# Help with understanding inductors

Discussion in 'General Electronics Chat' started by chipwitch, Mar 10, 2014.

1. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
I'm sorry to ask about this. I've read, googled, read and searched some more, but finding the bit of information that unlocks for me what I'm missing to get my head around inductors is like finding a needle in a haystack. I'm sure it's been discussed thousands of times.

Let me start with what I'm trying to do... almost everyone is familiar with having scavanged transformers on hand without identifying marks. I've found one I want to use for a project with the right wind ratio, but I don't know how much power it can be expected to deliver. I've read the blog by Jim Lux, if anyone knows it.

Anyway. I have one of those little testers for identifying caps, inductors, diodes, transistors, etc and from it am able to measure the inductance of my transformer. Given that, I'm able to calculate reactance. Is the power a transformer can deliver as simple as measuring the L of the secondary and calculating the P=V^2/w? Does the resistance of the windings play a role or is it reflected in the reactance?

If I'm just hooking up a voltage regulator to the secondary windings, does the phase angle have any significance? Not sure what I'm supposed to do with this information.

If I can simply calculate P=V^2/w, (secondary) and confirm that the wire size and insulation is sufficient for the load, then a similar procedure on the primary side would be all the analysis remaining to confirm the power handling ability of the transformer?

Oh, one more thing... just to confirm... if I measure the current in an inductive circuit, I'm measuring the RMS value? Only shifted to the phase angle indicated by the Impedance (in an RL circuit)?

2. ### crutschow Expert

Mar 14, 2008
13,508
3,384
The transformer inductance has nothing to do with the maximum power it can deliver. The inductance determines the magnetizing current, which must not exceed a value that will saturate the core. This determines the maximum input voltage that can be tolerated for a given input frequency.

Any secondary load current is simply reflected back to the primary with no effect from the primary inductance (any leakage reactance does have a small effect).

The main parameter determining the transformer power capability is the winding resistance and the heat generated by the I-squared-R loss in the windings. Conservatively you probably don't want the load current to cause more than about a 5% voltage drop in the winding resistance in both the primary and secondary windings.

Are you talking about an AC regulator or a DC regulator?

If you measure the inductive current the same way you measure the resistive current then you are measuring an RMS value. But the inductive part of the current, as determined by the power factor, dissipates no energy.

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3. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
The more I read, the more confused I seem to become. Can someone please confirm or refute the following? Naturally, the following assumes 60Hz A/C

1) The power meter outside my home is measuring real power.

2) Real power is Vrms * Irms. (is there no sub/superscript formatting for this forum?)

3) In an AC circuit, Vpeak and Ipeak play no role in power calculations and wire ampacity. Both are based on RMS.

4) A transformer's kVA is in Apparent Power.

5) Transformers don't like motor loads, unless their reactance is offset by capacitance.

6) The run capacitor in an induction motor is NOT for the purpose of offsetting reactance.

4. ### GopherT AAC Fanatic!

Nov 23, 2012
6,321
4,048
Chip witch

Transformers are more complex than you are assuming. You need to know what the core is made of (assuming silicon iron plates), determines the saturation current. The thickness of those plates and insulating material between plates determines Eddie current heating at a given frequeny and saturation (and waveform - don't assume sine wae as you approach saturation).

If you salvage a transformer, try to record some information off if the device (volt-amps or wattage) and operating voltage - even the voltage rating of the filter caps is a clue.

Jul 18, 2013
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The run or start capacitor creates a phase shift or split-phase between the two windings, otherwise the 1phase alone provides a magnetic field at 180° therefore no rotor revolution takes place.
Max.

6. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
Hi Crutschow, Thanks for the response.

Okay... "magnetizing current" is a concept I've completely missed. That gives me something to google. So, step one, determine maximum input voltage.

By "reflected," you're saying that a VA on the secondary side requires exactly one VA on the primary side, not counting leakage.

So, it's reasonable to simply "try" a transformer and as long as load/no load ratio is above .95, you're probably alright? Won't a 5% drop on the secondary automatically equate to ~5% drop on the primary?

DC, rectified, of course.

Because an ideal inductor uses no more power than it returns?

7. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
Gophert, I knew they were complex... just trying to get a handle on them. Surely there's a way to empirically determine their power output capability. Crutschow suggested a 5% drop as a rule of thumb. I'm okay with estimating output based on measurements taken.

I have a scope. If I use a variac to power a low voltage signal to the primary of a transformer, doing it at different voltages must somehow indicate something? You imply the waveform changes at core saturation. If so, can that be observed and used on a scope?

8. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
Reading about magnetic saturation... if one can measure quantitatively, the magnetic flux density of the magnetic field produced by a transformer, couldn't one determine rather accurately at what point saturation occurs by simply applying an incrementally increasing voltage in the presence of a magnetometer? Saturation is a relatively abrupt plateau which shouldn't be too hard to find, even with an inaccurate magnetometer.

If that were a practical approach, any reason one couldn't use a \$2 magnetic flux density sensor for an arduino to find at what voltage saturation occurs?

9. ### alfacliff Well-Known Member

Dec 13, 2013
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core saturation is a function of current. if the out put loads down 5% the primary does not oad down, it just draws more current. changing the primary voltage does not change any waveforms, just the secondary voltage. size determines the watage rating of the transformer, unless operated at higher frequency, then size is smaller, due to losses in the core. transformers designed for 400 hz are smaller than those designed for 60, and 60 hz transformers are a little smaller than 50 hz transformers. current rating is determined by size of wire. voltage is determined by insulation. if you want to check where saturation occurs, place a variable load on the transformer secondary, and watch for a sudden drop in voltage.

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10. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
I can't say I got all my questions answered but some of it is cleared up. Alfacliff, you've certainly answered the important/primary one. Now, assume my output is 10VAC. I determine the secondary voltage drops significantly when my variable resistance load reaches 10 ohms. I = 1A. Therefore, I can label the transformer 10VA? Now, exactly what does that mean? I know I could power a 10 ohm resistive load. But what about motor loads?

Where I'm now confused is the topic of reactance. Assuming I want to power a 10VAC fan, do I simply look at what the fan draws? If it's under an amp, it's all good? I'm thinking that since the motor will have a reactive load component, that the transformer won't handle a full amp motor load. Can you see why I'm confused?

Now, what about DC? If I use a full bridge rectifier and get an output of roughly 13 volts from that transformer. Since it's a 10VA transformer, does that mean I can now draw no more than .769A? Since it's DC, I gather it has no reactive load, but an impedance? Which can only be determined by V/R (can't be measured directly)? I feel like I'm mixing up different concepts that aren't related.

11. ### Veracohr Well-Known Member

Jan 3, 2011
559
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Transformers are designed to keep the magnetic fields within the core. You probably wouldn't measure a lot of flux from one.

12. ### crutschow Expert

Mar 14, 2008
13,508
3,384
And not counting the primary magnetizing current.
Yes, if you mean load/no load voltage ratio. And yes there is also a drop in the primary due to primary resistance which will probably (but not necessarily) be equal to the secondary drop. Both drops add together to give the total drop at the secondary since you can't directly measure the primary drop.
Then the phase shift is immaterial.
Yes. An ideal inductor stores energy and then returns it to the circuit without loss.

13. ### gnuuser New Member

Jan 17, 2013
13
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with any transformer (step up or step down) output is dependent upon the primary an secondary winding ratio
for example 120 volt @ 1 amp stepped down 10(primary) to 1 (secondary) will be 12 volts @10 amp
current transformation is inverse to voltage

14. ### gnuuser New Member

Jan 17, 2013
13
2
using an unlabeled transformer can be dangerous with out determining its voltage rating
generally salvaging a transformer one needs to pay close attention to the primary supply voltage
if this is known then a lower voltage can be applied and tested at the secondary to get a ratio
transformers are generally rated by kva