DM860T input voltage

120 x 1.414 = 169.7 yes BUT that's peak to peak. When you rectifiy that you effectively divide it in half but then you have twice as many half cycles per unit time, so it comes out to a little less than the initial ac value.
That's why most dc motors are rated at 90 volts to be run from a 120 volt ac line.
 
markba633csi said:
120 x 1.414 = 169.7 yes BUT that's peak to peak. When you rectifiy that you effectively divide it in half but then you have twice as many half cycles per unit time, so it comes out to a little less than the initial ac value.
That's why most dc motors are rated at 90 volts to be run from a 120 volt ac line.
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For 1/2 wave rectification that would be true since you are only using 1/2 of the AC cycle as the output. In motor circuits, this is only used for DC motor controllers.
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Full wave rectification would give you ~ 169 VDC output max, but more like ~140VDC useful output.

1518543677772.png
 
Hmm It's been a while since I studied this stuff- something still looks odd to me though- I shall return
Mark
ps dc motor controllers (of the twin diode, twin scr type) are full wave not half- I'm certain
 
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Ah I see where I goofed up, it's actually 338-340v peak to peak. So (according to wiki) dividing by pi gives about 108v average dc, without filtering.
So my original point is still valid, that the effective dc you end up with is just a little less than the average ac value you started with (120)
meaning the printed manual is the correct one
Mark
 
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I realize this post is somewhat old but I too am going through the same thing. Did you ever get this worked out? It's looking like I can use AC from the wall output but I didn't want to risk it. How did you end up making your input voltage connection?
 
ccomito1223 said:
I realize this post is somewhat old but I too am going through the same thing. Did you ever get this worked out? It's looking like I can use AC from the wall output but I didn't want to risk it. How did you end up making your input voltage connection?
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I ended up sticking with the 48v DC power supply. I would have preferred to go with AC power, but I didn't want to risk blowing the drive.
 
shooter123456 said:
I ended up sticking with the 48v DC power supply. I would have preferred to go with AC power, but I didn't want to risk blowing the drive.
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I was worried you might say that :-) I also have a 48V DC supply but would prefer to go with AC and do away with the supply all together. Thanks for the input though. I did email the manufacturer. I'm not sure if I'll get a response but if I do I'll post back here for future reference.
 
AC voltages are usually expressed as RMS which is the voltage that would give you the same power as a DC voltage of that value. Peak voltage is 1.414 times that and peak to peak voltage is twice that or 2.818.

When AC voltage is rectified, either with a half wave rec6tifier or a full wave rectifier, the peak DC voltage is 1.414 times the AC RMS voltage, providing there is no load. If there is a filter capacitor in the circuit, the capacitor will charge to the peak voltage, again with no load. As a load is applied, the output voltage will drop with increasing load. In real circuits, the DC output voltage is usually somewhere between the RMS and the peak voltage.

As for the OP's original question, it appears the 80 volts is a safe maximum voltage in either case. If I was concerned about doing damage, I would limit the voltage to that value until I had a definitive answer.
 
So to say the same thing in a way I understand it is, AC is measured in Vrms. When AC is rectified and filtered, the supply will charge up to the peak value (Peak=RMS*1.414). So if you take your 80V AC X 1.414 you get 113V DC or 113V DC X .707 = 80V AC. So,,, no you can’t use 120V AC rms, that will be 170V DC after rectifying and filtering. The short answer is this driver needs an 80V AC transformer, or a 110V DC power supply.
Thanks @RJSakowski, @JimDawson and @shooter123456 ..
 
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