# RL circuit practice

Discussion in 'General Electronics Chat' started by Elerion, Dec 19, 2015.

1. ### ElerionMember

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I got some inductances, and I just built a very simple RL (series) circuit.
The goal is to do a basic measure of real inductance.

L = 1mH (axial resistor-like inductance).
R = 100 ohm

Connected as:
voltage source --- resistor --- inductor --- ground

Approach:
A perfect voltage divider would happend when reactance equals 100 ohm. Right?
f = 100 / (1e-3 * 2 * pi) = 16 kHz

Simulation:
Peak-peak voltage across each element match. Current lags voltage.

Yellow : source voltage.
Blue: inductor voltage
Red: resistor voltage.

Real world:

1.- voltage on each element match at a much lower frequency (9.6k vs 16k).
2.- peak-peak voltage is not source/sqrt(2), but much lower (40% vs 70%).

I tried with 1 kOhm and 2 kOhm resistors too.
It seems that the gap between theoretical/real frequency narrows as R gets bigger.
It makes me think that the inductor has somekind of internal series resistance, but it seems far to big (I would hope to find something like 5 ohm ESR)

My multimeter measures 0.9 mH (within 10% tolerance).

I must admit this is all I can get.
Could anyone briefly explain what am I missing?

2. ### JimBSuper ModeratorMost Helpful Member

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Wrong.

The bit you are missing is that you cannot just add resistance and reactance (Inductive or capacitive).

When you have 100Ohms resistance and 100Ohms reactance, the overall impedance is 141.4 Ohms.

Impedance Z = ( R^2 + X^2)^0.5

Sorry about the notation, but to put it in words:

Impedance is the Square root of Resistance squared plus Reactance squared.

JimB

3. ### AnalogKidWell-Known Member

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Think about the diagonal of a rectangle or square. Same equation. Resistance is the distance along the bottom of the rectangle, and reactance is the height. Pure resistance, 100 ohms over, 0 ohms up. Pure reactance, 100 ohms up, 0 ohms over. The vector sum is 141 as above.

ak

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5. ### ElerionMember

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When I said "voltage divider", I meant to say that same voltage appears (peak-peak) across resistor and inductor, which in fact does happend when reactance is 100 ohm.

The question is, why real world circuit differs from simulation so much (in freq and amplitude)?

6. ### AnalogKidWell-Known Member

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Ususally because of the three baskc electronic components, the worst "real world" parts are inductors.

ak

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7. ### ElerionMember

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I measured 13 ohm (d.c) with my multimeter.
It seems to me that inductor Q is low.

I built a simple LC.

Resonance has a great bandwidth.
I'm able to see how capacitors starts drawing current at high freqs (its reactance is dominant).
For freqs below resonance my scope "sees" a near 15 ohm resistor in series with my generator output impedance (50 ohm) no matter how low freq is.

If I meant to build a simple LC tank circuit for AM radio tunning ( 500 kHz - 1.5 MHz), could I use this inductor??

8. ### RatchitWell-Known Member

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Your method looks solid to me. If the voltages across the resistor and inductor are the same, the reactance of the coil should be the same as the resistance of the circuit, even though the phases of the voltages are different. I notice that that the sum of the instantaneous voltages of the coil and resistor are equal to the instantaneous source voltage for the simulation. That is not true of the "real" circuit. Why don't the instantaneous voltages add up correctly? Where is that extra instantaneous voltage from the source going? What is the resistance of your voltage source? 50 ohms? That is half the value of your test resistance. See something wrong there? Don't try to measure the instantaneous voltage with your meter, you need a scope for that. No need to get tangled up in phase calculations of impedance either. Forget about resonance, you do not have enough capacitance in that circuit to make resonance happen. Use a 10k resistor instead of 100 ohms to swamp out the 50 ohm source resistance. Let us know what happens.

Ratch

9. ### ElerionMember

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Yes, I see. I'll try again today.
Also, I just realized another error on my side: scope's red trace is at 1V/division while the rest are 500mV.
This is just enough to repeat the whole process, although this time, with a 1k resistor.

What I double checked, for the LC (no R) parallel circuit, is that no clear narrow resonance could be achieved even though I tested three different capacitors. L = 1mH and C = 22pF, 560pF and 10uF (ceramic and tantalum).

pF range exhibit a greater peak voltage as freq goes up, as expected, and start declining only passed around 1MHz. But the max voltage was achieved at a much lower freq. I just couldn't see the capacitive and inductive regions by just changing the freq within the same decade. Intuitively I would think this is of no use as a passband filter (ex, AM radio tuning). Seems like Q is weel below 0.1.

10uF cap lead to peak on voltage (less current being drawn, closest to resonance) at around 1.5kHz, but below that freq the main contribution of the inductor was this "15 ohm" divisor I talked about (with my 50 ohm source); in other words, the peak voltage gets never high enough. I see the capacitive region above 1.5kHz, but lowering the freq exhibits a flat response (of a 15-50 ohm resistor voltage divider).

What would I need to do to get an LC parallel circuit with a higher Q ?
Is it the inductor which is limiting ?

10. ### RatchitWell-Known Member

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Eliminate as much resistance in the circuit as possible. What do you mean by inductor limiting?

Ratch

11. ### ElerionMember

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This is the good capture for 100 ohm + 1mH
At 16 kHz (theoric reactance = resistance; works on simulation, but on real circuit resistance is still bigger )

At 20 kHz we achive the point.

Next I tried with a bigger resistor. I'm using a pot set at 925 ohms.

200 kHz. Theoretical/simulation happends at 147 kHz.

The ratio of the "freq offset" increased.
From 20/16 = 1.25
To 200/147 = 1.33

Doesn't it suppose to decrease?
I mean, using a higher resistor, my generator output impedance is now negligible.

I'm using a pure LC, no resistor.

12. ### RatchitWell-Known Member

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How does frequency offset pertain to this problem? Look at the instantaneous voltages and see if they add up like they do in the simulation with no resistance in the source. If they add up correctly, then you have swamped out the source resistance and the reactance should be equal to the test resistor. There is no LC circuit that is completely free of resistance. The resistance comes from capacitor leakage and wire resistance of the coil. You cannot eliminate it completely. You asked how to measure the inductance of the coil. Why is a resonant circuit entering into this discussion? Did you get the correct inductance of the coil from your method? If so, we can go onto something else.

Ratch

13. ### ElerionMember

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No. I get 0,8 mH (for R=100 ohm) and 0.73 mH (for R=925) and not 1 mH.

What I meant by "frequency offset", is that I see the 3dB point (resistance = reactance) at a higher frequency than expected by formula f = R/(2*pi*L), which matches simulation.

14. ### RatchitWell-Known Member

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I wish you would have used a 10K test resistor like I suggested. OK, so (925+13)/(2 pi 200000) = 0.000746. The extra 13 ohms is for the resistance of the coil. Then multiply another (1+50/925+13)% because the test resistor is only 18 times greater than the source resistance of 50 ohms. That gives 0.000786. Next we multiply by 1.1 to account for the possible 10% variance of the inductance. That gives 0.865 mh. So that appears to be what the coil could be regardless what they labeled for its value. You can see how the values change if the test setup if not what you think it is.

Ratch

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15. ### ElerionMember

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Still quite big deviation from desired inductance.

I did not use 10k resistor because my signal generator cannot go up to the required 3dB frequency.

I'll buy some more inductors and repeat the process to see if I can get an expected disperssion in inductance value.

The real purpose of all this is building an LC tank circuit.
But first I wanted to start from the beginning, step-by-step, reviewing every concept.

16. ### RatchitWell-Known Member

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If you could get access to an impedance bridge, you could measure the inductance directly. I wonder if the 50 ohm impedance of your voltage source contains reactance. Try making a differential voltage measurement across the test resistor and the coil. Then it won't matter what the source impedance of the signal generator is. However, unless you have a 4 channel scope or two scopes, you will have to switch leads to read the voltage across each component. Or maybe you don't need a scope after all. Just use a voltmeter and check that each voltage component is equal. If you have a electronic voltmeter, make sure to use an isolation transformer so you don't ground the reference point.

Ratch

17. ### ElerionMember

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I tried with four different inductors, and got the same results for all of them. Too weird that all of them are approx 15% below nominal value. It seems that my measurements are, as you are pointing out, wrong in some way.

What do you mean?

18. ### RatchitWell-Known Member

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Didn't you try to Google "differential voltage measurement"? I did, and got 38,000,000 hits.

Ratch

19. ### specWell-Known MemberMost Helpful Member

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Hi Elerion,

I haven't analysed what's been said so far, but it occured to me that the reason why the resistance appears to be higher than you meaured at DC is because it is. As the frequency goes up skin effect, for one, increases the ESR and thet may account for the discrepency. To resolve your problem, you need to measure the effective series resistance (ESR) of the inductor at 10KHz, and that is what you have effectively done. The value you measure is probably correct.

What type of inductor are you using? 10KHz would be a heck of a high freqency for some inductors.

Remember though, that I am just thinking out loud. The way to check is to take a look at the inductor data sheet. If you post the inductor part number I will have a look.

spec

Last edited: Dec 26, 2015
20. ### ElerionMember

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Of course I did. It might be clear for someone who already knows what is looking for, but not for me. I know what differential voltage means, but that is what I was already doing ( the red trace on my scope plots are the difference between ch1 and ch2, and thus differential ), and so, I got confused, thinking that you meant some other thing I didn't know (for example, I thought the impedance bridge had some relation with your subsequent description, but now I think it didn't).

Now my question is, was or wan't I already making differential measurement? If not , what did I miss?

Spec, this kind of inductor (but color code brown, black, red, silver)
Don't have any datasheet or manufacturer. The store where I bought it didn't sell too many inductors,... and another one I went to, didn't sell any at all :-O
It seems that I will have to buy online,...

21. ### specWell-Known MemberMost Helpful Member

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If I am right, that may not be an inductor intended for low distorion, accurate, filters, tank circuits, etc. It is probably a lossy inductor (choke), intended for supply line filtering and the like. In that application the 'worse' the inductor is the better. As you probably know, inductors (and capacitors) have a Q (Quality) factor. which = XL/ESR. (XL= 2 * π * F). A Q of 100 would be typical for a precision inductor at its specified frequency. The best inductors are air cored, but they get physicallly large at high values, so a high permeability core is used. Air (vaccuum) is the reference with a permeability of 1.

UPDATE

Fairly sure now that what you have is a choke, most probably made by Taiyo Yuden.

Last edited: Dec 26, 2015
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