Amp Power output graph help

[joessportster]

Below is a screen capture, The way I read this is at 1.5ohm setting on the amp I would have 1 watt power at 32 ohms

 

Can someone else with more knowledge chime in 

 

[joessportster]

The amp has a lot of adjustments and output resistance is one of them, Options are 1.5, 32, and 69 ohms output resistance

 

Now with speaker amps I know it is desirable to match the speakers with the amps output IE...8 ohm speakers on the 8 ohm tap on the amp

 

 

Headphones dont seem to be locked into this matching, It is generally desired to have higher power on Planer Headphones, so though my cans are 38 ohms it appears setting the amp at 1.5 output resistance I will have 4X  more power at the headphones 38 ohms ???

Edited September 24, 2015 by joessportster

[joessportster]

Further can anyone expand on reasons why matching the resistance closer would result in better sound / performance  ??

 

 

I will say that the lower resistance of 1.5 seems to affect the bass a bit, Maybe a tad deeper ??

[joessportster]

This is a link to the amp in question

 

http://www.garage1217.com/garage1217_diy_tube_headphone_amplifiers_002.htm

[babadono]

Looks like at the 1.5 ohm output load setting you will get a little over 3 times the power out into 32 ohm headphones. That's about +5db. So a little more headroom. Bass is better, yea OK I can believe that. Why someone would not want that I don't know. Seems to me like we want to deliver the most power possible to the load, or at least have the ability to do so.

[Guest thesloth]

Now with speaker amps I know it is desirable to match the speakers with the amps output IE...8 ohm speakers on the 8 ohm tap on the amp

 

This is a hybrid amp with no output transformer. It uses SS for the output.

 

 

Tube amps with output transformers reflect the speaker load back to the primary to load the power tubes, which is why it is important to run them from correct speaker tap.

 

 

The reason they give an output resistance option is for tonal interactions with the impedance curve of the headphone driver. Drivers impedance no matter if they are in a loudspeaker of a set of headphones is dependant on frequency, it's not flat.

 

 

EDIT:  I should have stated that as output impedance increases the amp will start to behave less like a voltage source and more like a current source.

Edited September 24, 2015 by thesloth

[joessportster]

Higher output (69 ohms) will deliver higher current ?

 

I see alot of threads that state planers need more current which DRIVES them (controls the driver) Better, sound reasonable ??

 

 

I am trying to learn a little more about matching equipment do's and don'ts,

 

I just read an article that states for the best power to the speaker, High Impedance into the amp, Low impedance out of the amp going into higher Impedance speakers, was best ?? (so I am thinking this would be more voltage source)

[Guest thesloth]

Higher output (69 ohms) will deliver higher current ?

 

The output device can only supply so much current. The higher the output impedance of the device the more power will be lost in the output resistance and not in the load. So less power, to the load (speaker). One would model it as a simple potential divider.

 

In a perfect world the "device" would have zero output impedance and would be an ideal voltage source. But it doesn't so we call this output resistance and we model it with a resistor, it's in series with the load, which is another resistor.

 

 

Vin would be the perfect voltage source, R1 is the output resistance and R2 is the load (speaker), Vout would be the voltage at the speaker tap.

 

 

Vout =  (R2 / (R2+R1) ) * Vin    

 

 

So if the device puts out 5v and it has an output impedance of 1 and the load is 32 ohms then we will have 4.84 volts across the 32 ohm load.  Power would be (4.84^2) / 32 = .73 watts.

 

Now if you switch the output impedance to 69 ohms then you will have 1.58v across the 32 ohms. Power out to the load (speaker) is now 73mW or .073 watts.

Edited September 25, 2015 by thesloth

[joessportster]

Another Epiphany based on your replies, If you have higher current (higher impedance output) you will loose some DB / volume, in order to gain control

 

sound right ??

 

Please pardon my dumbing it down, I get a tad lost in some of the math 

Edited September 25, 2015 by joessportster

[joessportster]

Can you run those numbers for me using my Kenzie amp, 32Ohms out and 600Ohms out with 38 ohm headphones 

 

 

Dac is putting out 82 ohms into the kenzie, and at 32 ohms the kenzie is supposed to do 200mwatt

Edited September 25, 2015 by joessportster

[joessportster]

I just found at 69Ohms output I can run the amp into clipping (Not good)

[Guest thesloth]

Can you run those numbers for me using my Kenzie amp, 32Ohms out and 600Ohms out with 38 ohm headphones

 

Lets use 10 volts for our perfect voltage source to make the math easier.

 

38 ohms is the load (headphones) so it's "R2"

 

600 ohms out is;

 

10 * ( 38 / ( 38 + 600 ) ) = .59 volts

 

 

32 ohms out is;

 

10 * ( 38 / ( 38 + 32 ) ) = 5.42 volts
 

[Guest thesloth]

http://www.transcendentsound.com/Amplifier_Output_Impedance.html

 

Here is some good info.

 

 

The dampening has less effect on headphones because of the smaller mass me thinks, but the frequency response stuff is what I was trying to explain.

[joessportster]

 

Can you run those numbers for me using my Kenzie amp, 32Ohms out and 600Ohms out with 38 ohm headphones

 

Lets use 10 volts for our perfect voltage source to make the math easier.

 

38 ohms is the load (headphones) so it's "R2"

 

600 ohms out is;

 

10 * ( 38 / ( 38 + 600 ) ) = .59 volts

 

 

32 ohms out is;

 

10 * ( 38 / ( 38 + 32 ) ) = 5.42 volts

 

 

If I am reading this right we are talking volts which equals watts IE WPC Correct ??  these are not showing current or amps Correct ??

 

if the above are correct is there a way to figure the current / amps

[joessportster]

http://www.transcendentsound.com/Amplifier_Output_Impedance.html

 

Here is some good info.

 

 

The dampening has less effect on headphones because of the smaller mass me thinks, but the frequency response stuff is what I was trying to explain.

Damping referring to control or higher current ??

 

Thanks for the link headed there now  , I think I am getting  the idea if not the math 

[Guest thesloth]

That Kenzie amp uses output transformers. Which tap you use and headphone you use will have a greater effect due to it's reflective load nature. i.e. changing the tap and the load will change the output tubes operating conditions.

 

Doing power calculations for each amp are completely different.

 

With the Kenzie amp no matter what tap you are on if you go higher so say 56 ohm headphones on the 32 ohm tap you will get less power and less distortion. But if you ran 300 ohm headphones with the 600 ohm tap you will get more power out but with higher distortion.    Roughly.

[joessportster]

That Kenzie amp uses output transformers. Which tap you use and headphone you use will have a greater effect due to it's reflective load nature. i.e. changing the tap and the load will change the output tubes operating conditions.

 

Doing power calculations for each amp are completely different.

 

With the Kenzie amp no matter what tap you are on if you go higher so say 56 ohm headphones on the 32 ohm tap you will get less power and less distortion. But if you ran 300 ohm headphones with the 600 ohm tap you will get more power out but with higher distortion.    Roughly.

Thanks, I am going to keep playing with this math till I figure it out, the link is helping 

[Guest thesloth]

Thanks, I am going to keep playing with this math till I figure it out, the link is helping

 

 

 

I am not the best at explaining things so I do apologize. A good link is always the cure. Enjoy!

[joessportster]

 

Thanks, I am going to keep playing with this math till I figure it out, the link is helping

 

 

 

I am not the best at explaining things so I do apologize. A good link is always the cure. Enjoy!

 

No apology necessary, you jumped right in to help and I do appreciate it, 

[WMcD]

FWIW:

 

A voltage source is a theoretical construction.  No matter what the load and resulting current, the voltage remains the same.  Say 10 volts. 

 

(But note that the voltage source is also a perfect conductor.  It has no resistance at all.  All the current it puts out must also flow though it.)  In the d.c. world a car battery is a pretty good voltage source in that if you need 1 amp, it will do it.  10 amps -- fine. 100 amps, still pretty good and the battery is still putting out 12 volts.  (Of course that doesn't last forever.)  Note that this is a simple series circuit with, for example, the DieHard and the starter motor.  All the current which flows through the starter motor also flows through the battery.  So the car battery (call it a voltage source) must have a very low internal resistance.  Say 0.0001 ohms.  Keep that in mind.

 

Of course in real amplifiers we can't achieve that because otherwise you put a load of 0.000000001 ohms and get an enormous about of current and enough power to light up Cleveland. 

 

A current source is a theoretical construction too.  No matter what the load, the current remains the same.  So if you set the current to 10 amps and put no load on it (say just the air gap between the output terminals) it would produce enough voltage to cause lightning across the terminals at a current flow of 10 amps.  It is like an arc welder.

 

The theoretical description of an amplifier out put is a voltage source in series with a resistor.  That resistor is the output impedance.  You may read that SS amps have a very low output impedance, less than 0.01 ohms.  This really does not mean you can light up Cleveland.  It really means that as the speaker "ohms" load varies with frequency, say from 4 to 16 ohms, with a midpoint of 8 ohms, the voltage applied to it by the amp does not vary.  This holds until a current limit is reached which is determined by the amp's power supply and output transistors.  So only the most moosey of amps can drive a 4 ohm load without the voltage sagging.

 

There is another issue of output impedance.  Assume the diaphragm is moving and of course it is attached to the voice coil (input wires attached to the amp) and the voice coil is in the gap of the speaker magnet.  When the amp drives the speaker the speaker is a motor. But if the music goes silent and the diaphragm is moving, it acts as an electrical generator.  This is why a speaker and a dynamic microphone are two versions of the same thing.

 

But, if a generator (say the rotating type) is attached to an electrical load, the more the electrical load (which is to say a low resistance attached to it) the harder it is to turn the crank or mechanical drive.  So, when the speaker is moving and the amp is not sending it power (or not in proportion to the movement) the speaker (now generator) must send current (by reverse voltage) into the amp it is attached to. 

 

This means that the movement (unwanted movement) of the diaphragm is damped by the force needed to crank the generator.  The less ohms in the load, the more (harder to crank) damping force is being created. 

 

But what is the electrical load on the speaker turned generator.  It is putting current into the amp. The amp is the series circuit of the output resistance (which is the output impedance plus the resistance of the theoretical voltage source of the amp).  But remember we said the theoretical voltage source (DieHard) has no (or little) resistance.  Therefore the so called output impedance of the amp is now the load on the speaker turned generator.

 

Therefore, an amp with a low output impedance (or resistance) is better at damping out unwanted movement of the diaphragm.  To belabor the point, this is because it is more of load (lower resistance)  allowing current to flow and make the diaphragm harder to move when it should not be moving.  This is the "control."   You may also read about "damping factors" of amplifiers. This is "it."

 

The next issue is the amp as a current source.  Of course all suppliers of electrical power can create current when there is a load put across it.  A laboratory type or theoretical current source will adjust voltage so that the same current is always delivered as the load changes.  An audio amp is not set up to do that.

 

But now comes the series resistor which can be outboard of our amplifier and in series with the load.  We would like to achieve current which does not change even though the resistance (impedance) of the speaker changes from 4 ohms to 8 ohms to 16 ohms, or even more.   If fed with a constant voltage source, the current is going to change in relation to the resistance and Ohm's law.  At freqs where the resistance is 16 ohms we get some current, but at frequencies it is 8 ohms it is twice that, and at frequencies where it is 4 ohms he current doubles again.

 

What to do if we want the current though the speaker to remain constant (or nearly so)?

 

The answer is to use a 500 ohm resistance in series with the speaker where the amp is really a constant voltage source.  We don't monkey with the amp of course.  But now the voltage source has load of the 500 ohm resistor in series with the speaker.  In this situation, resistances add.  So the impedance "seen" by the amp is, 504 ohms, 508 ohms, and 516 ohms.  The load is not changing much, and therefore the current is not changing much either.  So we have almost constant current.  Not really but is accomplishing a lot with 500 ohm resistor.

 

Is this good for power to the speaker.  No.  We've very much restricted the current though the speaker.  There is a great voltage drop across the 500 ohm resistor.

 

You can apply this to the headphone amp when there is something like a 32 ohm output resistance and a 32 ohm speaker / earphone.  That is easy math.  Half the power is being lost in the output resistor.  We see that generally in the graph, but it does not work exactly. 

 

How is this for damping or control?  Not good.  The speaker / generator / earphone is putting current into the 32 ohm load (plus its own 32 ohms) the generator is not hard to crank and so the diaphragm is not damped or controlled.  Maybe that is good for bass freqs where the diaphragm is resonating.  The resonance is not controlled.  It may make the unit sound "warm" or even a bit tubby.  And you've lost 3 dB or so as compared to a near voltage source amp with a low (1.5 ohm) output impedance.

 

- -

 

I know this is getting long but we can tie up a few more concepts.

 

We see that pre-amps may have a "low" output impedance and they feed some device with a high input impedance.  This means that the pre-amp is acting closer to a voltage source (relatively) and the load is very light (high resistance) this means the pre-amp output voltage is never bogging down by the load. We see things that the output impedance of the pre-amp is 1000 ohms and the input of the tape recorder or whatever is 5000 ohms.

 

 An important concept here is that the voltage is the analog to the music.  We are not worried about power transfer at all.  Just accurate voltage transfer. Impedance matching does not come into play.

 

What could get in the way of this transfer or bog down the output of the pre-amp?  It is the "load" of the connecting cable.  The cable has "capacity" which is the same thing as a capacitor a/k/a "condenser.".  They are like little rechargeable batteries but instead of chemical reactions there are electric fields.  The wire has "capacity" which must be filled up (the electric field has to be filled up).  But if the output impedance of the pre-amp is low, it can supply the current necessary to charge up the electric field (and actually, discharge it to (damp it). There is no voltage drop across the output resistor of the amp.  So we get an accurate transfer of voltage.  (Inductance can also get in the way as we have to charge up the magnetic field. Same concepts.)

 

- - - -

 

Again far afield from earphones and amps, but we're talking about constant or near constant current sources created just by the use of a series resistance. We see it in trying to find the T-S parameters of a speaker. Therefore we should look at the this other big use of a constant current source and see how things are different aspect of the same principles.

 

The big mountain in impedance is the resonance Fs of the moving system.  How far down the resistance falls determine the Qts. By using weights or a box we can find the Vas. These are the important small signal T-S parameters and we can't always get them from a manufacturer.

 

In testing a woofer for T-S parameters we are really trying to determine the resistance of the woofer's electrical input as it varies with frequency.  We know that resistance is the voltage across the speaker input terminals divided by the current though the speaker.  So we could hook up a volt meter and a current meter and do the math as we use a signal generator to vary frequency. 

 

Is there an easier way?  Suppose we could keep the current constant as the speaker resistance changes.  Again, a constant current source.  Remember above we saw that if we use a series resistor of a high value, the total resistance in the loop does not change much.  504 to 508 to 516 ohms in our example above.

 

To make the math easier let's use a 1000 ohm series resistance so we've got a loop resistance varying from 1004 to 1008 to 1016 ohms.  The resistance and therefore current is hardly changing all and the current though the woofer is not changing much at all even though it's own resistance is varying from 4 to 8 to 16 ohms (or higher).  But we need to know the actual numbers of speaker resistance and want to determine them. 

 

Just for the sake of math we're going to have an amp outputting 1000 volts. (We can scale that down of course.)  So now the current though the speaker is just about always 1 amp despite its own resistance changing with frequency from 4 to 8 to 16 and higher.  The loop resistance changes from 1004 to 1008 to 1016 ohms.  (Are you sick of reading, smile!)  The current is fairly constant at 1 amp.

 

What does that do for us?  Now we can measure the voltage across the input terminals of the speaker.  We know that the actual resistance is the measured voltage divided by 1 amp.  All of a sudden, we are able to measure the actual resistance of the speaker with a volt meter.  The voltage drop IS the resistance because we have a near constant 1 amp in the divisor.  So we can vary the frequency of the signal generator and take out resistance measurement in the form of voltage.

 

- -  -

 

Anyway, I hope the above has put a lot of the different aspects of voltage sources and current sources and other concepts in the same realm.

 

WMcD