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RE: TL497A How Can I get 18 volts and 1.5 Amps current from a 9 volt Lithium ion battery Using the TL497A chip

chudy ihedinmah
Prodigy 20 points
Other Parts Discussed in Thread: TL497A

Hello all,

My design requires a minimum of 1.5 A current, I want to be able to boost my 9V lithium ion battery to about 18 Volts output voltage and get a minimum of 1.5 A output current from the TL497A chip. I have tried using the equations on page 5 of the TL497A datasheet all to no avail. Can anyone help please

Some of the problem I am having is that I don't have a value for the ripple voltage I have tried using the following equations:

For a full-wave rectifier:

For a half-wave rectification:


 they both require a capacitance value. There are 2 capacitors on the EXTENDED POWER CONFIGURATION on page 5 of the datasheet. CT and Co  I am not sure which should be in the Vpp equation above.
On bullet number 4:

R1 = (Vo -1.2V)K; is the Vo the output voltage that I want,in this case 18V?

on bullet number 5:

RCL = 0.5V/I pk; is I pk the output current that I want in this case 1.5A ?

On bullet number 1 and 5:

What is Io, is it a fixed value or is it the output current that I want, in this case the 1.5A?


The connection between pin 14 and 10  and also between the base and the emitter of the transistor shows a wire with a resistor, is that a resistor or the resistance of the wire


Finally what type of Transistor and diode can I use? will any NPN BJT work for it ?


I need help in calculating the values of all the resistors, capacitors inductors


Can anyone please help  based on my requirements
thanks in anticipation
  • 0
    TI__Guru*** 147906 points

    Chudy,

    The ripple formulas are based on analog rectified AC and 'C' is the output capacitor that is recharged every cycle, 1/f , and discharges over the cycle, rate of decrease proportional to Iout. 

    The switch and diode in TL497 are 500mA each. You need to boost 9V to 18V. So the inductor will charge for over 50% of the time and supply output current for less than 50%. Therefore to get an average 1.5A output current will require the inductor, switch, and diode to pass over 3 amperes of current. TL497 is not the best solution for this requirement.

    I suggest you try Webench tool on ti.com main page. It will give you complete schematic and parts list. It is a very cool tool.