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 ...? Thanks, Bob

# Full wave rectified source for DC motor

Started by ●January 6, 2021

Reply by ●January 6, 20212021-01-06

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

Reply by ●January 6, 20212021-01-06

On 2021-01-06, Bob Engelhardt <BobEngelhardt@comcast.net> 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 ...?When the unfiltered rectified AC voltage is lower than the motor back-emf no current flows through the rectifier, therefore there is no electrical drag during that part of the AC cycle, so the motor is only slowed by magnetic ans mechanical losses during that time. Motors are like capacitors, so it's unsuprising that you effectively get more voltage than the RMS of the AC after rectification. Try addign a load parallel with the motor. and same load parallel with the rectifier input: see the difference. -- Jasen.

Reply by ●January 6, 20212021-01-06

Phil Allison wrote: ===============> Bob Engelhardt wrote:> > ** 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. >** Update: I did a test with a 12V, 5 pole motor with regulated DC and raw rectified AC. For the same average DC value, there was no change in motor speed. As Jason posted, the motor acts much like a filter electro - boosting the *average* DC value to near the AC peak value. ..... Phil

Reply by ●January 7, 20212021-01-07

On 1/6/2021 4:15 PM, Jasen Betts wrote:> When the unfiltered rectified AC voltage is lower than the motor back-emf no current > flows through the rectifier, therefore there is no electrical drag > during that part of the AC cycle, so the motor is only slowed by > magnetic ans mechanical losses during that time. > > Motors are like capacitors, so it's unsuprising that you effectively > get more voltage than the RMS of the AC after rectification. > > Try addign a load parallel with the motor. and same load parallel with > the rectifier input: see the difference. >Thanks Jasen, that makes sense. And just to confirm it, here's the current through the motor: https://imgur.com/vunNd8Q It is off for about 40% of the cycle. Which means that the motor is coasting during that time. And you could say that the motor is being driven by 60% duty cycle pulses.

Reply by ●January 7, 20212021-01-07

On 1/6/2021 2:53 PM, Phil Allison wrote:> Bob Engelhardt wrote:[snip]>> 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. [snip]> ** 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.We're actually saying the same thing: my 88% of RMS is .88 x .707 x peak which is 0.622 of peak (OK, it would be 90% of RMS to be 0.637). And "average value" is just what the DC component of the Fourier series is.

Reply by ●January 7, 20212021-01-07