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 24th August 2012, 08:58 PM #2 diyAudio Member   Join Date: Aug 2012 I asked an electrical engineer at my Tech. School and he said that buffers don't necessarily increase power or deliver more current. What they can do is increase the current capacity beyond what the original source can deliver, and, somewhat more importantly, through having 0 output impedance they prevent the circuit from loading the source and "increase the voltage drop over the circuit"? So since more voltage goes to the circuit there are more amps. The internet could try to explain buffers better.
diyAudio Member

Join Date: May 2010
Location: Skokie Il
Quote:
 Originally Posted by crimsonokami I asked an electrical engineer at my Tech. School and he said that buffers don't necessarily increase power or deliver more current. What they can do is increase the current capacity beyond what the original source can deliver, and, somewhat more importantly, through having 0 output impedance they prevent the circuit from loading the source and "increase the voltage drop over the circuit"? So since more voltage goes to the circuit there are more amps. The internet could try to explain buffers better.
I think I understand your confusion.

A "unity gain" buffer delivers no voltage gain, but can deliver current gain. This results in a net power gain.

Your Ipod can deliver a limited amount of current output. It also has a very limited voltage output. This voltage output is sufficient to drive low impedance headphones to a loud level, but the output may be limited by current capacity. A unity gain buffer will be able to deliver enough current to drive a low impedance load to the full voltage output of your Ipod, which will result in more power delivered to the load. Another big advantage of a unity gain buffer is that it decouples the reactive load and back emf of a typical dynamic driver from the feedback loop, reducing distortion. In fact, the Nakamichi Stasis amps exploit this by employing a unity gain output stage that is out of the global feedback loop. I own an old Stasis reciever that I refurbished and the power amp section is quite excellent.

If you are going to use high impedance phones, then your Ipod will not drive them. You need an amp that can deliver a high output voltage. If you build an amp that could drive 10 volts rms @ 200 mA, it would drive any set of headphones you desire with ease and even overkill. Even 10 volts @ 50 mA would blow your Ipod headphone amp away.

diyAudio Member

Join Date: Sep 2010
Quote:
 Originally Posted by crimsonokami I asked an electrical engineer at my Tech. School and he said that buffers don't necessarily increase power or deliver more current. What they can do is increase the current capacity beyond what the original source can deliver, and, somewhat more importantly, through having 0 output impedance they prevent the circuit from loading the source and "increase the voltage drop over the circuit"? So since more voltage goes to the circuit there are more amps. The internet could try to explain buffers better.
Here are some schematics and plots that may help out. Every amplifier output stage, or current buffer, has a "hidden" resistor inside that is in series with the output. That is the output resistance of the amplifier/buffer. You can model it as in the schematics below, as an ideal voltage source for the amplifier (outputs voltage and has zero internal resistance because it is a fictional ideal concept) in series with a resistor. That combo in turn drives your headphones, which is modeled as a resistor with some capacitance across it for the cable in the most simple terms. In this case the ideal voltage source isn't constant, like a battery (yes, even a battery can be modeled as an ideal voltage source in series with internal resistance inside the battery), but rather outputs a 2kHz voltage sine wave.

Like Fast Eddie D wrote, something like an iPod has a fairly large output resistance. I don't know what the iPod's Zout actually is, but in the first plot I have it at 32 ohms. I wouldn't be surprised if it were higher, like 100R, 250R, or considerably more. But look what happens then. That internal 32R output resistance (R = ohms here, by the way) forms a 50/50 voltage divider with the 32R headphones. So like your EE at school said, half of the voltage from that internal ideal voltage source (I have it at 1.4V peak, which is 1.0V rms for sine waves here) winds up across that 32R internal resistor! Only half the voltage winds up across the 32R headphones.

But the second schematic is a current buffer, which has a lower internal output impedance. Here I've set it at 2 ohms. This could also just be the output stage of a headphone amplifier with high current capability. For example, the O2 amplifier in a thread in this section, has about 0.5 ohms output resistance looking back into the amplifier. The output chips can handle around 140mA per channel and they are wired up as current buffers. Both the high current amplifier output and the current buffer will have low internal output resistance. The big difference in terminology is that ideal voltage source, for the case of a current buffer, has a voltage gain of 1. The voltage out equals a control voltage input [sidenote, technically that ideal voltage source should be a "dependent ideal voltage source" controlled by an input voltage, but this is simpler to start with]. In the case of an amplifier though (that has voltage gain) the ideal voltage source will output a multiple of the control voltage input, like 3 times, 7 times, etc.

So with that second schematic, you can see from the plot that the current buffer with the low 2 ohm output resistance now passes most of the ideal voltage source's voltage on to the headphone load. Relatively little voltage is dropped across that low 2 ohm output resistance.

In the LTSpice plots below, blue is the output voltage of the ideal voltage source while green is the voltage across the 32R headphone. All the plots are done at 2kHz frequency. The left scale is voltage. Red is the current through the 32R headphone and the right scale is current. See how the current buffer, with the lower 2R internal output resistance allows 40mA to flow, while the higher output resistance 32R amplifier only allows 22mA to flow, yet both are being fed with exactly the same sine wave voltage from the ideal voltage source?

I hope this helps! Some stuff to ponder and Google at any rate.
Attached Images
 Amp with 32R Zout into 32R - circuit.png (28.7 KB, 163 views) Amp with 32R Zout into 32R - plot.png (30.1 KB, 159 views) Amp with 2R Zout into 32R - circuit.png (28.4 KB, 158 views) Amp with 2R Zout into 32R - plot.png (31.2 KB, 158 views)

Last edited by agdr; 25th August 2012 at 03:29 AM.

 25th August 2012, 05:13 AM #5 diyAudio Member   Join Date: Jul 2005 Location: Cambridge, England Look at the output section of the O2 amp designed by NwAvGuy. Using a couple of 200K resistors across the input (I do not remember what he used, but the value is not critical), two batteries, and a JRC4556AD connected as a buffer, you have a pretty exceptional amplifier for about \$5. You can try to parallel another IC like in the O2 amp, or just use a single one. It will provide all the current you need for a variety of earphones. You can try different input capacitors. I am using this with a Behringer UCA202, and no capacitors are needed until I bypass the ones in the UCA202. Great fun for minimal expensive and very quick results. Your simple amp will be comparable to any CMOY. The low impedance of the output of this IC gives a huge reduction in distortion. NwAvGuy has a link to an article that discusses in detail the effect of headphones with high impedance outputs. Recommended reading. Mine is built using point to point wiring on a small piece of perforated board. After such a low cost start--you can branch out and spend more money, time and effort into more complicated circuits, and compare the results. Great fun, this is what DIY is all about. Last edited by Cornelis Spronk; 25th August 2012 at 05:20 AM.
diyAudio Member

Join Date: Aug 2012
Thank you all for helping me understand amps, I think I'm finally getting it even though I was never good with circuits to begin with. I might have a talent for Software Engineering, but for Electrical Engineering I do not .

I didn't think about an O2 to begin with because i thought they were \$150~, but for parts I could make it for only a little more than the modified Cmoy i had in mind. I might also wait for NwAvGuy to release his O2 based desktop only version.

I think I also heard somewhere about an amp that had dual rail splitters, an output op amp buffer, and two op amps each doing half of both channels; but it probably wouldn't be any better than an O2 anyway

Quote:
 Originally Posted by Fast Eddie D Even 10 volts @ 50 mA would blow your Ipod headphone amp away.
From the NJM4556 datasheet, unless there's something limiting voltage, I think a cmoy with dual rail splitters plugged into a wall wart can manage 10v @ 70 MA.
Which should barely manage to drive very insensitive 32ohm ~ 600ohm headphones to normal db's. But it probably won't do anything for self-energizing electrostats and earspeakers, but what a rare breed those are.

Quote:
 Originally Posted by agdr I don't know what the iPod's Zout actually is.
Just for future reference, Many of apple's portable devices seem to vary in Zout but most fit within 5 ~ 10 and .7 surprisingly for the newest iphones i hear.

Quote:
 Originally Posted by Cornelis Spronk Look at the output section of the O2 amp designed by NwAvGuy. Using a couple of 200K resistors across the input (I do not remember what he used, but the value is not critical), two batteries, and a JRC4556AD connected as a buffer, you have a pretty exceptional amplifier for about \$5. You can try to parallel another IC like in the O2 amp, or just use a single one. It will provide all the current you need for a variety of earphones. You can try different input capacitors. I am using this with a Behringer UCA202, and no capacitors are needed until I bypass the ones in the UCA202. Great fun for minimal expensive and very quick results. Your simple amp will be comparable to any CMOY.
Sounds almost like a generic datasheet circuit for op-amps, would definitely be cost effective if I just wanted to try and see if I even need an amp!

 25th August 2012, 07:52 AM #7 diyAudio Member     Join Date: Nov 2005 Location: San Antonio re: the edit quandry A way I find to easily visualize it is this: Imagine the voltage follower as a "black box" with a large value resistor in parallel at its input and a small value resistor likewise at its output. Not at all hard to imagine! The output voltage follows the input voltage, and so Ohm's Law demands that there be a current increase at the output. Therefore output V*I must be greater than input V*I. __________________ It is error only, and not truth, that shrinks from enquiry. - Thomas Paine
diyAudio Member

Join Date: May 2010
Location: Skokie Il
Quote:
 Originally Posted by sofaspud re: the edit quandry A way I find to easily visualize it is this: Imagine the voltage follower as a "black box" with a large value resistor in parallel at its input and a small value resistor likewise at its output. Not at all hard to imagine! The output voltage follows the input voltage, and so Ohm's Law demands that there be a current increase at the output. Therefore output V*I must be greater than input V*I.
That's very practical. One more step and we have the Thevenin equivalent circuit. It is very handy for analysis of circuits.

More on Thevenin equivalents here. Thévenin's theorem - Wikipedia, the free encyclopedia

diyAudio Member

Join Date: Aug 2012
Quote:
 Originally Posted by sofaspud re: the edit quandry A way I find to easily visualize it is this: Imagine the voltage follower as a "black box" with a large value resistor in parallel at its input and a small value resistor likewise at its output. Not at all hard to imagine! The output voltage follows the input voltage, and so Ohm's Law demands that there be a current increase at the output. Therefore output V*I must be greater than input V*I.
I think I understand that, but the current increase is only inside the voltage follower and relative between the input and output. It doesn't physical deliver more current to the secondary circuit if the primary circuit has low enough impedance and high enough current handling, right? If the primary circuit already had those a voltage follower would be pretty useless, since they're used to optimize the interaction between two circuits.

 25th August 2012, 08:54 PM #10 diyAudio Member   Join Date: Jul 2005 Location: Cambridge, England The output of the O2 is a generic opamp buffer circuit. You will find that when opamps are used, they are designed to be used in a simple way. Extra circuit tweaking is done for discrete designs. The 45556AD is designed to be better that most other opamps to give the amount of current needed for a headphone. If you need a more accurate amplification at a high output resistance, other opamps would do a better job. Put your opamp and a DIL socket, and try different ones. The buffer circuit is always the same.

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