I wan't to make a headphone amplifier that can drive any type of headphones

Hi. I've gotten a real interest in headphone amplifiers. I've made and sold some Cmoy amps, but I wanted something more powerfull to drive my heavy AKG K-701 headphones. I bought some LM386 and made a circuit which is supplied by a 12V wall-wart. I even implemented a bass-boost feature and with this activated, it can really play with my AKG's

But I still don't feel I've crossed the finish line. I wan't to make and sell a headphone amp which will be able to drive any headphones to their limit, but my LM386 amp will have a hard time with headphones with an impedance of 600Ohms or so. I need something that can output a voltage of atleast +-13V and the LM386-4 can only output maybe +-8V. I've looked at chips like the TDA2004 and LM1875. These chips will definitely do the job, but these require a heatsink which will require more space and ventilation. These chips will also consume alot more power which will make running the amp off batteries impossible. I was thinking that battery operation could be an option, allthough getting 26V from batteries is hard, a 9V battery will probably do even though those with 600Ohm headsets would need an external power supply in order to get full power.

Are there any chips out there like the LM386 but with a higher supply voltage? I wan't my design to be simple, cheap and involve few components.

Dear Plecto
Hi
Try TEA2025 ( at bridge arrangement) or TDA2008 or TA8221 ( it need preamp )
Best Wishes
Goldsmith

The TEA2025 looks interesting. Does bridge arrangement mean that each amp does half of the job (amp1 the positive side of the sine-wave, and amp2 the negative side)? Will this make an ideal op-amp output +-12V even if the supply is 12V? Can't I use the LM386 the same way?

The TDA2008 and TA8221 seems to dissipate alot of heat aswell. The power dissipation graph of the TA8221 seemed to start at around 4-5W :S

No , my mean by bridge was not that . see it's datasheet , please . it means , that you use your headphone , floating with two IC .
and about TA8221 , it can give you , the powers up to 30W ! ( real power )
Best Lucks
Goldsmith

Those big TO-220 etc. chips is not an option, unless there is a type that don't require a heatsink. I see the bridge application on the TEA2025, but I have a problem understanding how this bridge application works, how does it make the potential output voltage higher? And aren't there really a chip out there with a power rating of 500-1000mW and with a max applied voltage of atleast 26V? How is the LM386 and the TEA2025 when it comes the sound quality (rather than power)?

Dear Plecto
Hi again
If you use bridge arrangement , your load will be float ( no problem ) . and when your output A , become high , your out put b will become low . thus , your out put wave will be v-(-v)=2v and you'll have twice voltage at out put .
And the other advantage is removing noises .
BTW : you'd better to use the same ICs as a bridge amplifier .and about TEA2025 out put power :it can give you the powers up to 5 watt , simply .
Best Wishes
Goldsmith

Plecto said:

Does bridge arrangement mean that each amp does half of the job (amp1 the positive side of the sine-wave, and amp2 the negative side)?

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No, both amplifiers conduct for the whole cylce. See the diagrams below.

The first diagram shows the two possible output connections of a "normal" amplifier. In both those cases, the maximum voltage swing across the loudspeaker can't exceed the total value of the supply voltage.

Assume, in the first example, there's a single supply of 12V. The voltage across the loudspeaker can swing between 0V and 12V (12Vpp)

In the second example there's a dual supply of plus and minus 6V (total12). The voltage across the loudspeaker can swing from -6V to +6V (12Vpp)





In the second diagram, there are two amplifiers, both with a single supply of 12V.

One of them inverts the signal with respect to the other.

Now, while the output of one amplifer is at a maximum positive excursion, i.e. +12V, the other output is at a 0V. On the next half cycle, the first amplifier's output will be at a maximum negative excursion, e.i. 0V, while the other will be at maximum positive, i.e. +12V. The total voltage swing across the loudspeaker in this case is 24V.

https://nwavguy.blogspot.com/2011/08/o2-summary.html

Best diy headphone amp I've seen so far. Built one myself and it sounds amazing!

Thank you for the brilliant explanation syncopator, I totally get it now. I have some problems with it though. One thing, this won't decrease current or wattage drain from either amp, right? I've been using some weaker NE5532 amps for my Cmoy projects, was thinking if bridging them will give me twice the power

The other thing is the ground connection. Now that either end of the speaker element is connected to each output of the op-amps, there is no "common ground", right? Do I just connect my two elements like this?:



In that case it will cause a problem if the two signals aren't in phase, but I guess they allways have to be sinse the input ground is common?

Can bridge mode be used with any op-amp? As long as the two are the same? Thing is that I allready have a bunch of LM386 so it would be better to use these rather than TEA2025. Even though the TEA2025 has more power, the 1W the LM386 gives each element of the headphones is allready enough to blow it up a couple of times But then again, if the sound quality of the TEA2025 is superior to the LM386, I would have to consider some TEA2025 instead I have heard people say that the LM386 is "ancient" and "toss out that crap"

Plecto said:

I totally get it now. I have some problems with it though.

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The two statements are mutually exclusive.

The other thing is the ground connection. Now that either end of the speaker element is connected to each output of the op-amps, there is no "common ground", right?

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The most misused word, and, I suspect, the least understood, in the field of electronics, is "ground".

The 0V line is usually called "ground", both in systems with a single voltage rail or with a dual (pos and neg) rail. It is also the case in equipment with multiple voltages, both pos and neg.

You can, if you so desire, connect either the 0V or the positive rail to "ground" in either of my diagrams.

Can bridge mode be used with any op-amp?

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Bridge mode can be used with any amplifiers, but the configuration, as far as audio is concerned, is beneficial only when the available voltage is a limiting or inconvenient factor in developing power in the load.

Your diagram shows two sets of amplifiers with their outputs in parallel. You can't do that.


Look at my second diagram again. There are two power amplifiers, one of them inverting, one of them not. That gives twice the "normal" voltage swing across the load. The price you pay is that power supply must now deliver twice the current that it would "normally" have to supply, and the power amplifiers now have to be able to carry that currrent too.
In view of that, it is unlikely that normal i.c. op-amps will find much application driving bridged loads.


I have seen wiring diagrams of a torch (flashlight), or a cell plus resistor plus l.e.d., where the author talks about "ground". It's a word which most people use, and most of those people do so without understanding what they are saying.

vis-à-vis the LM386, I don't see that it warrants the label of "crap". It's still widely used. Another one which might interest you is the TBA820M. I wasn't familiar with the TEA2025 until you drew my attention to it. It looks fine too.

That circuit I posted was wrong, I was aiming for something like this:

**broken link removed**

Allthough I'm having a problem picturing how that will work, but then again makes me wonder how any stereo signal with a common ground works The ground on minijack connections are connected together so theres no way around this problem.

The price you pay is that power supply must now deliver twice the current that it would "normally" have to supply, and the power amplifiers now have to be able to carry that currrent too.

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Why will the current double? The current through the load has to be the same in order to produce the same power in the form of sound, right? And this will make the total power equal if the chips themself aren't dissipating alot of heat.

There is an inherent limitation on the power a normal op amp can deliver to a load. When driving a sine wave and using the full rail to rail supply voltage, the max theoretical efficiency is 78.5%. In reality it will always be less because you can't operate completely rail to rail, so you might get ~65% efficiency at best (for example when driving a 13V sine wave off a +/-15V rail, you'll get 68%). The only way to get better efficiency is to use a diffent class of amplifier, like D, G, or H.

That circuit I posted was wrong, I was aiming for something like this:

Click to expand...

Unfortunately no valid atachment, so we still need to guess about your solution. One thing can be said however: A bridge circuit can't work with a popular 3-poles stereo jack connector. Their outer contact has to be connected to circuit ground - or whatever you call it.

Plecto said:

That circuit I posted was wrong, I was aiming for something like this: **broken link removed**

Click to expand...

The link doesn't work.

makes me wonder how any stereo signal with a common ground works The ground on minijack connections are connected together so theres no way around this problem.

Click to expand...

A sterophonic system has channels of amplification and, therefore, two outputs. Each output has two connections - four in total. Normally, i.e. in a non-bridged configuration, one connection of each output is to the system common (the bit which is often called "ground"). The other connection of each ouput can be considered the "signal".

Now, since one output connection of both channels is connected to "ground" it is only necessary to provide three wires to an output jack. One common to both channels, the other two carrying its respective signal (left and right).

As you seem to have realised, you can't do this with bridged loads. Both connections to each output carry a signal. i.e. neither of them has a connection which is common to the other channel, and no connection between the two outputs can be made. The outputs must have four wires - two to each 'speaker or earpiece.

Unless you have a set of 'phones with two conductors to each earpiece - very unlikely - you are faced with either replacing the 'phone's lead, or abandon the idea of bridgeing the two loads (the two earpieces).

Why will the current double?

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Because the voltage applied to the load has doubled. In a given load the current is directly proportional to the applied voltage. Double the voltage and the current doubles. Reduce the voltage by a third and the current reduces by a third. etc.

The current through the load has to be the same in order to produce the same power in the form of sound, right? And this will make the total power equal if the chips themself aren't dissipating alot of heat.

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Sorry, I don't follow your reasoning.


Your 'phones, the AKG K-701, have an impedance of 62Ω and a maximum input power of 200mW (presumably each earpiece). To develop that power they need about 3.6Vr.m.s. across them, that's 10Vp-p. Which makes me think that the LM386, or the TBA820M, would be satisfactory - if you use a 12V battery. This can be made of, for instance, two PX28L batteries in series. They're about 12mm in diameter and 25mm long, so they would occupy less space than the common rectangular 9V battery.


It occured to me to look at Rod Elliot's site as I seemed to remember he had some headphone amplifier informtion. This might interest you **broken link removed**

Unfortunately no valid atachment, so we still need to guess about your solution. One thing can be said however: A bridge circuit can't work with a popular 3-poles stereo jack connector. Their outer contact has to be connected to circuit ground - or whatever you call it.

Click to expand...

Strange that the attachement just works for me, I uploaded it using this forum, but I'll try imageshack or something the next time The circuit I uploaded was just the two loads have one of their ends connected together, as you would get when using a minijack.

I need something to work with a minijack along with most headphones out there so then bridging is out of the question? If so, what do I do?

Because the voltage applied to the load has doubled. In a given load the current is directly proportional to the applied voltage. Double the voltage and the current doubles. Reduce the voltage by a third and the current reduces by a third. etc.

Click to expand...

I am aware of Ohms law and Watt's law. Let's say the listener of this amp is looking for a sertain volume level (a sertain wattage), if he plugs in his 62Ohm headset and adjust the gain so it matches what he want's. He then plugs in his 600Ohm headsets and adjust the gain to get the same volume level. To get IE. 100mW at 62Ohms you would need 40mA. To get 100mW at 600Ohms you need 12.9mA. I really can't see why the current would increase. If the extra voltage is needed, the load would have to be alot bigger, and a bigger load calls for less current :S Unless someone want's to drive a speaker rather than a headset

Although mentioning battery operation, Plector also requested +/- 13V output. Obviously, these requirements don't go together well. If high voltage output is a serious objective, I would personally go for DC/DC converter supply from 2 or 4 AA batteries.

Referring to LM386, TBA820 and similar classical small audio amps, I think they serve their purpose for moderate quality requirements. But for a headphone amp, I won't like the > 0.1 % THD at low sound levels. I don't intend to start a big debate who'll hear it at all. I also presume that THD numbers aren't the best criterion for perceptable sound quality. But you can take it at least as an indicator for negative quality impact.

For state-of-the-art sound quality with a headphone amplifier, I would rather think about TI TPA6120, or the high output current audio OPs that have been previously mentioned in this thread. See TP6120 THD versus output level. THD is about factor 1000 lower than with LM386:

Although mentioning battery operation, Plector also requested +/- 13V output. Obviously, these requirements don't go together well. If high voltage output is a serious objective, I would personally go for DC/DC converter supply from 2 or 4 AA batteries.

Click to expand...

I was thinking that battery operation could be an option, allthough getting 26V from batteries is hard, a 9V battery will probably do even though those with 600Ohm headsets would need an external power supply in order to get full power.

Click to expand...

The TPA6120 looks absolutely beautiful, exactly what I am looking for if I'm reading everything correctly. The info in the datasheet was also brilliant and easy to understand. This can deliever 80mW into a 600Ohm load, which I believe to be enough (80mW is pretty loud, right?), and I see the graphs showing an output of over 3W into a 32Ohm load, which means that it would have no problem blowing up any low-impedance headset I plug into it

There is alot of mentioning of oscillations in the datasheet, will this amp oscillate easly compaired to IE. the LM386? Is the best option to give this a dual sided supply? And in that case, will a virtual ground do? It's hard to get a pure dual sided supply from a wall-wart

The "typical applications" in the datasheet shows very few external components which I find very delightful Will the following circuit really do the job?



With a dual sided supply like shown it the upper circuit here:

https://tangentsoft.net/audio/cmoy-tutorial/misc/cmoy-tangent-sch.pdf

And with 2x0.1uF caps in parallel with the 220uF caps perhaps?


The datasheet said that a feedback resistor of more than 10kOhm was illadviced (for some reason), but if you have a load of 600Ohms you will need a gain of far more than 10 to really push the headset to it's limits, right? I have a gain of 20 on my LM386 amp and with my 62Ohm headset this gain is a little too much, but it's not enough to kill my headset.

Final question. How do I go about if I wan't to implement a bass boost with the TPA6120?

TPA6120 is a high gain, high bandwidth amplifier which can more easily fall into parasitic oscillations when operated incorrectly. Besides good bypassing, the suggested output series resistor should be connected to isolate capacitive loads.

When operated with direct output coupling, a DC path of circuit ground to mid supply is absolutely required.

Plecto said:

I am aware of Ohms law and Watt's law. Let's say the listener of this amp is looking for a sertain volume level (a sertain wattage), if he plugs in his 62Ohm headset and adjust the gain so it matches what he want's. He then plugs in his 600Ohm headsets and adjust the gain to get the same volume level. To get IE. 100mW at 62Ohms you would need 40mA. To get 100mW at 600Ohms you need 12.9mA. I really can't see why the current would increase. If the extra voltage is needed, the load would have to be alot bigger, and a bigger load calls for less current :S Unless someone want's to drive a speaker rather than a headset

Click to expand...

It was assumed that if someone wants to drive a certain load at more power than can be easily delivered by a normal amplifier, then a dual amplifer driving a bridged load would be a possible solution.
The two-amplifer/bridged load can produce twice the voltage across the load than one amplifier can when run from the same supply, as has been explained.

The assumption is also made that the load would be the same for both cases.

With a 12V supply the best a single amplifier can give is (almost) 12Vpp, 4.2Vrms. If the load is, say, 32Ω, the maximum power developed would be 550mW. The maximum current would be 130mA.

With a 12V supply the dual amplifier (one of them inverting) can give (almost) 24Vpp. With the same load, 32Ω, the maximum power developed would be 2.2W, and the maximum current 260mA.

If you start thinking about different imedances you have to do the arithmetic for both cases to see what, if any, advantage you might realise from the bridged load configuration.

It was assumed that if someone wants to drive a certain load at more power than can be easily delivered by a normal amplifier, then a dual amplifer driving a bridged load would be a possible solution.
The two-amplifer/bridged load can produce twice the voltage across the load than one amplifier can when run from the same supply, as has been explained.

The assumption is also made that the load would be the same for both cases.

With a 12V supply the best a single amplifier can give is (almost) 12Vpp, 4.2Vrms. If the load is, say, 32Ω, the maximum power developed would be 550mW. The maximum current would be 130mA.

With a 12V supply the dual amplifier (one of them inverting) can give (almost) 24Vpp. With the same load, 32Ω, the maximum power developed would be 2.2W, and the maximum current 260mA.

If you start thinking about different imedances you have to do the arithmetic for both cases to see what, if any, advantage you might realise from the bridged load configuration.

Click to expand...

Bridge configuration is not an option since I wan't it to be compatible with regular headphones that use a mini jack connection.

When operated with direct output coupling, a DC path of circuit ground to mid supply is absolutely required.

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Will the dual supply with a virtual ground (as I linked) work?