Choosing the Proper Headphones for your Metal Detector
Headphone Theory: Freq., Impedance & Sensitivity

 

STEVE GHOLSON
 
* The following article is a reprint of two posts made on the AZO Forum by Steve Gholson on May 20th and June 6th, 2003.

There are numerous things that contribute to the performance of a metal detector and its’ ability to achieve maximum efficiency. We are all aware of the mechanical things that we can do to “connect” us to the detector. For instance, we can add a bungee, lengthen or shorten the shaft, make adjustments of the arm strap, and change the way in which we carry the battery to name just a few. These are all things that we do to mechanically interface with the detector. We depend upon our sense of sight and touch for these things, but how is it that we connect the other important sense - our hearing?

The only true interface that we have which is hardwired to us is the headphones (I exclude speakers and meters that are used on some detectors because these are generally built into the detector and cannot be interchanged). So in a nutshell, we are depending wholly upon one sense that we have to connect us electrically, if you will, to the detector, and that is our sense of hearing. We can look at the detector as the transmitter and the headphones as the cable or interface to the receiver, which is none other than your ears. Our goal is to make sure that we get maximum transfer of energy or power from the detector to our ears.

The subject of headphones might on the surface seem like a simple subject, but quite the contrary, this is a very complex issue and requires more than this post to really get your arms around it. I will touch on some of the more dominant points that should give you the basics and some food for thought.

Aside from the obvious things like cost, comfort and seal (from ambient noise) there are a few things that you should be aware of before you purchase a set of headphones. If you look at the manufactures’ specifications, there will be a long list of things that describe the headphones. Assuming that the only purpose for the headphones is to use them with your detector, you can eliminate all but about three of these specifications. As an example, you don’t need to worry about distortion because we are dealing with an essentially pure sine wave within a very narrow range of frequencies. If the headphones were to be used for music or speech reproduction, this would be a very important parameter. So, what do we look at on the list of specifications?

Frequency range is one of the parameters that you should take note of. Even though a healthy ear has a range from about 60 hertz to 18,000 hertz, the audio frequency range of your detector is probably somewhere between 200 hertz and 2,500 hertz. If the headphones cover this detector range, they will work just fine regarding this parameter.

Headphone Impedance & Headphone Sensitivity – Because these two parameters are related, I would prefer to discuss them mutually instead of exclusively. Let us look at a mock example of two different headphones and their specifications versus the detector requirements. Most detector manufactures will specify the recommended headphone impedance in ohms. This may or may not be the actual impedance of the detector at the audio output plug (headphone jack). For our example, we will call the impedance at the headphone jack 600 ohms. Now the two different headphones that we are looking at have the following specifications:

Headphone #1 – Impedance = 600 ohms, Sensitivity = 98 dB
Headphone #2 – Impedance = 60 ohms, Sensitivity = 98 dB

If we had to choose one of these headphones, we would pick headphone #1 which has an impedance of 600 ohms and matches our detector output impedance. Why is this so?

In classical impedance matching theory, we know that we get maximum transfer of power from one circuit to the next when these impedances are equal (matched 600 ohms to 600 ohms). So, as a first general statement, all other things being equal you want the impedance of your headphones to match the audio output impedance of the detector. If neither of your choices match the detector impedance, and they have the same sensitivity (dB) pick the one that is the closest in impedance value to your recommended headphone impedance. What if the sensitivity is different between the two headphones (i.e., 80 dB vs. 100 dB)? Hang with me a while longer and we’ll get back to this question.

Let’s get back to the example. To prove that we get maximum transfer of power (signal) from the detector to the headphones if we use the 600 ohm headphones vs. the 60 ohm headphones, a little math is in order.

We will call the detector output impedance (600 ohms) Rs and #1 headphones R1. Because Rs and R1 are in parallel, we can use the following formula to determine the total impedance (Rt1):

Rt1 = Rs x R1 / Rs + R1
Rt1 = 600 x 600 / 600 + 600
Rt1 = 360,000 / 1200 = 300 ohms

Now for headphone #2:

Rt2 = Rs x R2 / Rs + R2
Rt2 = 600 x 60 / 600 + 60
Rt2 = 36,000 / 660 = 55 ohms

Now if we say that the audio output voltage of the detector is 2.2 volts (E) and the current is say, .0037 amps (I), we can calculate the voltage that will be delivered to the headphones.

Voltage (E) = I x R (ohms law)
Voltage across headphone #1 = .0037 x 300 = 1.1 volts
Voltage across headphone #2 = .0037 x 55 = .2 volts

At a quick glance, you can see that headphone #1 is getting about 5 ½ times as much voltage across them as headphone #2.

But what I really want to know is how much power is being delivered to each of the different headphones. We can use the following formula to calculate the power in each case:

P (power in watts) = Voltage^2 (voltage squared) / R1 or R2

Headphone #1 = 1.1 x 1.1 / 600 = .002 watts (or expressed as milliwatts multiply by 1,000 = 2 milliwatts (mw)
So headphone #1 is being driven by 2 mw of power.

Headphone #2 = .2 x .2 / 60 = .0007 watts
And headphone # 2 is being driven by .7 mw

Again you can see that the headphones with the matching impedances of 600 ohms (#1) will have much more power delivered to them than the headphones with 60 ohms impedance (#2). The point that I am trying to make here is that you can take any headphone impedance above or below the output impedance of the detector and none will have as much power delivered to them as the ones that have been matched to the impedance of the detector. Remember, maximum transfer of power takes place when the impedances are matched in a circuit.

Please understand that the math examples are theoretical in nature and can be used as a guideline if you really want to dig into this subject.
In reality, for you to make the calculations you will need to know something about the circuit (audio output amp, etc.) that is driving the headphones. With the proper equipment, you can measure the output voltage, impedance, etc. otherwise you should be able to understand enough about headphones to make general comparisons.

To digress for a moment, when we speak of audio output voltage from the detector to the headphones we must keep in mind that this voltage will have some minimum and some maximum value. For instance, with the Minelab GP Extreme, the output voltage will be at a minimum with the detector threshold set to fully counter-clockwise and the volume knob set fully CCW and no target under the coil. The audio output voltage will be at a maximum value with the threshold set to fully CW position, volume knob set fully CW, boost in the deep position, and a substantial target under the coil. I believe that the GP generates a maximum audio voltage of approximately 2.2 volts. So in general, the audio output voltage will start from some minimum value, threshold only no target, and increase in value from a small target to a large target (I am sure this is not a linear function). We can now move on to headphone sensitivity.

Headphone sensitivity (or loudness if you will) is generally specified as dB’s (decibels) at 1.0 milliwatt input power at 1.0 kilohertz (1,000 hertz) frequency. An example would be, 98.0 dB/ 1 mw, 98.0dB spl (don’t worry about this) / 1 mw and 98.0 dB/ 1mw/ 1 kHz. etc.

First and foremost, what is a decibel (dB)? A decibel is nothing more than a power ratio and was derived from the last name of the inventor of the telephone, Alexander Graham Bell. The fundamental unit is the Bel and a decibel (dB) is 1/10 of a Bel. The original purpose of the Bel was to conveniently calculate the losses in electrical power of telephone signals transmitted over long lines. As was mentioned, the dB is a power ratio and a convenient way to compare one signal level against another signal level or against some standard reference signal. The formula for calculating dB’s with reference to power levels is:

dB = 10 x log(10) P1 /Po

This formula say’s, dB is equal to ten times (10 because we are using decibels and not Bels) the log (base ten) of the ratio of P1 divided by Po.

Let’s go back to our headphone examples. Headphone #1 had a delivered power of 2 milliwatts, which is P1 in the formula. Po is equal to 1.0 milliwatt (this is our standard reference). Plugging the numbers into the formula we get:

dB = 10 x log10 (2.0 / 1.0)
dB = 10 x .301
dB = 3.01

This 3.01 dB is the value (providing we have 2.2 volts at that moment) that will need to be added to the specified sensitivity of 98.0 dB, which is equal to 98.0 + 3.0 = 101 dB. This is the sensitivity of the headphones #1 that have an impedance of 600 ohms.

Headphone #2:
dB = 10 x log10 (.7 /1.0)
dB = 10 x (-1.5)
dB = -1.5 dB

Since this number is negative, we will need to subtract it from the original sensitivity of 98.0 dB which would be 98.0-1.5 = 96.5 dB. This 96.5 dB is the sensitivity of headphone #2 that have an impedance of 60 ohms.
The difference in sensitivity of headphone #1 and headphone #2 is equal to 4.5db. NOTE - since 1.0 milliwatt is used as the standard, to be technically correct, we should say dBm and not dB.

The question is can we hear this difference of 4.5 dBm when comparing headphone #1 to headphone #2? The answer is a little complex, but the bottom line is yes, we can hear this difference. With a healthy ear, we can detect changes of about +/- 1.0 dBm with steady state tones. Our number of 4.5 dBm is 4 ½ times this amount. By the way, for music and speech we need changes of about +/- 3.0 dBm. In either case, we should be able to hear the difference in loudness between the two set’s of headphones.

As you can see, Impedance and Sensitivity are related and to the trained ear, we would notice a difference in the headphones used in our example. Let me explain what I mean when I said the answer is a little complex.

As I stated earlier, the audio output voltage from the detector to the headphones has some minimum and maximum range. In the example we used 2.2 volts as our maximum starting point, but instead let’s call this our mid-range of the voltage swing. If we plug in both headphones at the same time (you can’t do this because you will change the total circuit impedance), you would notice that headphone #1 would sound louder. If you could remember exactly how loud the signal from headphone #1 was (you couldn’t) and unplug them leaving only headphone #2 connected to the detector, you could increase the threshold output voltage of the detector making headphone #2 just as loud as headphones #1 were providing you were originally in the low to mid-range and had more voltage to give them. In our example the specifications for sensivity (98 dBm) was the same for both sets of headphones. Now suppose that in another example with different headphones the impedance of both were 600 ohms (as required for best match since our detector output is hypothetically 600 ohms). If headphone #3 had a specification of 98 dBm sensitivity and headphone #4 was 80 dBm you would naturally choose headphone #3 which would give you more power output. If you are not comparing apples to apples and both headphones have different sensitivity levels and different impedance levels (other than the detector output impedance) the only way to know which pair will work best is to know the circuit parameters and do the math. If you have access to both headphones you could also just give them the old taste test (i.e., put them on and just listen).

In closing, I suspect that you will notice the biggest difference in headphone loudness (because of the difference in impedance and sensitivity) with your threshold set to some nominal value (same threshold position for both headphones) and no target under your coil (i.e., background and ambient noise only). In my opinion, you should choose a set of headphones that are the closest in impedance to the detector output impedance with the highest sensitivity specification as possible. We should also be aware of the fact that some manufacturers might specify the headphone impedance at a higher level than the detector output impedance in order to conserve power or voltage consumption. If we match the detector impedance to the headphone impedance (which is preferred), you will immediately drop one half of the audio output voltage across the headphones, but you will have maximum transfer of power to the headphones. Take a look at our first example using headphone #1. We stated with 2.2. Volts and with the impedance matched, we dropped 1.1. Volts across the headphones. Whether you will notice this power consumption as related to detecting time I suspect not, if you are using the large gel cell type of battery - but this is a different topic that I don’t want to make general statements about. With a post of this nature some of the math or statements might not be error free. Any and all corrections or input would be most welcome.

The following information is included as points of interest:

By doubling the power in a circuit you get an increase of approximately 3 dB.

By cutting the power in half in a circuit you get a decrease of approx. 3 dB.

By doubling the voltage in a circuit with equal impedances you get an increase of approx. 6 dB.

By cutting the voltage in half in a circuit with equal impedances you get a decrease of 6 dB.

Prolong expose to 90-100 dB signals could be harmful to your hearing.

The dynamic range of our hearing is approx. 120 dB.

Loudness is very subjective. Complicated systems have been devised to quantify loudness.

The ears are logarithmic in nature.

The ears have uneven sensitivities to sounds at various frequencies.

Some frequencies in the range of 100-4,000 hertz appear louder than other frequencies.

The threshold (minimum value) of our hearing is 0 dBm.

It takes approx. 1 dB change in power with a steady state tone to notice a change in loudness from one signal to the next, but we can hear a signal with less than 1 dBm of power.

As a person ages it is quite normal for them to lose a little of their hearing, especially on the high frequency side. For a young person in their mid- twenties, a frequency range of 100 – 18,000 cycles per second (hertz) would be normal. As you age, the upper limit starts to fall off, and I would expect that by mid-life the upper end would fall to around 13,000 cycles per second.

The question of whether you need to worry about missing targets because of this hearing loss is in my humble opinion…. no.

After conducting some rather simple and crude tests on the Minelab GP Extreme, I have the following to share with you.

As you know, the variable tone control works in conjunction with the threshold. As you increase or decrease the tone setting you in fact change the audio frequency of the threshold either up or down.
With a steady threshold and background noise only (no target) the audio frequency range that the detector will generate is from approximately 550 – 950 hertz. With the tone control set to fully counter clockwise the frequency is about 550 hertz, mid- range is 700 hertz and fully clockwise it’s about 950 hertz. The threshold and the tone control work as a pair.

The variable signal control is used for changing the audio frequency that you hear when you hit a target. Just as in the tone control, by increasing or decreasing the signal control, you adjust the target audio frequency either up or down. The signal control is dependent upon the tone control and the following applies;

A) Tone control fully CCW (counter clock wise)
Signal control fully CCW yields a frequency of 750 hertz
Signal control mid- range yields a frequency of 900 hertz
Signal control fully CW yields a frequency of 1550 hertz

Tone control mid- range
Signal control fully CCW yields a frequency of 900 hertz
Signal control mid- range yields a frequency of 1050 hertz
Signal control fully CW yields a frequency of 1650 hertz

C) Tone control fully CW
Signal control fully CCW yields a frequency of 1100 hertz
Signal control mid- range yields a frequency of 1300 hertz
Signal control fully CW yields a frequency of 1950 hertz

The volume control has no effect upon the audio frequency and is used to set the amplitude limit (loudness) that is obtained from hitting a rather large target, kind of like a limiter.

As you can see from the above data, the highest audio frequency output from the detector is approximately1950 hertz and the lowest frequency being about 750 hertz (with target). This high and low limit should be well within the range that even the most stubborn ear can hear. It is the amplitude of the signal that gives most people a problem and not the frequency of the signal.

This information is academic in nature, but nonetheless maybe worth understanding. For those not interested in the numbers, just set your detector for a smooth and pleasing sound for your ears and that’s all that is required!




Thanks to everyone that has acknowledged the post and given some feedback. Kimble asked an interesting question about using an audio amplifier to help compensate for different headphone impedances, sensitivity, etc.

Kimble, you are correct when you say that you could make up the difference from one headphone to another headphone by using an amplifier. There is however something that we need to be aware of when we use an audio amp. In electronics or most anything for that matter, you do not get something for free; you have to pay the price. In using an amplifier, you are getting an increase in output voltage or power to the headphones, the price you must pay for this will be in the signal to noise ratio that the circuit has. In other words, the amplifier not only increases the voltage that is delivered to the headphones, it also increases any other noise that is generated within the circuit. There is absolutely nothing wrong in using an amplifier provided it is really needed. Let’s take a look at one example where headphones would be beneficial.

As we know, the threshold knob will control the audio voltage that is supplied to the headphone jack. This voltage will have some minimum and some maximum value. On the GP Extreme for example I have measured this value at the headphone jack without headphones to be 0 – 600 millivolts with no target under the coil. In my particular case, I need to set the voltage that is being supplied to the headphones (Koss TD80’s in this test) to about 150 – 200 millivolts to get the proper threshold level in the headphones. Now if Mr. X who had hearing that was much worse off than mine, he might not hear anything at all! If fact, his hearing is so bad that he can’t hear the threshold even with the voltage set to maximum. What to do? Well I know that with an amplifier used in the circuit and the threshold set to maximum we can increase this voltage level to about 2,000 millivolts or 2 volts.
Now Mr. X can set the voltage level he needs which might be say 900 millivolts and hear the threshold just fine. Naturally the target signals will be louder also. The most important thing is to just hear the target, not how loud it is. The amplifier is correcting for the poor hearing in this case. Use an amplifier if you feel you need one to help with your hearing or you think your headphones are not up to par with what they should be. You should not use an amplifier for fashion or just because… We are all different and have different hearing abilities.

Something that must be stressed at this point is the signal level at which you run your detector threshold. You must set your threshold to the absolute lowest signal level possible to give yourself the best chance of hearing those faint signals. As we all know, a big percentage of all large gold found starts out as a faint whisper (it’s normally deeper in the ground). To improve your chances of hearing these signals, set your threshold so that it is not breaking up but giving you a steady and smooth
sound. Again, set the threshold signal level to the lowest level possible. Let me show you a little simple math that will verify this and make the picture a little clearer.

Let us look at four different threshold levels (E1) for our example.
Threshold #1 = 100 millivolts, #2 = 200 mv, #3 = 400 mv, #4 = 800 mv
The formula we use for calculating dB when using voltages is;
dB = 20 x log10 E2 / E1

Remember that our ear needs at least 1.0 db change for us to detect any changes in level. If we take threshold #1 which is 100 mv and hit a target that raises it to 112.5 mv we will get a change of 1.0 db.
Db = 20 x log10 112.5 / 100 = 20 x log10 of 1.125 = 20 x .0512 = about 1.0 db. So the ratio of 112.5 to 100 is 1.125
Now let us use this ratio of 1.125 on the other threshold levels and see what happens.

#2 = 20 x log10 225 / 200 = 1.0 db (notice 225/200 is the same ratio as #1 which is 1.125).

#3 = 20 x log10 450 /400 = 1.0db

#4 = 20 x log10 900 / 800 = 1.0db

Notice that the signal of #1 is 12.5mv above the reference, #2 is 25mv, #3 is50mv, and #4 is 100 mv.

What this says is, that if you hit a target that is 12.5mv above your threshold level #1 of 100mv you would get a 1.0 db change in signal level and notice there was a target. But if you had your threshold level at any of the other values, you would not notice any change what so ever (with all things being equal!). Let’s look at #2 with a signal level that is only 12.5mv above the reference level of 200mv and see what happens.

#2 = 20 x log10 212.5 / 200 = 20 x .0263 = .53 db
As you can see, there is only a change of .53db, which means you would not detect any change.

The moral of the story is, KEEP THE THRESHOLD AS LOW AS POSSIBLE…