I thought if you were to put a full bridge rectifier on a 120 home outlet and smoothed it with a capacitor you would get a DC output close to the peak voltage of about 170 V.

I used 6 HFA15TB60 diodes in my rectifier.

Originally Posted by brandons

I thought if you were to put a full bridge rectifier on a 120 home outlet and smoothed it with a capacitor you would get a DC output close to the peak voltage of about 170 V.

I used 6 HFA15TB60 diodes in my rectifier.

That 170 VDC voltage is correct if the capacitor is large enough at the required current level.

The formula is CV = IT, for a constant current.

So for single phase, T = 0.0083 seconds.

Assume 1000 uF capacitor and one Amp load.

V = I*T/C
V = 1*0.0083/1000E-6
V = 8.3 Volt change between cycles.

But for 3 phase DC, the phases overlap providing fairly clean DC with no capacitor at all.

Originally Posted by brandons

I would have thought that my open circuit voltage already takes the diode voltage into account already. My open circuit Line to Line voltage was about 23 VAC, so dividing by 2 the peak value was about 11.5 V. Taking the diode voltage drop from this value we get about 10V which lines up reasonably close to my open circuit voltage.

Voltage drop over a diode is not a constant. It varies as a function of current. For open circuit voltage it may only drop a tenth or so of a Volt. Under a 3A load it may drop a full Volt.

In any case line voltage should be open circuit voltage minus voltage losses. Those losses would include diode loss plus the vector magnitude of copper loss and iron loss (resistance and reactance). For voltage drop on the diode at a given current you have to consult a current versus Vf chart. You should be able to download a data sheet for the rectifier you are using which will contain that chart. Iron and copper losses are usually pretty close so adding fifty percent to winding resistance times current will probably cover the voltage drop for both.

Once you calculate that information you'll be able to arrive at a maximal allowed resistance for your windings. If you can't get Kv low enough without going over that maximal resistance you'll have to gear up the generator for higher Kv or use a bigger generator with heavier windings.

Originally Posted by vollrathd

That 170 VDC voltage is correct if the capacitor is large enough at the required current level.

The formula is CV = IT, for a constant current.

So for single phase, T = 0.0083 seconds.

Assume 1000 uF capacitor and one Amp load.

V = I*T/C
V = 1*0.0083/1000E-6
V = 8.3 Volt change between cycles.

But for 3 phase DC, the phases overlap providing fairly clean DC with no capacitor at all.

Ok that makes sense, I see what you're getting at now.

One question that I forgot to ask was that the waveform of my motor line-to-line voltage was a pretty clean sine wave, but when I hooked up a 7.7 ohm solenoid as a test load my voltage looked more trapezoidal. I was wondering what would have caused this change?

Thanks,
Brandon

The brushless motors we use normally create a trapezoidal voltage wave. With no load the inductance of the motor becomes more significant since it's like a voltage divider with one side of very high impedance inserted by the scope. The divider makes inductance more prominent which smooths out the trapezoidal wave to look more like a sine. Higher end scope leads have adjustable impedance to compensate for test lead insertion, but it won't eliminate that completely. You need to normally check output under operational load.

Originally Posted by CraigHB

The brushless motors we use normally create a trapezoidal voltage wave. With no load the inductance of the motor becomes more significant since it's like a voltage divider with one side of very high impedance inserted by the scope. The divider makes inductance more prominent which smooths out the trapezoidal wave to look more like a sine. Higher end scope leads have adjustable impedance to compensate for test lead insertion, but it won't eliminate that completely. You need to normally check output under operational load.

Thats not how scopes work. Their imput impedance is orders of magnitude greater than the motor windings. And their 10/1 probes makes the scope impedance even higher.

(I have four scopes in my workshop, two good for 100 mhz)

Originally Posted by CraigHB

. . . . With no load the inductance of the motor becomes more significant since it's like a voltage divider . . . .

With no load the inductance becomes irrelevant since and inductor has no effect until current flows through it.

Quick update on my project, I was able to get my hands on some electrical steel and one of my coworkers cut out some laminations for me with enough to make a stator that was twice the thickness, which should decrease my KV by about 30-40% assuming a rotor that was twice as thick as well, correct? I haven't been able to test yet, need to make a new rotor.

I think I need to make a larger motor though, not enough power at 1000 RPM, might try a 24 arm 28 pole motor.

I got low vf schottky diodes and i still see a voltage drop that i can't account for when my load is connected, iron losses maybe?

So my description of the waveform earlier was not entirely accurate, when I hooked up a light bulb as a resistive test load my waveform was trapezoidal. The waveform looked more like a stepped square wave than trapezoidal for the solenoid, and I still can't figure out what causes this. Maybe some RLC function?

Thank you everyone, any further advice would be greatly appreciated,
-brandon

Originally Posted by brandons

Quick update on my project, I was able to get my hands on some electrical steel and one of my coworkers cut out some laminations for me with enough to make a stator that was twice the thickness, which should decrease my KV by about 30-40% assuming a rotor that was twice as thick as well, correct? I haven't been able to test yet, need to make a new rotor.

I think I need to make a larger motor though, not enough power at 1000 RPM, might try a 24 arm 28 pole motor.

I got low vf schottky diodes and i still see a voltage drop that i can't account for when my load is connected, iron losses maybe?

So my description of the waveform earlier was not entirely accurate, when I hooked up a light bulb as a resistive test load my waveform was trapezoidal. The waveform looked more like a stepped square wave than trapezoidal for the solenoid, and I still can't figure out what causes this. Maybe some RLC function?

Thank you everyone, any further advice would be greatly appreciated,
-brandon

what you mean thickness? thicker laminations on the stator shouldnt effect kv. thicker rotor back iron would hold more magnetic field and bring the kv down
i have a bunch of 47 diameter stators i want to get rid of if youre interested: 52mm long. .2mm laminations of kawasaki electrical steel.

careful with your stator there and it's got a lot of exposed steel that can be shorted through easily

What Hummina said! On those turns that are riding on the sharp corners of a bare metal stator at the start and ends of the stator arms can easily be shorted out on the stator. Adding a strip of insulation to the stator arms will usually keep it from happening. That green coating that is seen on most stators is a baked on insulation that is not easy to do yourself.

I have used the Tyvek paper from CD-ROM/DVD sleeves for insulation, also bought some sheets of insulating paper on eBay that let me cut long narrow strips. I make them wide enough that the fold over the corners on the arms as the turns are applied.

The images show a bare stator with good coating in good condition and a motor that has a strip of insulating paper added.

Jack

Jack the date of Brandon's post guys, that was also his last post on RCG.
Last Activity: Jul 10, 2018 08:26 PM

Vriendelijke groeten, en wees voorzichtig, Ron