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Question about multi-tap transformers

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hkBattousai

Member
**broken link removed**

I have a multi tap transformer like in the picture above. I want to learn what happens when I don't use a winding between two taps, and the power distribution when I use multiple windings.

Question #1:
Suppose that I'm using the windings B-C and C-D. Is there any power wasted on the winding A-B when;
a) the transformer is ideal,
b) the transformer is realistic?

Question #2:
Sin: Power drained from the AC source by the primary windings.
Vxy, Ixy: Respectively the voltage and current between windings X and Y.
Is there a relationship like
Sin = Vab.Iab + Vbc.Ibc + Vcd.Icd
a) the transformer is ideal,
b) the transformer is realistic?

Question #3:
When the transformer is ideal, is there any power drained from the AC source when there is no load connected on the secondary side? If no, what is the explanation?

Question #4:
Do the wave forms on A-B and B-C corrupt if I drain only the positive alternance of the sine wave on the windings C-D?
 

Diver300

Well-Known Member
Most Helpful Member
**broken link removed**

I have a multi tap transformer like in the picture above. I want to learn what happens when I don't use a winding between two taps, and the power distribution when I use multiple windings.

Question #1:
Suppose that I'm using the windings B-C and C-D. Is there any power wasted on the winding A-B when;
a) the transformer is ideal,
b) the transformer is realistic?

With an ideal transformer there is no power lost in winding A-B
In a real transformer the extra losses from unused windings is minimal, and only comes from second-order effects.

The only current flowing in an unused winding is due to the capacitance of the winding, so the current is tiny. The power loss due to that would be so small that it would be very difficult to measure.

Unused windings make a transformer larger than it would otherwise be, so the leakage inductance on the other windings will be larger.

Question #2:
Sin: Power drained from the AC source by the primary windings.
Vxy, Ixy: Respectively the voltage and current between windings X and Y.
Is there a relationship like
Sin = Vab.Iab + Vbc.Ibc + Vcd.Icd
a) the transformer is ideal,
b) the transformer is realistic?

In an ideal transformer there are no losses. However, there will still be power absorbed magnetising the core, that is given out later.

If there is no load, the input inductance means that there will be input current, so the instantaneous power is not zero, although the average power over a cycle will be zero.

So for instantaneous power the equation is incorrect because the magnetising power is not included.

If there are loads, the loads might not be resistive. If the loads are not resistive, the average power is the voltage * current * power factor.

So for average power the equation is incorrect because the power factors are not included.

For a real transformer, the power losses should be added to the equations.



Question #3:
When the transformer is ideal, is there any power drained from the AC source when there is no load connected on the secondary side? If no, what is the explanation?

There is no average power taken by an ideal transformer, but there is current, and instantaneous power, because of the inductance of the primary winding.


Question #4:
Do the wave forms on A-B and B-C corrupt if I drain only the positive alternance of the sine wave on the windings C-D?

In a real transformer, especially smaller ones, the input winding has significant resistance, so when current is taken, the effective input voltage drops, and that will be shown in the unloaded windings.
 

hkBattousai

Member
Actually, what I'm planning to do is a power supply. There will be a bridge rectifier after the transformer. The tap-A will always be connected to one of the input ports of the bridge rectifier, and the other taps will be connected to it over three different relays. A control circuit will control which tap (B, C or D) to be connected to the bridge rectifier by switching these relays according to the desired output voltage level. I'm doing this to minimize the voltage drop on the linear voltage regulator; so that the overall efficiency will increase.

However, my lack of knowledge about transformers led me to a confusion. I couldn't make sure if this design would be efficient or not. What I understood from your message is that I can do this design without too much worries. Is that so?
 

Diver300

Well-Known Member
Most Helpful Member
Actually, what I'm planning to do is a power supply. There will be a bridge rectifier after the transformer. The tap-A will always be connected to one of the input ports of the bridge rectifier, and the other taps will be connected to it over three different relays. A control circuit will control which tap (B, C or D) to be connected to the bridge rectifier by switching these relays according to the desired output voltage level. I'm doing this to minimize the voltage drop on the linear voltage regulator; so that the overall efficiency will increase.

However, my lack of knowledge about transformers led me to a confusion. I couldn't make sure if this design would be efficient or not. What I understood from your message is that I can do this design without too much worries. Is that so?

That would be a normal way to do it, and it would be far more efficient at low voltages than using a linear regulator supplied from a high voltage.

The transformer that you are showing does not look like a standard transformer. The 24VA rating might not be for any one winding, but for all the windings put together, so you have to make sure that you do not overload any one winding.
 

hkBattousai

Member
The transformer that you are showing does not look like a standard transformer. The 24VA rating might not be for any one winding, but for all the windings put together, so you have to make sure that you do not overload any one winding.

That image was only for explaining my problem. It is not the transformer I am going to use. It will be at least 100VA and will have 8 or 16 taps to control with a power shift register.
 

carbonzit

Active Member
One concern about your power control scheme: while it seems to make good sense to do things the way you're proposing (by changing transformer taps rather than adjusting a linear regulator), have you taken into account the possible surges and spikes you're likely to induce when you change from one tap to another? It's going to be as if you're turning your power supply completely off, then on again, with a possible inrush current as the supply works to charge the filter capacitors. (I suppose this would be minimized by a very fast changeover cycle, but it's still a concern.)
 

hkBattousai

Member
One concern about your power control scheme: while it seems to make good sense to do things the way you're proposing (by changing transformer taps rather than adjusting a linear regulator), have you taken into account the possible surges and spikes you're likely to induce when you change from one tap to another? It's going to be as if you're turning your power supply completely off, then on again, with a possible inrush current as the supply works to charge the filter capacitors. (I suppose this would be minimized by a very fast changeover cycle, but it's still a concern.)

Hmm, a good point you have there. Is there anything you could suggest me to minimize this risk?
 

Reloadron

Well-Known Member
Most Helpful Member
If you want a variable voltage, rather than screw around with taps and bridge(s) why not consider what we call a variac or autotransformer which is one of these. You get one for the VA and voltage range of your choice like 0 to 250 volts out and run with that into a full wave bridge. I use one for assorted DC applications. Works fine & last a long time. :)

Ron
 

carbonzit

Active Member
I think the O.P. wants the output voltage level to be somewhat automagically selected by the circuit. They said:

There will be a bridge rectifier after the transformer. The tap-A will always be connected to one of the input ports of the bridge rectifier, and the other taps will be connected to it over three different relays. A control circuit will control which tap (B, C or D) to be connected to the bridge rectifier by switching these relays according to the desired output voltage level.

So I don't think a Variac would work here. (Well, unless they rigged up a stepper or servo to turn the knob, based on the voltage selected by the circuit ...)
 

crutschow

Well-Known Member
Most Helpful Member
......... In an ideal transformer there are no losses. However, there will still be power absorbed magnetising the core, that is given out later.....
I would argue that an ideal transformer would have infinite inductance, thus the magnetizing current would be zero.
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
Just buy one. **broken link removed** uses this design. It works really well.

**broken link removed**
 
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Reloadron

Well-Known Member
Most Helpful Member
I think the O.P. wants the output voltage level to be somewhat automagically selected by the circuit. They said:

There will be a bridge rectifier after the transformer. The tap-A will always be connected to one of the input ports of the bridge rectifier, and the other taps will be connected to it over three different relays. A control circuit will control which tap (B, C or D) to be connected to the bridge rectifier by switching these relays according to the desired output voltage level.

So I don't think a Variac would work here. (Well, unless they rigged up a stepper or servo to turn the knob, based on the voltage selected by the circuit ...)

Then I would maybe be inclined to just use relays. I would suggest AC SSRs but the lowest AC voltage versions I know of are like maybe 12 volts. If using relays maybe place snubbers (RC Type) in there to suppress the inductive kicks when switching taps.

This may go much better if there was an actual data sheet for an actual transformer to be used so voltages would be known. That data would likely open more possibilities.

While a motor driven variac is very doable it is not quite inexpensive.

<EDIT> After seeing the post, I vote for the KISS solution. :) </EDIT>

Ron
 
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