Transformer and rectifier math

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SceneryDriver

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It is. With a lot of smoothing you might get get closer to 1.42Vac

Shouldn't the average DC voltage be higher (assuming appropriate filtering) with three phase rectification?
The value of:

VDC = (VAC * 1.65)

makes sense intuitively, as there is inherently less ripple in rectified three phase, as opposed to rectified single phase AC. Is my intuition wrong here?



Thanks,

SceneryDriver
 

Besoeker3

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UK
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Retired Electrical Engineer
Can you determine the peak diode current? That's one very important parameter for diode rectifier design. Both at stead state, and also when the power is first applied and the caps. are completely discharged.

Although, I still don't know how we are getting from a 3 phase supply to a single phase output, so maybe you don't know what the complete circuit is either.
I have dealt with a lot of rectifiers, mostly at high/relative high currents. Up to 70kA but more usually in the 20-40kA range.

Typically, these were 12 or 24-pulse arrangements made up of two or four units operating in parallel. The DC usually had a choke (inductor) in series with the output of each. This made the current quite smooth. Peaks were not an issue for such applications.
 

SceneryDriver

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Scenery - how much is the load?

You can buy off the shelf 5 kW Vienna bridge 3 phase PFC rectifiers a lot cheaper than designing a straight rectifier with big caps.
About the size of 2 packs of playing cards, that should fit, eh.


Post your amperage needs and can probably give you a part number even. Lots of sources, Vicor, Murata, Kepco, Aveox, etc....

I had not heard of Vienna bridge rectifiers until I googled them. Very interesting. Active rectification crossed with SMPS tech. Very cool. Looking at needing 60A @ 70VDC peak. Much lower currents most of the time.


SceneryDriver
 

Besoeker3

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Location
UK
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Retired Electrical Engineer
Shouldn't the average DC voltage be higher (assuming appropriate filtering) with three phase rectification?
The value of:

VDC = (VAC * 1.65)

makes sense intuitively, as there is inherently less ripple in rectified three phase, as opposed to rectified single phase AC. Is my intuition wrong here?



Thanks,

SceneryDriver
For a 3-phase, full wave rectifier Vdc = 1.35Vac.
Others here have shown the mathematical derivation.
 

steve66

Senior Member
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Illinois
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Engineer
I had not heard of Vienna bridge rectifiers until I googled them. Very interesting. Active rectification crossed with SMPS tech. Very cool. Looking at needing 60A @ 70VDC peak. Much lower currents most of the time.


SceneryDriver

You may also want to consider a switched mode power supply. They are usually more complex, and normally include an inductor, but they are typically more efficient and they reduce the input current spikes.

With a rectifier circuit, input current only flows when the input voltage is higher than the output voltage across the capacitor. In the case of a simple half wave rectifier, the input current may only flow for a few milliseconds, but it must charge the capacitor enough to sustain the load current for the rest of the cycle. That means the peak input current may be several times the output current. As the output capacitor is increased in size, the output voltage doesn't droop as much, and the capacitor charging time actually becomes shorter, resulting in higher input current spikes.

Of course, using a bridge rectifier, or a three phase input, or almost any of the more complex circuits adds more diodes to conduct, and more conduction time. So these all lower the input current spikes. But if you don't calculate the spike peak current, there is a good chance you are going to kill the diodes or something else in the circuit.

Switched mode power supplies can also reduce the input current spikes.
 

GoldDigger

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I'm not trying to be snarky, but that is not a good response. I want my money back ;) That was basically my question: why not apply the RMS principle to DC like we do with AC. Nothing says RMS is for AC only.


As long as the ripple percentage is low compared to the DC component, the difference between average and RMS will be small, and may not justify the extra expense of a true RMS meter or the additional complication to the math.
 

paulengr

Senior Member
Agreed that current is torque, no question about it. As long as voltage is unlimited this is true. A DC drive in current regulation mode is a torque regulator until we hit maximum voktage. Similarly if we need a speed controller we regulate armature voltage (to a point). Physics says it's all current and this is great when we are trying to understand DC motors and drjves.

So what happens if we are voltage limited using the current? No way to tell because we are affecting voltage, not current. What if the bus voltage fluctuates faster than the drive response time? Torque varies. How much? Hard to say except if we apply Ohms Law.

I think we can agree torque is proportional to current minus counter EMF that we can hopefully ignore in practice.

https://www.motioncontroltips.com/torque-equation/

If I have a drive involved that adjusts current faster than supply voltage fluctuations below maximum torque, torque is approximately constant. But if the drive is not involved or too slow, torque will vary as described. If this was not a concern we would not control DC bus ripple at all.


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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190712-1415 EDT

paulengr:

You are making something simple look complicated. Don't bring in extraneous factors.

Again you are mixing counter EMF with current.

The basic simple factor is that current and torque are directly related when the magnetic field is constant in which the current is flowing. If you can not supply the current required for a given torque, then you can not produce that torque.

If a motor and its source voltage can cause enough current to flow with a given load, then there is some speed balance point where the needed current will flow to produced the needed torque at that speed balance point.

At that speed balance point the Vsource - Counter EMF divided by the armature resistance will equal the current required to produce the needed torque at that balance point.

.
 
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