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Anonymous Poster

lm2907

03/17/2007 10:32 AM

Hi,

i dont quite understand how to calculate the max aloowable ripple voltage or is one supposed to just choose a reasonable value?

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#1

Re: lm2907

03/17/2007 11:01 PM

you feed from a 120 cycle full wave rectified source. With zero capacitance your output would ramp from zero to your set voltage and then flat at that voltage until the input fell below the set point, and then to zero.

With a capacitor before the regulator you make it big enough so that the load never sees the voltage drop, = zero ripple.

http://www.pilom.com/BicycleElectronics/DynamoCircuits.htm

http://www.google.com/search?hl=en&q=lm2907+%2Bapplication&btnG=Search

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Anonymous Poster
#2

Re: lm2907

03/18/2007 1:09 AM

You can calculate the amount of p-p ripple for a given capacitor size using the equation: C = I dT/ dV , where C is in Farads, I is the current your load is drawing in amps, dV is the maximum p-p ripple you want to see. Use 6.3 msec for dT if your are full-wave rectifying.

This equation is really assuming a constant current draw out of the capacitor during the time that the rectifiers are not conducting, and that's a pretty close assumption if the load on the output of the LM2907 regulator is constant.

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#3

Re: lm2907

03/19/2007 8:32 AM

The datasheet shows that the maximum allowable input voltage is 0 - 28 VDC for LM2907 and +/- 28 VDC for LM2907-8 or LM2917-8.

Minimum input is +/- 10 mV, typical is +/- 25 mV, and max is +/- 40 mV, all at 1 kHz, which means that it will operate at +/- 40 mV input but might operate @ as low as 10 mV input.

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