Full wave rectified source for DC motor

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

 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 

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.

 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

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.

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.

The flaw in my analysis was using superposition when it didn't apply. 
Or using it without considering the back emf as one of the elements to 
be superimposed.  I don't know how I would do that, but I'm sure that 
it's possible.