General magnetic induction question

[Njguy]

I am trying to get a better understanding of magnetic induction. Lets say you have an electromagnet wrapped around an iron rod (primary coil). On the other end of the rod you have another coil(secondary) hooked up to a meter. If you simply pulsed the primary coil with DC would this produce a lower voltage in the secondary coil if instead you instead ran AC through the primary coil?

 

[shortbus=]

This is how an ignition coil in a car works. Other than the ignition coil is a 'step up' coil.

 

[KeepItSimpleStupid]

probbaly, your real issue is understanding "voltage". The voltage across an inductor is Ldi/dt and that relationship is always valid, so you can think of v(t)=L di/dt or, the inductance times the first derivative with respect to time. If you don't yet know what a derivative is, think of the instantaneous slope.

or you can think of the distance x(t), and dx/dt is velocity or speed.

Voltage can be measured in RMS or p-p or pretended to be a sine wave and shown as RMS.

A DC pulse would have a much higher peak voltage at the instant applied. AC however, would depend on where in the cycle it was applied initially.

 

[crutschow]

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A DC pulse would have a much higher peak voltage at the instant applied.
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If I understand what you are saying that's not a proper statement. A transformer with an applied pulse gives an output voltage equal to the turns ratio, essentially the same as a sine-wave (ignoring any frequency response limits of the transformer). The high voltage generated by a transformer in a flyback configuration, such as an ignition system, uses the inductance of the transformer to generate a high voltage pulse. Those are two quite different modes of operation.

The initial application of DC to a transformer primary will give an output voltage equal to the turns ratio times the DC voltage (ideally). But if you then suddenly interrupt the DC after some current has started to flow in the primary, you will get a large voltage spike both at the primary and secondary due to the inductive energy of the transformer being suddenly released. Again the primary spike voltage appears at the secondary multiplied by the turns ratio.

 

[KeepItSimpleStupid]

Tough one to put into words. What I really wanted to emphasize is that talking about voltage depends on how it's measured and applied.

 

[BobW]

As to which method gives the most voltage, it all depends.

For one thing, it depends on whether you are looking for maximum rms voltage, or maximum peak voltage.

Also, as the others have indicated, the secondary output voltage depends on the rate of change of primary current. So, "pulsing the primary" voltage doesn't really say what's happening to the primary current. If the primary is pulsed on long enough for a reasonable amount of primary current to develop, and then open circuited so that the primary current if forced to zero almost instantly, then you can develop some very high voltage spikes on the secondary. However, if you simply apply square wave pulses to the primary from a low impedance source such as a signal generator, then the current will not be interrupted when the voltage goes to zero, and you won't get the same high voltage spikes.

 

[crutschow]

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Also, as the others have indicated, the secondary output voltage depends on the rate of change of primary current. So, "pulsing the primary" voltage doesn't really say what's happening to the primary current. If the primary is pulsed on long enough for a reasonable amount of primary current to develop, and then open circuited so that the primary current if forced to zero almost instantly, then you can develop some very high voltage spikes on the secondary. However, if you simply apply square wave pulses to the primary from a low impedance source such as a signal generator, then the current will not be interrupted when the voltage goes to zero, and you won't get the same high voltage spikes.

Again, it should be noted that talking about primary current (other than magnetizing current) is referring to flyback type operation of a transformer. It's not pertinent to normal transformer action where the output is (ideally) equal to the input voltage times the turns ratio, independent to the rate of rise of the voltage (i.e whether it is a pulse or a sine-wave).

 

[BobW]

Agreed. However the OP is referring to an "electromagnet", a primary and a secondary coil wound on an iron rod. The coupling will be much looser than with a normal transformer, which throws another complication into the problem. So, I don't think you can use the simpler transformer theory in this case.

 

[steveB]

The OP's question is very poorly worded which I think is leading to confusion about how to answer. Here is my take on the question.

The first sentence is "I'm trying to get a better understanding of magnetic induction.".

This is kind of like if someone asked, "I'm trying to get a better understanding about string lengths. Lets say you take a string and wrap around your hand a certain number of times. Which string would be longer, a thick one or a thin one. "

Both questions could be taken in under ideal simplified assumptions, or under more realistic assumptions. But, which is case is truly being asked.

The simple case of a string says, both lengths are the same, but the non-ideal case implies that the string has thickness and the thicker string will end up longer.

The transformer case can be taken as ideal or non-ideal. Which aspect is the OP trying to understand? I have no idea from the wording of the question. No details are given. The simple ideal case says the voltage is the same, at least relatively speaking input to output. But, non-ideal cases would perhaps assume that the pulse is fast and the AC is slow and then we need to talk about non-ideal aspects of frequency and flux leakage.

In other words we have one of those famous impossible question that is often answered with the wise crack, "how long is a piece of string?".

 

[crutschow]

Agreed. However the OP is referring to an "electromagnet", a primary and a secondary coil wound on an iron rod. The coupling will be much looser than with a normal transformer, which throws another complication into the problem. So, I don't think you can use the simpler transformer theory in this case.

The coupling may be looser but that just means the effective turns ratio will be much less than the ideal and there will be much more leakage inductance. But it should still act as a transformer.

 

[dr pepper]

There are 2 ways in which a transformer can operate, and this device is a transformer, and thats as a pure transformer or a flyback.
You could split hairs and say a flyback transformer is in fact a coupled inductor.
If you are pulsing, as in connecting to a power source and then disconnecting the primary (as in open circuit) then your operating the device as a flyback, in which case the output voltage forumula - turns ratio x the primary voltage doesnt apply, its more to do with the amount of energy the primary stores up during the on cycle (assuming it doesnt saturate).

 

[MrAl]

I am trying to get a better understanding of magnetic induction. Lets say you have an electromagnet wrapped around an iron rod (primary coil). On the other end of the rod you have another coil(secondary) hooked up to a meter. If you simply pulsed the primary coil with DC would this produce a lower voltage in the secondary coil if instead you instead ran AC through the primary coil?

Hi,

You may want to reword your forth (last) sentence there because it is not written correctly. It is written like a statement but sounds like you really wanted to make a comparison and ask a question about that comparison. In other words one interpretation would be:
"If you pulsed the primary coil with DC would this produce a lower voltage than if you instead applied regular AC to the primary?"

The short answer is that you can get a higher voltage from using pulsed DC than you can with regular sinusoidal AC because if the driver is designed for flyback operation then you can get a high pulse output. But it's worth looking into it a little more because there are constrains that apply for both.

The answer for AC alone is fairly simple because we assume the turns ratio governs the output voltage, so with a turns ratio of 10:1 and 100 volts AC on the primary we'd see 10 volts AC on the secondary. With a turns ratio of 1:10 however with 100 volts AC on the primary we'd see 1000 volts AC on the secondary. So that depends mostly on the turns ratio.

The answer for DC pulsed is more complex, because it involves having knowledge of the driver circuit. We have to know what kind of driver it is and what mode the designer intended.
Mode 1 might be called the "standard" or "non flyback" mode and Mode 2 would be called the "flyback" mode.

With Mode 1 we would apply a symmetrical DC pulse (positive and negative half cycles have the same amplitude and pulse duration) and that would be almost the same as the AC mode above, where we use an AC signal. The output then matches the turns ratio in the same way.
This also requires knowledge of the transformer inductance (or use a ready made formula) so that we dont saturation the core. But assuming all that has been worked out, we get basically the same as we get as with the AC wave except the waves are rectangular not sinusoidal.
As i said though, there are conditions that must be met to get this kind of operation.

With Mode 2 we would apply a non symmetrical DC pulse and when we turn it off we would expect a large inductive kick back in the primary. This means that the inductance causes a very high voltage pulse for a shorter time than the applied pulse. Although the voltage will be much higher than the applied rectangular signal the volt seconds of both pulses will be equal. This higher voltage pulse then appears on the secondary according to the turns ratio. So now we might apply a 100 volt pulse, turn it off, then see a 500 volt pulse, and that 500 volt pulse gets stepped up by the secondary according to the turns ratio. This kind of design allows for a very very high output voltage, but higher than with an ordinary AC signal on the primary. The driver must be designed with the knowledge that the primary voltage itself will shoot up to a much higher value once the pulse is turned off, The turns ratio then boosts this voltage to an even higher level.
The two reasons this happens are:
1. The inductive kick back on the primary generates a voltage amplitude that is higher than the applied pulse amplitude.
2. The turns ratio is a step up, which raises that high kick back pulse even higher.
We dont get something for nothing however, as the volt seconds for both half pulse cycles have to be equal on the primary side. This means that for a high voltage output pulse of a given duration we have to have an applied pulse on the primary of longer duration, and it may have to be much longer if the inductive kickback is very high. The kickback however allows us to get a very very high output voltage and sometimes that's the most important thing for the application.

Square wave inverters use the standard method where a DC symmetrical pulse is used to generate a secondary voltage. Automobile ignition coils use the flyback method to get a very high output voltage pulse.

There are other modes too that are a bit more specialized such as a "resonant" mode where the whole thing resonates and produces an AC output from a DC pulsed input. In this case the whole construction acts like a transformer plus a filter.

As long as we are talking about flybacks and associated voltage boosting, we might as well mention that we can get a high voltage without using a secondary coil in this same manner. By pulsing an inductor we can get a very high output voltage because of that inductive kick back. This is how many boost circuits are made.

The driver circuits for the regular mode and flyback mode are different because for regular mode we want to force the primary to be a certain voltage for both half cycles. For the flyback mode we only force the voltage to be a certain level during the driver 'on' time, and when the driver turns 'off' it really turns off and so it relinquishes control solely to the inductance of the primary.