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enter image description here

The image above shows how a typical AC to DC adapter works, high voltage (230V) at low current is fed into the primary coil and then gets stepped down to a lower voltage with higher current, afterwards it gets rectified to DC and finally a capacitor turns this DC to pure DC.

The power from the primary coil has to be the same as the power at the secondary coil (assuming no power loss). Let's for instance use an example, a phone charger, most phone chargers normally output 5V@1A DC which means 5W. Since the voltage from the wall is 230V this means that for there to be 5W there has to be 5W/230V = 0.0217A which means that the resistance at the primary coil is 230V/0.0217A = 10599\$\Omega\$

This is the part that I don't get, how can an inductor (coil), something with almost no resistance, have so much resistance. At first I thought that they would put a resistor before the coil but then I figured out that all the power would be used up by the resistor and wasted as heat instead of reaching the coil.

So what is really going on? Do they make these coils with materials that have really high resistance or am I missing something? Probably the latter is true.

Any help is appreciated.

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    \$\begingroup\$ Welcome to EE.SE. It is NOT being used as an inductor. That is where you are being miss-led. \$\endgroup\$ – Sparky256 Sep 1 '18 at 3:06
  • \$\begingroup\$ That's how ac to dc converters used to work, but now, especially things like phone chargers are generally switch mode supplies. \$\endgroup\$ – Colin Sep 1 '18 at 18:10
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The following approach may also help a little

The ohmic resistance of an ideal coil is 0 ohm in DC, but in AC the coil resistance is called inductive reactance and its amplitude can be calculate by the following formula:

\$X_L=2 \pi f L\$

Where:

  • \$X_L\$ is the Inductive Reactance of the coil (in your case the primary of the transformer) in Ohms
  • \$\pi\$ =3.14
  • f= frequency of the AC signal (probably 50 Hz in you case)
  • L=inductance of the coil measured in Henries

An inductor or coil is a device that, when subjected to an increasing electrical current flow, generates a back voltage that opposes this current. Inductance quantifies how much energy an inductor can store

Please note that heat losses in a non ideal transformer are made by the wire resistance and by parasitic currents the transformer core and not by XL.

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Semi-ideal transformers assumed.

So what is really going on? Do they make these coils with materials that have really high resistance or am I missing something?

You are over-complicating things by thinking about a transformer plus bridge and capacitor. Think about just the transformer with nothing connected to the secondary. Then, after thinking about it you should realize that if the secondary is not connected to anything, it can be removed. This leaves you with the primary connected to your AC mains circuit and that primary is now just an inductor.

So, if the inductor is (say) 10 henries in value, it will have an impedance at 50 Hz of \$2\pi f L\$ = 3142 ohms and will take a current of 73.2 mA when the applied AC voltage is 230 volts.

Additionally, because this current is created by the inductance of the primary, there is no power associated with it i.e. the current waveform is lagging the voltage waveform by 90 degrees: -

enter image description here

If it were a resistor of 3142 ohms the current would be in-phase with the voltage thus: -

enter image description here

Picture source.

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You are correct that in an ideal transformer the coils have only inductance and zero resistance. However in a transformer a resistor in the secondary is transformed on the primary side to a resistance equal to the resistor value times the square of the turns ratio. For your example, we will assume that the secondary voltage is about 5 vrms which is sufficient to produce a 5 volt DC output with a bridge rectifier. That means that the turns ratio is 230/5 or 46. Your load resistor is 5 ohms (5 volts/1 ampere). Thus the resistance seen by the primary source is 46^2 X 5 or 10580 ohms, close to what you calculated. This resistance value determines the primary current. However all of the power is dissipated in the 5 ohm load resistor and none in the transformer.

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  • \$\begingroup\$ The problem with this answer is that it does not explain why transformer with disconnected load does not go up in flames \$\endgroup\$ – Maple Sep 1 '18 at 6:34
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I'll add this in the hope that it helps. You've clearly started learning about electricity, and you appear to understand roughly up to the start of inductance and capacitance. If it helps you, I think it's usually taught in roughly this order:

  • Nature of electricity
  • Ohm's Law
  • Series/parallel resistance
  • Watt's Law
  • Kirchoff's current law
  • Edison 3 wire
  • Skin current and the relationship between AC RMS and DC calculations

So you've probably seen everything up to about here. Next you might want to check out AC RLC circuits.

  • Capacitors
  • Faraday's Law
  • Lenz's Law
  • Inductors
  • Series/Parallel capacitance
  • Series/Parallel inductance
  • Capacitive and inductive reactance
  • Inductive/Capacitive lead and lag
  • Resistance, reactance and impedance
  • Active or real Power, Reactive power and Complex and Apparent power
  • Power factor and power factor correction

By the end of this you'll be able to do all the basic AC calculations and predict what happens in an RLC circuit, plus the way current affects wire size and basic transformer/ motor/ resistive behavior as supplies and loads. This is also necessary to calculate for multi phase systems. If you want to find course material and learn it in order, that should help.

The image above shows how a typical AC to DC adapter works,

It may be more accurate to say "simple" as this may no longer be typical, with many units now adding a switching converter to improve load regulation.

high voltage (230V) at low current is fed into the primary coil and then gets stepped down to a lower voltage with higher current,

Ok so what you actually have in this transformer is two inductors with different numbers of turns sharing a magnetic circuit. This makes a voltage transformer specifically. Note that AC current is necessary as magnetic field lines caused by a current only move when that current is changing.

When a voltage is applied to the high side, the inductor resists increase in current by storing energy in it's magnetic field. The field is proportional to the current flowing, and you can think of it as constantly trying to collapse, with the magnetic flux lines only being held away from the wire by the current flowing. The more current, the more field, and the harder the field pushes back, When the voltage decreases again through the AC cycle, and is no longer enough to increase or maintain current, the magnetic field collapses back into the inductor, returning the stored energy and "resisting" the decrease in current. Because the device resists changes in current and also stores the energy and returns it to the circuit, we differentiate this from pure resistance and call it reactance.

Because the energy is stored and not actually used, we refer to this portion of the power as reactive power(Q), in volt-amps reactive or var, and only the power being used by the losses of the circuit as real power, measured in W. When we put a meter on an RLC AC circuit and measure rms voltage and current, we can multiply them together to get the apparent power. You can see from this diagram of the mathematical relationships that when we measure apparent power, we get a value determined by the impedance, Z, not the resistance.

Impedance and apparent power triangles

You can see the mathematical relationship between the different "types" of resistance and the different "types" of power in the diagram. Using trigonometry and pythagorean theorem, you can use two sides of the triangle or one side and an angle to calculate the other values. You use a right angled triangle with reactance pointing upwards for inductance and downward for capacitance. Resistance and reactance, respectively, are the horizontal and vertical components of impedance.

If you have multiple devices of varying impedance like a two different motors or a motor and a capacitor bank in series, and you want to add up their values and find out how much current will flow, you can't just add up the values like you would with resistance if the angles don't match. Instead you must find the resistance and reactance values, add them, and calculate the value of the resulting vector.

The angle between power and apparent power is referred to as power factor angle, it is the cosine of the power factor, which is power/apparent power.

Magnetic field created by a inductor is proportional to the permeability of it's magnetic circuit and the current flowing through it up to the amount of current that saturates it's magnetic field. The inductance, and inductive reactance of a transformer are designed to be high enough that if no load is attached, the current that flows will be negligible at it's operating frequency. When a load is attached to the output of a transformer, a circuit is created and a current is permitted to flow in the direction favored by the field of the primary winding passing through the secondary. This current has it's own field, which cancels out part of that which created it, decreasing the inductive reactance and impedance of the primary and allowing more current to flow.

afterwards it gets rectified to DC and finally a capacitor turns this DC to pure DC.

Yeah.

The power from the primary coil has to be the same as the power at the secondary coil (assuming no power loss).

I think we've basically covered this. You're talking about apparent power, in VA to be clear, note that because you are assuming no power loss, and no resistance, you could also refer to this as purely reactive power, in var.

Let's for instance use an example, a phone charger, most phone chargers normally output 5V@1A DC which means 5W.

We'll go ahead with the example, but note that usb standard devices require better load regulation than can be provided by a transformer and rectifier. Also be aware that a phone charger is likely a switching converter, and care may have been taken to match the reactance of the output inductor and capacitor to cancel out power angle that would be created by just using a transformer. This means if you measure 5V@1A apparent power off the output of the device, no extra reactive current would be flowing and you could trust that you were actually getting 5W of true power flowing.

Since the voltage from the wall is 230V this means that for there to be 5W there has to be 5W/230V = 0.0217A which means that the resistance at the primary coil is 230V/0.0217A = 10599Ω

So now you should see that the value you've calculated is impedance, which is not the same as resistance.

This is the part that I don't get, how can an inductor (coil), something with almost no resistance, have so much resistance. At first I thought that they would put a resistor before the coil but then I figured out that all the power would be used up by the resistor and wasted as heat instead of reaching the coil.

I'll just add that because of reactance impedance, inductors can be used to create a series voltage drop in a DC circuit without actually using the reactive power being stored, using the same effect that limits current flow in a transformer.

So what is really going on? Do they make these coils with materials that have really high resistance or am I missing something? Probably the latter is true.

Hope this helped.

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  • \$\begingroup\$ Very interesting, thanks. From reading all the other answers I've concluded that the total resistance in the primary coil is the sum of the impedance at the primary coil and the resistance of the secondary circuit * (number of turns on the primary / number of turns on the secondary) ^2 correct? \$\endgroup\$ – rr1303 Sep 2 '18 at 22:09
  • \$\begingroup\$ That answer seems sound to me, and that side of your question had already been covered, so I decided to go into depth with "So what is really going on?". The reason my answer is so long is that I feel like the individual parts here are much easier to understand with a look at the big picture. I'm not going to derive it right now, but the reason that looks like an acceptable field calculation is that it accounts for the ratio of primary to secondary, and the ^2 would account for the I^2 in I^2R. \$\endgroup\$ – K H Sep 2 '18 at 23:28

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