# Inductors - linear or exponential?

Discussion in 'General Electronics Chat' started by ithinkican, Apr 6, 2013.

1. ### ithinkican Thread Starter New Member

Apr 6, 2013
6
0
Hi there -

I was shown this site by a friend and WOW what a great resource for those of us learning in the field of electronics!

I'm hoping that someone cna help with what seems a simple question, but it probably has a complicated (and interesting!) answer :-D

OK, so inductors "charge" and "discharge" exponentially as shown in many text books - example: http://hades.mech.northwestern.edu/index.php/RC_and_RL_Exponential_Responses

BUT, inductors are shown as having linear behaviour whenever you read about how SMPS work - example here: http://www.maximintegrated.com/eequiz/eeanswer.htm

So which is it???

While i wait for insights from the community here I am of course trying to explain this to myself... so far i am thinking that the exponential behaviour is how an inductor behaves when connected to a voltage step in an isolated case, but in an SMPS the other parts in the circuit are changing the behavior....like maybe the input cap on the SMPS?

Is the exponential discharge of the cap on the input of an SMPS matching the exponential charge of the inductor and giving a linear current rise?

thanks for any help in my learning!

2. ### ithinkican Thread Starter New Member

Apr 6, 2013
6
0

I understand that for an inductor on a ferrite or other material core, as the core material starts to saturate it can not support any more increase in flux. This appears as a sudden increase in the current slope according to all i have read.

My question is - at the point of saturation and in fact past it, does the inductor now have NO inductance i.e. it is just a piece of wire? Or does it still have it's same inductance but no more flux can be created because the core is saturated?

3. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,813
1,731
Did you see somewhere that an exponential response isn't linear? ;-)

The big difference between these two circuits is one has a big fat resistor between the inductor and the voltage source. That means the inductor doesn't see the full supply voltage as part gets dropped across the resistor, and that drop (when solved using a bunch of linear differential equations) (umm, that's calculus) causes the driving voltage seen by the inductor to change, which makes the rate of current change change, which makes the "exponential" curve out of linear parts.

4. ### ithinkican Thread Starter New Member

Apr 6, 2013
6
0
ah - the resistor makes a bit of sense, i am just trying with a scope and power supply and inductor to get some waveforms to explain this to myself and I already noticed the change of waveform depending on the value of resistor i am using to see the current with on the scope!

I also read in a Maxim whitepaper that the linear current rise in a SMPS (buck converter for example) is the first part of the exponential curve but we never see the exponential part.

5. ### WBahn Moderator

Mar 31, 2012
19,121
5,170
An inductor is a "linear device" meaning that it obeys superposition. That's all that it means for a device or circuit to be "linear". This has nothing to do with the shape of the waveforms in a particular application.

So what does it mean to obey superposition?

Let's say that we have some component or circuit and the voltage at some point of interest is a function of the current at some point of interest:

$
v(t) = f(i(t))
$

Now let's say that the current is expressed as the sum of two scaled currents

$
i(t) = Ai_1(t) \; + \; Bi_2(t)
$

The component or circuit is linear if the output can be written as the similarly scaled sum of the voltages that would result from each of the separate currents by themselves.

$
v(t) = Af(i_1(t)) \; + \; Bf(i_2(t))
$

Let's look at the defining relationship for an inductor:

$
v(t) = L\;\frac{di(t)}{dt}
$

If we replace i(t) with the alternate form, we get

$
v(t) = L\;\frac{d[Ai_1(t) \; + \; Bi_2(t)]}{dt}
v(t) = L\;\ \left( \frac{d[Ai_1(t)]}{dt} \;+\; \frac{d[Bi_2(t)]}{dt}\right)
v(t) = L\;\ \left( A \frac{di_1(t)}{dt} \;+\; B\frac{di_2(t)}{dt}\right)
v(t) = A \: L\frac{d[i_1(t)]}{dt} \;+\; B \: L\frac{d[i_2(t)]}{dt}
v(t) = Av_1(t) \; + \; Bv_2(t)
$

So the behavior of an inductor is linear. If you were to perform the same exercise for a resistor or a capacitor, you would find they are linear, too. But if you were to do the same exercise for a diode, it would not be linear.

Now, if you have an inductor that you have driven past the point of saturation, then you have to use a different equation and you would find that it is NOT linear.

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6. ### ithinkican Thread Starter New Member

Apr 6, 2013
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0
WBahn, thank you for your explanation. Am I correct in suddenly "getting it" that the inductor's linear behaviour is modified to seem non linear when fed from a voltage source with significant resistance (or a resistor in series) because the voltage drop across the resistor changes with the inductor current, thus the slope of the inductor current changes with the inductor current - i.e. we see the exponential curve?

I'm trying to put your explanation and ErnieM's together in my head..... so far i have this:

- the inductor current ramp would be simple and linear if fed from a voltage source with zero output impedance.

- in real life a suitable cap is reasonably close to a zero impedance voltage source, so in for example a buck or a boost converter we do see the mostly linear current ramp

- if we put in a series resistor or use a higher output impedance voltage source, we see an exponential current waveform at the inductor because the votage across teh resistor changes with current through the inductor, whcih in turn changes with votage across the resistor.... and this breaks the linearity of the inductor?

actually on that last point - i suppose it's more correct to say that the inductor is still a linear device, but now the CIRCUIT we have created is non linear.

I have a cheap "Rigol" scope - not sure if anyone has even heard of it, but it was cheeeeeeep! and some resistors and inductors salvaged from old electronics, and the will to learn, so i am going to go to the garage and try and rationalize this all to myself some more

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
So ithinkican - linear AC circuit analysis deals with sinusoidal excitation. By your definition of linearity a sinusoidal response is non-linear. Do you think that is the case?

8. ### crutschow Expert

Mar 14, 2008
14,770
3,770
At the point of saturation the incremental inductance starts to drop drastically. In full saturation the incremental inductance is not zero but is very small, likely close to what an air core coil of the same dimensions would have (air or a vacuum never saturate from a magnetic field).

9. ### WBahn Moderator

Mar 31, 2012
19,121
5,170
The shape of the signal and the linearity of the device are two separate things. That's the point you are losing yourself on. The DEVICE or the CIRCUIT is linear if superposition applies. It doesn't matter if the SIGNAL is linear, exponential, sinusoidal or what have you. The linearity of the DEVICE or CIRCUIT is based on the following notion. I have a SIGNAL. I can analyse the RESPONSE of the DEVICE/CIRCUIT to the SIGNAL as a whole. Now, I can also decompose that SIGNAL into the sum of several different signals and find the RESPONSES to each of those constituent SIGNALS. IF, and ONLY IF, the sum of those constituent RESPONSES is equal to the RESPONSE to the full signal taken all at once is the DEVICE/CIRCUIT linear.

So the LINEARITY of a DEVICE/CIRCUIT has to do with how it responds to inputs and whether those inputs can be broken into pieces, run through the system separately, and added together at the end to get the same response as when the signal is run through it all at once. It has nothing to do with a waveform being a straight line.

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10. ### YokoTsuno Member

Jan 1, 2013
41
9
Both circuits are linear.

BTW, the method you use isnt the most convenient way to determine if a component/circuit is linear. If you get hold of (a cheap Rigol) function generator and apply a sinusoidal voltage to your RL circuit you can determine if the circuit is linear if the scope shows a sine wave if you measure the voltage across the coil or resistor leads.