[enemigo]

To really understand what we are building here, I would like to read more about the maths behind an amp. It will be easier to experiment with different component values if one know what will be affected. So I drew what I hope is a correct schematic over one channel of the amp section for a Cmoy. I tried calculating a bit, not sure if I got it right though.

EDIT: Put in new picture and changed alot of calculations

I need to know the voltage of my source of course. Should have used an osciloscope, but haven't got acess to one right now. So I measured AC voltage with my meter, not very accurate. It showed 0.18V max over one channel, so I assume that is the total p-p voltage!?

Values for the OPA2134PA at 15V:
Input impedance: 10^13 Ohm
Output, closed loop: 0.01 Ohm
Max output current: 40mA

I am assuming the voltage potential over the opamp is approaching 0V when AC voltage source is not connected. So the voltage potential equals the AC voltage.

Code:

[left]Total output voltage: V_Out = Gain * V_In = 11 * 0,18V = 1.98V Total output current, I use RL = 100 ohm (32ohm headphone + 68ohm R5): I_Out = V_Out / RL = 1.98V / 100ohm = 19.8mA Then I can calculate the output power for one channel: P_Out = V_Out * I_Out = 1.98V * 19.8mA = 39mW[/left]

19.8mA current output for each channel is probably the absolute maximum for this opamp, as it max current out is 40mA rated at 15V. Might be experiencing clipping here. So I take it a load of 100 ohm should be minimum here.

A power output of 2 x 0.039W is not very impressive. Provided I've interpreted th V_in p-p correctly.

An increased power output could be acheived by increasing the gain. An R4 of 15 kOhm will give a gain of 16, which is probably totally inecessary:


Code:

[left]Total output voltage (R4 = 15 kOhm): V_Out = Gain * V_In = 16 * 0,18V = 2.88V Total output current, I use RL = 100 ohm (32ohm headphone + 68ohm R5): I_Out = V_Out / RL = 2.88V / 100ohm = 28.8mA Then I can calculate the output power for one channel: P_Out = V_Out * I_Out = 2.88V * 28.8mA = 83mW[/left]

The opamp has no chance of delivering this much power, still we only get 83mW per channel. Are my calculations correct? I believe I've read that this amp should be able to put out around 0.5W.

Knut

 

[silvervarg]

I guess your note is slightly confusing and that is the reason you havn't got any answers so far.
The GIF you posted have both In- and In+ on the OP-amp connected to VGND, so you are sure not to get any signal through. You should draw where the input signal comes in. You typically want some kind of protection. If you just connect it straight to In+ (as I guess was intended) you could damage the input of the OP-amp.

Quote:

P=U*I --> P_out = 0.01 Ohm * 40 mA = 0.4W

You replace U with 0.01 Ohm. That is surely wrong...
And 0.01*0.040 becomes 0.004 on my calculator.

Using the components in your image:
R2 (from source signal to In+) just limits current into the OP-amp.
The gain of the OP-amp (assuming infinate closed loop gain) is set to R4/R3+1.
In yout example this is 10K/1K+1=10+1=11 times.
I believe the maximum output of that OP-amp is 40mA, so any load on the output for a certain voltage output that would exceed 40mA will cause clipping to happen, and that means major sound distorsion. To avoid this R5 is typically put in series with the load on the output.

Not really sure what else you want to know. Are you trying to keep the OP-amp out of the equations?
Also note that the voltages you drive the OP-amp with relative to the imput signal is really important. If you use +9/0V instead of +4.5/-4.5V you need to use a cap on the input to stop DC flow.

 

[enemigo]

silvervarg:
You must have one of 'em "scientific" calculators, mine use different laws of physics. Either that or I read the results wrong

Seriosly... Thanks for your reply, I probably should have spent some more time thinking before posting, as I should have seen some of the facts you pointed out. Anyway, I've edited my post, and it should probably be a bit closer to anything correct now. Sorry to make your reply look "meaningless"

Knut

 

[breez]

Question: do you need 0.5W output? Many headphones deliver 90-100dB with just one mW.

 

[enemigo]

No, I definately don't need that high output. I'm just trying to understand how this works, and was surprised at the output power, since I vaguely recall reading about several 100 mW. But I might be confusing the figures with a "larger" amp.

Knut

 

[enemigo]

Hmm, I really thought there would be someone here interested in this aspect of amps.

Anyway, I'd appreciate some confirmation as to wether the calculations above are correct.

Knut

 

[tangent]

Quote:

I would like to read more about the maths behind an amp.

Then you should read Op Amps for Everyone, available for free in PDF, or for pay in paper form.

Quote:

It showed 0.18V max over one channel, so I assume that is the total p-p voltage!?

Assuming you measured a sine wave, that's the RMS voltage, regardless of whether you used an RMS-reading meter. This is a little low, but typical for a portable source.

Quote:

I use RL = 100 ohm (32ohm headphone + 68ohm R5)

That's only valid if R5 is outside the feedback loop, which is usually not what you want. If it's inside the feedback loop, effective output impedance should still be under 1 ohm, but you'd have to test it to find out for sure.

Quote:

So I take it a load of 100 ohm should be minimum here.

This is based on the faulty assumption that you need 2V to reach reasonable loudness. With most 32 ohm phones, 0.5V should be enough.

 

[enemigo]

Hey, thanks for replying

Will read "Op Amps for every one". My portable source don't go very loud, so I'll assume 0.18v is pretty correct then, until I get my hands on an osciloscope.

Quote:

That's only valid if R5 is outside the feedback loop, which is usually not what you want. If it's inside the feedback loop, effective output impedance should still be under 1 ohm, but you'd have to test it to find out for sure.

Regarding the R5. I can't see how placing it within the feedback will create a different RL. Any reasonable R5 value will have practically no effect on the feedback current and resistance since the R4 is 10 kOhms. And wouldn't the RL be the product of loads in series regardless of the feedback?

Quote:

This is based on the faulty assumption that you need 2V to reach reasonable loudness. With most 32 ohm phones, 0.5V should be enough.

Not sure if we understand eachother correctly. I did not intend to say that 2v is needed to reach reasonable level for a 32ohm load when calculating this. I was just trying to calculate what the amp was capable of. And I assumed that 100ohm a minimum load to avoid clipping at theoretical max output.

Knut

 

[tangent]

Quote:

I can't see how placing it within the feedback will create a different RL.

It's not that it changes RL, it's that feedback almost "removes" it from the circuit. There are error voltages across it, but the op-amp compensates because feedback is taken from after the resistor. If you place R5 outside the feedback loop, this does not happen, so its value is simply added to RL.

Quote:

I was just trying to calculate what the amp was capable of.

You aren't given enough information in the op-amp datasheet to calculate this to any reasonable degree of accuracy. Just taking the maximum output current spec, that only applies into a dead short, typically, and it's probably done while the op-amp is kept at 25C. In the real world, the max output current will be different.

The way to find out what you want to know is to simply test it. Using a scope and a dummy load, you can find the clipping point, and then you'll have your answer.

 

[enemigo]

Quote:

Originally Posted by tangent It's not that it changes RL, it's that feedback almost "removes" it from the circuit. There are error voltages across it, but the op-amp compensates because feedback is taken from after the resistor. If you place R5 outside the feedback loop, this does not happen, so its value is simply added to RL.

Hmm, it still puzzels me that the R5 would have any effect on the feedback... I'm sure that it has though, since you say so

I'll try reading some about this, so you don't have to explain it all to me from the bottom. Hopefully I'll find some on this in "Op Amps for everyone"? That was a big document btw

Quote:

Originally Posted by tangent You aren't given enough information in the op-amp datasheet to calculate this to any reasonable degree of accuracy. Just taking the maximum output current spec, that only applies into a dead short, typically, and it's probably done while the op-amp is kept at 25C. In the real world, the max output current will be different. The way to find out what you want to know is to simply test it. Using a scope and a dummy load, you can find the clipping point, and then you'll have your answer.

I'll have access to a scope next week probably, so I hope to get some answers then. But as for a theoretical indication to what the amp is capable of, it's ok to calculate the way I've done? And take into concideration that the max values from the datasheets are under ideal circumstances... So, as long as I am within reasonable values for the opamp, my calculations will be fairly realistic?

Knut

 

[tangent]

Quote:

as for a theoretical indication to what the amp is capable of, it's ok to calculate the way I've done?

I haven't examined your calculations closely, but if you just apply Ohm's law and keep the datasheet limits in mind, yes, you should be able to get well within an order of magnitude of the real value. But if you want to get within 3dB, ya gotta test.