rate is enough?

It's enough to make the circuit work somehow, but it's one of several points that causes bad performance of the boost converter design.

 

[quaternion]

Is a chip as uc3843 more suitable to make this regulator -i.e. more power saving and efficient and of course have the ability to operate as a boost sps- ?
Or power dissipated in current sense resistor is unavoidable?








Thanks alot FvM & Raza , at last I found a good solution in the local market it is MC34063.

 

[quaternion]

Raza you are right.

I made the circuit with MC34063 and the result is:
With a new battery (just get out of its plastic cover) is used , i adjusted the output to 19.3Volt without a load when loading with 670ohm resistor the voltage drops to 19.1Volt so now I get more than 0.5Watt(not bad for now)[I heard the sound of magnetic core oscillation as I used a torroid and wound it with nearly 140 turns of 0.4mm enameled wire],
using an older battery(which went old very fast during testing these circuits) the voltage drops to around 12Volt[no sound heard]

Now ,I concluded that concerning to the 1st circuit I actually needed a higher inductor-as FvM said- and a schottky diode such as 1n5817 (which i used with the MC IC)as BY399 fast diode has double Vf , & after making the 2nd circuit i concluded that it is not possible to get 1Watt from such battery for considerable time.

Now , what bothers me really is that i couldn't calculate this before doing all of that, Could anyone explain it -more clearly- numerically for me [I mean battery power calculations]