On 12/23/05, John Waters wrote: > Hi All, > > I'm try to use LM380 to drive a load but encountered some problems. The > circuit is a very standard one with a capacitor in the input and a capacitor > in the output to block the dc, I used a function generator to apply a > sinewave to the input, but found that the output would very easily get > clipped (unless the input signal is very very low in amplitude). the maximum > p-p amplitude of the output signal is 4 volts while my Vs is at 12V dc. I > tried all frequencies from 20Hz to 1KHz, the result is the same. Is there > any way I can get an output sinewave without clipping for higher than 4 volt > p-p, preferablly close to the supply voltage (i.e. 12V)? > > Thanks in advance! > > John > John, Look at the datasheet and check out the chart labeled "Device Dissipation vs Output Power -- 8 ohm Load". The lowest curve is 12V supply. The maximum output power appears to be 1.5W into 8 ohms. To calculate the voltage swing, remember that for sinusoids, Vp-p=Vrms*sqrt(2) for and power can be expressed as (Vrms^2)/R. So Vrms = sqrt(8*1.5) = 3.46V and Vp-p would be 3.46*sqrt(2)=4.89 V. The performace you see is about what the datasheet promises. If you need more help, describe what your circuit is trying to accomplish and what your load is. My first instinct would be to choose a different IC, but you might be able to bridge-tie two 390's across the load, which would double Vp-p that the load sees. Regards, Mark markrages@gmail -- You think that it is a secret, but it never has been one. - fortune cookie -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist