Pickup Inductance

You may have come to this page from the “GFS Pickups” page, if not it is still fine to start here. The issue is, as stated elsewhere, the inductance of pickups, especially humbucker pickups is somewhat complex, where single coil pickups are by nature somewhat simpler to deal with in this respect. If we should recap the dilemma stated on the above mentioned page, the problem was what if you measure the inductance of the two individual coils in a humbucker, their inductances do not add up to the total inductance, something that was discussed on the main “Guitar Electronics” page where you would typically measure the total inductance and then apply half the inductance to each coil in the equivalence diagram used for simulation of the pickup performance. As was brought up on the first mentioned page, why is the sum of the two individual inductances not equal to the inductance for the complete humbucker. That is exactly what this is all about, so if you have been wondering the same, you have come to the right place. Also, for this investigation, I picked a humbucker, of course, with both coils equipped with slugs to make the two sides as even as possible, same coil resistance and inductance. The reason why I did not use a half slug and half screw is because they may have the same resistance, but they have different inductances. There is an example on the main page concerning this, where I demonstrate how much inductance can vary based on the magnetic media in the pickup, so it is no mystery that even with the same number of turns in the coil (yielding same resistance) the important parameter, inductance, is different for the two coils just because one is screws and the other one is slugs .

In electronics, inductance is everywhere, sometimes for the good and a lot of times for the bad. In the case of guitar pickups, it is there and it greatly affects the output of the pickup since it, along with a compound capacitance, determines the frequency characteristics of the pickup, something we have discussed quite a bit elsewhere. Right here, we are going to look into the details of inductance and specifically the measurement of inductance.

The best way to explain how an inductance works is that if you apply a constant voltage across an inductor. it will instantly accept that voltage without any change in voltage value. The current through the inductor, however, will built up slowly starting from 0 current and the inductance and the value of the instantaneous voltage will determine the rate of rise of this current, measured in Amps per second, also written as dI/dt. So if the constant voltage placed across the inductor is called U and the inductance of the coil is L, we can set up the following equation:

Now we know what the voltage is and we can measure dI/dt, therefore we can calculate L, our inductance. Next step is to determine how to measure dI/dt and we will find that the method is dictated by the circumstances, especially how to measure current without altering the result of the measurement. I have personally measured inductance of a countless number of inductors in my time and what I will describe later is one method I have used on most occasions.

The whole thing started with what was discussed above, almost immediately leading to the question about my digital LCR meter which I consider being of high quality without being a fancy expensive one. This meter has been used for both resistance and inductance measurements on pickups. As I have thundered elsewhere it is IMPOSSIBLE to measure capacitance of a pickup coil with such a meter (please see “Resonance Measurement”). Do not believe anyone that tells you otherwise! Anyway, I had used the meter and suddenly I saw what has been described above. That was leading to me asking the question, how does this meter measure inductance?

The first problem in answering this question is that the meter only has two wires sticking out of it, so some of the measuring is internal to the meter. Only thing I could do was to hook a pickup up to the meter and try to measure what was going on with my oscilloscope. As it turned out, if was not that easy, because part of the measurement circuitry was inside the meter, so I had to add something similar to the outside to measure current and my conclusion was that the meter is applying voltage pulses to the output, expecting there to be an inductance of a certain size and what it looks like the meter takes the measurement of the signal (current or equivalent) and the end of this pulse and the value is used to calculate the inductance L. And I believe that the value determined depends on the size of the inductor and the scale the meter is set to. The fact that I can only speculate, which is something I always shy away from, I cannot get much further with the meter, so let me continue with some other ways of measuring or looking at the inductance.

I have used this tool several times before when analyzing pickups or other circuitry, in this case I want to use simulation to demonstrate the measurement setup.

What we see in Fig 1 is the voltage waveform impressed on the pickup circuit which is using values from a measurement of the coils individually, the orange waveform is a 4kHz, 10V square wave with 50% duty cycle. The blue is the measured waveform across a resistor in series with the pickup under measurement. Using the resistor for measurement is because you can easily calculate the current through it (same as current in the pickup) by dividing the voltage with the resistance. So we can consider the blue curve as an equivalent current, as we shall see later on with actual measurements. Let us concentrate on the blue curve for a moment.

In Fig 2, we see the current equivalent alone, in form of the voltage across the resistor, the spikes you see, two positive and one negative are indicating the current through the capacitance of the pickup during the switching of the orange waveform in Fig 1, maybe I should not bring this up, but you could technically determine the capacitance by analyzing these spikes, but we will not do that for various reasons, accuracy for one.

Now, Fig 2 contains a lot of information repeated, we only need what is shown in Fig 3, a small piece of the waveform in order to determine the inductance. Please notice the equation on the chart and the dotted line that is approximating the curve which as you can see is not a straight line. We will use the dotted line to determine the dI/dt that we need from Eq. 1. Before we do the calculation, we go back to the curve itself, since it is not a straight line, it means that the inductance changes as the current (or in the curve, the voltage (equivalent)) increases. Later on we will look at actual measurements of the pickup in question that is using a different method than the meter.

Back to the equation of the Fig 3 chart, y = 17260x – 87.273, where y describes the function (straight line) indicated by dots, that we will use for our inductance calculation. If you are familiar with math and differentiation of a function, you will know that dy/dx = 17260. This is in Volts per Second and the same as dV/dt, but we need dI/dt, so we divide 17260 with the resistance value of the resistor used for the measurement, 1000 Ohms, so we get dI/dt = 1.726 A/s (Amps per Second). Since the voltage used is 10V, our inductance is L = 10/1.726 = 5.79 H. This is the the inductance of the complete humbucker, using similar methods, the individual coils can me calculated to 2.98 H and 2.93 H. Adding the two, we get 5.81 H which is close to the calculated value of 5.79 H. It should be mentioned that the similar measurements on this pickup using my LCR meter came out to 2.92 H for one and 2.89 H for the other measured individually. These values are within the tolerance of being equal to the simulated values. Now, here is the issue, the measured total inductance is 6.76 H which is as much as 1 H larger than the sum of the two individuals and also the simulated /calculated value. This is an issue, of course, if the larger value is used for capacitance calculation, which is why we are going through all this. Next, we will look at the inductances measures by other means than the LCR meter.

For the measurements I used the same pickup, an oscilloscope and a function generator. In the meantime, I had experimented with how big a value measurement resistor I could use in series with the pickup to get low noise and still an accurate result, it turned out to be 10k. The circuit as used in the simulation in Cir 1 was the test bed for this determination of the resistance value

Fig 4 shows the from earlier familiar waveforms, except these are actual measurements, the orange square wave is the signal provided by the waveform generator, 50% Duty, 4kHz, 2.6 V. The blue is the voltage measured across the 10k resistor.

Again the waveform is amplified in Fig 5. I have to mention, that the time is still in “scope” units that we have to convert to real time before we can us the curve for inductance calculation.

Again, we focus on the equation in Fig 6, that we need for calculation. If I did not mention it before, the dotted line represent the values as a “mean” value of all the measured points, the trend line. So using this line for our inductance calculation we get what could be characterized as a average value. If we take a closer look at Fig 6, we see that the measured values are mostly below the dotted line in the beginning of the graph, they are above the line in the middle and finally below the line at the end of the curve. To summarize this, at small currents through the coil the inductance is smaller than the average, at the high end the inductance is larger whereas the inductance is average in the middle. So the inductance will vary through the measurement range, so by now it is clear what kind of dilemma that puts a meter in, where on the curve do we measure the inductance and how? I do not have that dilemma, I can vary everything and get me the inductance that is suitable for the current level in my circuit. Here, however, I have chosen to use a trend line to give me an middle value, so to speak, basically for the sake of clarity and ease of comparison. Later on, I will show what it means to pick a specific section of the curve and calculate the inductance from that.

Inductance Calculation Example

From earlier we remember that the voltage was 2.6 V and the resistance for measurement was 9.81k so the dI/dt = 8061.8/9810 = .8218 giving us an inductance of Lwr = 3.16 H. That again is using the entire curve in Fig 7. Now, to drive the earlier argument home, we will focus on the top end of this curve, see Fig 8.

In this case we also go by the trend line and we see that the dy/dx = 5407.6 and as we have seen dV/dt = dy/dx and dI/dt = 5407.6/9810 = .5572 resulting in L = 4.72 H which is the inductance of the same coil (Lwr), but obviously much larger than 3.16 H, giving us a perfect example of how the inductance varies depending on where on the original curve we are.

You might say that calculating the inductance at the top end of the curve is a bad idea, true, it should have been calculated at the bottom end of the curve in Fig 7. This was done to show how much the inductance can vary over the course of the time of measurement. We can indeed do the same for any segment of the curve and find out that this will yield different values of L.

Correct, I would be remiss unless I do the same thing at the bottom end, as hinted. So here it is, keeping the original time stamp so you can see where on the curve in Fig 7 it was copied from. As that figure clearly shows, the slope of the curve is different, higher dy/dx that all previous ones, see Fig 8.

If we apply the same method of calculating Llwr, we find that Llwr = 2.79 H, which is almost what was measured by the meter, if you refer to Table 2, it is Lwr = 2.92 H.

Now, this brings us to lean in the direction that the meter measures at low currents which can be accomplished simply by applying short pulses, now the problem is that the meter has to do this for a wide range of inductances so we will just see this inaccuracy at other, possibly higher values of L. The good part is, that the pickup operates at low currents, this will be to our advantage in most cases. At least we got to a point where the discrepancy can be explained. I hope this was helpful in understanding how far from simple this really is.