From: pallison49@gmail.com   
      
    Bob Engelhardt wrote:   
   >   
   > I need a sanity check - I think I understand, but there's doubt.   
   >   
   > If I rectify AC with a bridge and use the full-wave output to power a DC   
   > motor, it's not the same as using "pure" DC. There is an AC component   
   > to the full wave. (The motor is a brushed PM if it matters.)   
   >   
   > If I take the Fourier series of the 1/2-sinusoid and consider each   
   > component separately & then superimpose them, I should get the behavior   
   > of the motor on the full-wave source. The Fourier series consists of a   
   > DC component and the even harmonics of 120Hz. The DC will simply drive   
   > the motor as would a battery. The AC, however, will not have a net   
   > affect on motor's output: for its positive 1/2 cycle it will contribute   
   > to the output and on the negative 1/2 it will oppose it. So the   
   > superimposed result is that the useful motor output is due to the DC   
   > component only and the AC components only produce a modulation (240,   
   > 480, ... Hz "buzz") on the output.   
   >   
   > I long ago lost any ability to do the Fourier calculation, but somewhere   
   > on the web (source lost), I found that the DC component (a0) is 88% of   
   > the RMS AC input to the bridge. (If it's not too much trouble, could   
   > someone confirm this?)   
   >   
   > Now here's the problem: reality contradicts theory (I hate when that   
   > happens!). The theory is that if I apply 20v AC, for example, to a   
   > bridge & use the output to drive a DC motor, that motor will run at 88%   
   > of the speed which it would if it was driven a regulated DC source of   
   > 20v. (DC motor speed is linearly proportional to voltage.)   
   >   
   > In a test, it doesn't - it actually runs faster on the rectified AC than   
   > on DC!!! That's impossible! What's wrong - my understanding of the   
   > theory, or my test? Or both? Or ...?   
   >   
      
   ** DC motor speed *approximately* follows the average value of the input   
   voltage, not the RMS.   
      
   For 20VAC, this is 0.637 times the peak or 18.0V minus diode losses, so about   
   16.5V   
      
   I suspect your test is flawed.   
      
      
   ..... Phil   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)