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Rectifier Electrical / Electronic Question

Vanman

Subscriber
Last Subscription Date
07/10/2019
Suppose I have a 60 volt, 50 ampere, single phase AC source, and I want to supply a 30 volt, 100 ampere DC load.

The source and the load need not be isolated.

Center tapped transformer secondary windings in conjunction with two diodes are frequently employed so as to achieve full wave rectification without the need for a four diode full wave rectifier.

I wish to use a single winding transformer with a center tapped 60 volt winding, and two diodes, connected across the source.

The question is: What is the required ampacity of the transformer winding?

I believe it would be 50 amperes. Is this correct?
 

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Thaumaturge

In Memory Of
Age
68
Last Subscription Date
07/12/2019
Roughly... rule of thumb is 20% overhead on desired rating. Critical design spec for transformers however isn't amps, but VA, or basically continuous heat load on windings without release of magic smoke.

For the kind of load you are looking for though, I'd go for a switching supply.
Doc

If you could live with 220VAC input TWO of these in paralell would cover it for roughly $200 https://m.banggood.com/AC-200V-250V-To-DC-48V-50A-2400W-Power-Supply-For-ZVS-High-Frequency-Induction-Heating-p-1279188.html?gmcCountry=US&currency=USD&createTmp=1&cur_warehouse=CN&utm_source=googleshopping&utm_medium=cpc_elc&utm_content=frank&utm_campaign=usc-nobrand-elc2-us&gclid=EAIaIQobChMIq9yC7sPm3gIVxBx9Ch1glQE3EAQYAyABEgJShvD_BwE
 
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Bob Willman

Registered
Is your input voltage RMS AC or a peak voltage? What your diagram shows is not a standard transformer which would have 2 windings, a primary and a secondary. If you are going to draw 100 A from the winding, the wires in the winding must be capable of 100 A. The diodes will also need to be rated for 100 A. Without some capacitors after the diodes the output will be pulsed DC not a true DC voltage. I'm not sure your circuit will do what you want it to.

Bob
WB8NQW
 

Vanman

Subscriber
Last Subscription Date
07/10/2019
The AC voltage of the supply is RMS. The circuit is just going to be charging a big battery, sometimes with a load on it. So it doesn't need to be clean or smooth or even regulated DC.
 

armandh

Sponsor
Last Subscription Date
09/02/2010
do it in Watts

100x30=3000 watts

3000/117 about 26 amps
3000/60 is yes 50 amps

RMS volts not peak
 

DMeed

Subscriber
Last Subscription Date
07/12/2019
Think about when the diodes are conducting. Each diode is conducting for 1/2 the AC waveform and off for the other 1/2 the AC waveform. So each diode is acting alone. For 100 amps output, you will need 100 amps through each diode and from the transformer windings as well. You are getting the 100A*30V or 3000 watts alternately from each half of the transformer. So the transformer needs to be rated for 6000 watts.

Are the 60V CT transformer windings fed from a separate 60v AC source or are they are the sole source of current for feeding the diodes. It might make a difference if the current through the diode can come from the 60V AC source separately from the transformer winding (and the windings are just splitting the centre. Might work with a 50 amp winding in that case (?someone else will have to chime in on that one?) There was a thread showing how to feed a single phase split 120/120v circuit from a 240v Generator with a transformer rated 1/2 the generator wattage. In any case the diodes will have to be 100 amp anyway.

Thinking further - with diodes in there steering the current each 1/2 cycle I don't see how it would balance the current over both windings.
 
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G.M.Johnson

Subscriber
Last Subscription Date
11/05/8018
If it doesn't have to be exactly 30 volts a 36 volt fork truck battery charger will do the trick. You can find them in 208/240 single phase input versions. If you are stuck with exactly 60v input then scratch that idea.
 

armandh

Sponsor
Last Subscription Date
09/02/2010
Think about when the diodes are conducting. Each diode is conducting for 1/2 the AC waveform and off for the other 1/2 the AC waveform. So each diode is acting alone. For 100 amps output, you will need 100 amps through each diode and from the transformer windings as well. You are getting the 100A*30V or 3000 watts alternately from each half of the transformer. So the transformer needs to be rated for 6000 watts..
I'm not sure that is correct

if the 60 Volt winding is producing 3 KW, [50 A]
each half can produce 1.5 KW. [50A at half the voltage]
summed together after the diodes still 3 KW [half the voltage at 100 A]

the half wave, the diodes are alternately not conducting, does not reduce the total power transferred [IMHO]


the kind of counter EMF kick one can get here it might not be a bad idea to put a MOVs or neon bulbs across the diodes
 
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DMeed

Subscriber
Last Subscription Date
07/12/2019
I'm not sure that is correct

if the 60 Volt winding is producing 3 KW, [50 A]
each half can produce 1.5 KW. [50A at half the voltage]
summed together after the diodes still 3 KW [half the voltage at 100 A]

the half wave, the diodes are alternately not conducting, does not reduce the total power transferred [IMHO]
In an AC system the top of the coil is alternately +30v in relation to the centre tap and -30v in relation to the centre tap. (NO I'm not worrying about peak voltage, just the rms).

So on the +ve half of the wave form I'm looking at 100 amps leaving the top of the coil, travelling through the diode, out through the load and back to the centre tap of the coil. Now where does that 100 amps go? It is +ve in relation to the bottom of the coil, so it wants to complete the circuit by travelling through the lower half of the coil to the -ve side of the AC. I just can't see how it is going to split and go 1/2 toward the -ve side of the coil and 1/2 toward the +ve side. Willing to be wrong, but not seeing it yet. :shrug:
 

Vanman

Subscriber
Last Subscription Date
07/10/2019
Thank you all for the replies.

This hypothetical transformer is just two 30 volt windings in series. The supply is 60 volts AC from a generator.

In thinking about it furfher, it appears that the transformer still functions as an autotransformer, just as it would if it were supplying 30 volts AC @100 amps. In that scenario, without the diodes, each 30 volt winding carries 50 amps. One serves as the primary, the other as the secondary, so the transformer KVA required would be 1-1/2, ie 30 volts x 50 amps.

The same is true with the diodes back in the circuit, except that the windings switch roles every half cycle. The winding supplying the conducting diode is the secondary, and adds 50 amps to the 50 amps from the supply. The other winding is the primary, taking 50 amps from the supply. Next half cycle the windings reverse roles. The 100 amps on the common lead splits, half going in one winding, the other half to the other winding.

If there were a capacitor and NO load on the DC side, it is true that the DC voltage would indeed climb to ~42 volts. But any load would drag this down. With no capacitor and just a resistive load, and the rms DC voltage would still be 30. Charging a battery would fall somewhere between the two. A DC reactor in series with the battery and rectifier would smooth out the current, making less heating in the transformer and generator.

Diodes themselves are a peculiar thing as their forward voltage drop is practically constant, regardless of current. As such their maximum peak current is many times greater than their continuous current. The average current is what must not exceed the continuous current rating. So a 50 amp diode would serve perfectly well here. Half the time it passes 100 amperes, but the other half is zero. The average is thus 50 amperes.
 

armandh

Sponsor
Last Subscription Date
09/02/2010
Willing to be wrong, but not seeing it yet.
"once more into the breach" with pictures

picture it as two separate supplies each supplying 1.5KW 30VDC +

each feeding a load. you are still transferring 3KW of power it does not change

it is the same total area under the curve, see attached
 

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pegasuspinto

Registered
3kva is the heat rating of the transformer. When making 3kva of output, it can dissipate the resulting heat without failing. There should be a separate amp rating on the transformer. If you exceed that rating, you risk just blowing the wire apart like a fuse. Your drawing up to 100A from each winding, even if only for a split second.

What the heck battery are you dumping 3kw of charge into? Surplus submarine batteries? That would be pretty severe for any single string of automotive size battery.
 

Vanman

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Last Subscription Date
07/10/2019
Haha, no, not car batteries. :) A submarine battery would be sweet, but too big and heavy for this application. :D The battery will be an electric forklift battery. Way bigger than a car battery, but not even as big as a single submarine battery cell. :crazy:

Here are a couple of diagrams depicting current flow during each half cycle of the supply, of course ignoring peak vs RMS etc etc.

In the upper drawing, the upper winding is the primary, absorbing 1-1/2 kva, and the lower winding is the secondary, delivering the 1-1/2 kva.

In the lower drawing, the lower winding is the primary absorbing 1-1/2 kva, and the upper winding is the secondary, delivering the 1-1/2 kva.
 

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Vanman

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Last Subscription Date
07/10/2019
30v*100A is 3000 VA. 60v*50A is 3000VA. The Supply and load is 3kva.
Yes, but since an autotransformer is used instead of an isolating transformer, a portion of the power is delivered directly to the load, with no tranformation required.

In this case, since the voltage ratio is 2 :1, half the power is transformed, and the other half is delivered directly from the supply (after dropping 1/2 of it's voltage in the primary).

In the more typical example below, a 120 volt, 50 amp load, 6 kva, is supplied from a 240 volt, 25 amp source via a 3 kva autotransformer. The upper winding is the primary and is in series with the line and the load, dropping 120 volts at 25 amps. The secondary is in paralell with the load, and adds 25 amps at 120 volts to the 25 amps from the supply.
 

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Rich Mc

Registered
If the 60 volt winding was two 30 volt windings in series and the windings were changed and paralleled then one could then deliver 100 amps DC thru a full wave bridge rectifier, else I would agree with Dmeek that a 6Kva transformer is needed.
 

Vanman

Subscriber
Last Subscription Date
07/10/2019
A full wave bridge has the disadvantage of dropping twice as much voltage, about 4 volts instead of 2. :brows: You'll find most things designed to deliver low voltage DC use two diodes with a center tapped transformer secondary.

The 60 volt supply is just a 120 volt, two wire generator running at ~1/2 speed. So the fact that the circuit halves the voltage makes it perfect for me.

Because of this, the frequency will be 30 cycles. Therefore I need a transformer with twice as much iron, ie designed for twice as much voltage. A 60 cycle transformer designed with two 60 volt windings (that I will run on 30 volts, 30 cycles), each capable of 50 amps, would be perfect. If it existed lol. Will likely end up winding my own. :shrug:
 
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