Headphone amp output confusion (Tori Amp)

After reading the posts on headphone amps and power and what to use as an output series resistor I am confused. I built the Tori amp for my 32 ohm phones and assumed 'rload' was the load (my headphones). Should that be a 32ohm resistor? Or a 29 ohm resistor? The amp already sounds great by the way.

THanks, JEff

link to Tori Amp

Whatever. It works fine without any added output resistor.

I'm going to build this, my friend peter just had a listen to Jeff's and was really impressed, no easy task there. this is going to be built into my monitor controller

I've got loads of 5532s so I'll be using that just south of an ina134 line unbalancing circuit.

3 questions about the design.

1-I'm pretty sure there won't be dc leakage as I'm running off the ina134.
I think I'll add the offset trim circuit from the datasheet, but just for kicks,
how to test for DC leakage? just put a meter on the output of the ina134 and test for any DC voltage present, or do I need a certain test signal or input loading to be in place?

2-the tip 41s have a range of collector base voltage
from tip41a 60v , tip41b 80v, tip41c 100v -suggestions?

3-what am I looking to do with R2,I2 and TIP41,
this combination sets the bias voltage yes?

We know that in a transistor operating in its active mode, collector current is equal to base current multiplied by the ratio β. We also know that the ratio between collector current and emitter current is called α. Because collector current is equal to base current multiplied by β, and emitter current is the sum of the base and collector currents, α should be mathematically derivable from β. If you do the algebra, you'll find that α = β/(β+1) for any transistor.

Click to expand...

so I should be adjusting to set my idle current as α in relationship to my supply voltage which I'm thinking wiil be +-15?


Thanks
Kelly

Re questions: use a DVM and look to see if d.c. is present, and remember that a few millivolts won't hurt anybody much. Much more than that and something is wrong.

Any of the voltage ratings for the TIP41 will work fine with these rails.

Feedback via the R3 R2 network, and the offset voltage referred to the input of the op amp, determines the output d.c. voltage. It will be close to zero unless there is an input d.c. component. With feedback the output voltage of the op amp automatically assumes the value required to have the emitter of the TIP41 at about zero, provided that the op amp can supply enough current, which at quiescence is I2/(β+1). See PRR's earlier thread remarks about choosing a resistor or an active current source for I2.

Thanks Bcarso,
one more question- I've been looking around and reading a bit more but can't find exactly what is meant by the term active current source.

Would this be something like another transistor in a current mirror configuration or a second tip41 as a current sink like NYDave mentioned in the original thread???

for me, I think I best just stick with a resistor in there, but I wouldn't mind knowing what to look for... I'm starting to make some bit of sense out of these various transistor topologies.

Thanks for the info
Kelly

Kelly, an active current source (sink, although the term source is used often irrespective of the polarity of the current) is some combination of components that achieves a high incremental impedance at the designated output terminal. In some cases it's a two-terminal device whose terminal currents are nearly constant once there is sufficient voltage across them. You can buy these or make them out of one or more JFETs and sometimes an added resistor. Usually within a circuit it is more convenient to use bipolars and resistors. The circuits do resemble current mirrors sometimes, although those are optimized to function as such, whereas constant-current generators are meant to remain constant.

An example to get a handle on the sorts of numbers involved: suppose I have ten volts or so available in a circuit and want to have a given current flow from one node to another at about that ten volt difference in potential. Say I want a milliampere. I can put a 10k resistor there and get that current. But if my voltage varies with signal or whatever, the current will vary directly with the voltage by 100uA per volt change.

This may be just what I want it to do, but if not---if I substitute a bipolar transistor with a series emitter resistor and apply a voltage between the base and the other (supply rail typically) end of the resistor, accounting for the roughly 0.6V-0.7V base-emitter voltage magnitude plus an additional amount that will appear across the emitter resistor, I will have a current output from the transistor collector of about that excess over 0.6-0.7V voltage divided by the emitter resistor. Depending on the size of the resistor and the transistor characteristics, and the correct polarites for the transistor type and with sufficient magnitude of collector-emitter voltage, the collector current will be nearly constant despite the change in collector voltage. The number describing this output impedance will be given by the ratio of the change in collector voltage to the change in collector current. It is easy for typical situations to achieve megohms of impedance even though currents like a milliampere are flowing. It is as if you had a few-megohm resistor connected to a few-kilovolt power supply, and clearly a lot more efficient and convenient.

In the case of NYDave's suggestion, another TIP41 could be used with a suitable resistor from its emitter to -12V, and some bias network with fairly low impedance at the 41's base. For 200mA you could have a volt across the emitter R of about 5 ohms, thus needing about 1.7V between the base and -12V. The output impedance of this circuit will be fairly low, maybe around 10-20k, but still much higher than the resistor of 60 ohms that would give the equivalent current at quiescence, i.e., at zero volts out.

In fact the intrinsic properties of the collector of a bipolar in the common-emitter or common-base configuration are to have a relatively high impedance all by itself. When you put resistance in the emitter it introduces a local feedback effect that raises the impedance at the collector. For a while the relationship holds: new Z out = (Z out without R) times (1 + [device transconductance times R]). At some point other effects prevent a further increase in current generator output Z. Then one must resort to other techniques...

And don't forget pentodes, I hear our bottlehead brethren saying.

Brad

> the ratio between collector current and emitter current is called α.

Uh, yeah, but I've never found Alpha to be much use in BJT design.

> this combination sets the bias voltage yes?

No. Look at the whole circuit. If the op-amp "+" is at ground, and the op-amp is the least bit happy, the op-amp "-" input also has to be at ground. The op-amp "-" input is set by the voltage divider R3 R2. One end is grounded. The only way for the center (the op-amp "-" input) to stay at ground is for the other end (tied to the output) to be at ground. If it is not, the op-amp will waggle its output until it is happy.

The actual output will be "ground" plus op-amp error (few mV) and input signal. Assuming R2=10K, 2:1 divider, if the op-amp has 1mV of error the output will sit at 2mV. NBD. If the input signal is 1V (and neglecting op-amp error), the output must sit at 2V. That assumes infinite gain in the op-amp; a real-world result might be 1.9999V if the op-amp's gain is "only" 10,000.

> I should be adjusting to set my idle current

The idle current IS the "IDC" current. If IDC is 200mA as shown, then the transistor emitter current is 200mA. (Yes, due to Alpha the collector current is only 198mA for Beta=100, another NBD.) The Base current is about 2mA, but could be 10mA to 0.2mA depending what transistor you use. As long as the op-amp output can supply this current, the exact value does not matter.

So there are no bias adjustments. The output voltage IS zero. The transistor current IS IDC.

The choice of IDC is a matter of opinion and is a no-win situation. This amp is VERY inefficient: 5 Watts power for ~0.2 Watts output, ~4% efficient. I have another amp that can touch ~20%-40% efficiency, and of course any Class AB chip-amp can exceed 50%.

The minimum value of IDC is the maximum peak load current. If you want 3.2V peak in 32Ω, 100mA peak, then IDC must be at least 100mA. That gives over 100mW, which should be generally ample. If you actually get up to 90-99mA peaks there will be some rounding as the transistor current falls very low. Since this thing is a power-pig to begin with, I arbitrarily wrote 200mA, just to be sure of very-clean at 100mW-200mW power levels, and 500mW before THD degrades. But this would be 6 Watts per channel total dissipation, which is hard to manage. My 100mA idle headphone amp uses one heatsink from a 70W speaker amp, and still runs quite warm. Gonna need a lot of aluminum to run 200mA at 30V total supply. 100mA should be very ample and half the heat.

But headphones come in a wide range of impedances. For the same power, 300Ω phones need 1/3rd the current of 32Ω phones. If you only drove 300Ω cans, 30mA-60mA migth be a wiser bias, and a lot less heat.

Also note that for the same power, 32Ω needs 1/3 the Voltage of 300Ω phones. +/-15V is gross overkill for 32Ω headphones. +/-7V is plenty. So for a 32Ω-only amp, lower supply rails mean much less heat for the same performnce.

If we want to drive "any" headphone, we need both high current and high voltage. But we do not need both at the same time. Therefore the "IDC" part does not have to have low voltage drop at high current. In fact it can be a simple resistor. I like that: less to wire, and I trust resistors much more than active current limiters. More reliable, and less likely to have odd little nonlinearities.

Using the suggested +/-12V supplies, a 50Ω resistor in place of IDC gives 12/50= 240mA idle current. To estimate the peak output current, kill the transistor and see where the output sits. 12V * (32Ω/(32Ω+50Ω)) or 4.6V 146mA peak, 330mW RMS max output. This will generally be more than ample. For 300Ω loads, 12V * (300Ω/(300Ω+50Ω)) or 10.2V 34mA peak, 170mW RMS max output, which is still generally ample. Total dissipation is 5.76 Watts per channel. If you use +/-15V supplies, I would use a bigger resistor to keep the 32Ω power about the same yet not let total dissipation rise much. 75Ω or 100Ω are resonable values.

The classic current limiter these days is an LM317 with a resistor up its tail. 6Ω to 20Ω are good values depending how hot you want to run. The LM317 is a very complicated high gain amplifier with all the high frequency nonlinearity that implies; it probably is completely swamped by emitter action but I still like a simple resistor load rather than complicated amplifiers.

> I'm pretty sure there won't be dc leakage as I'm running off the ina134.

The INA134 itself adds a small and negligible DC error. I would not worry.

But the same thought applies: is any DC crap leaking out of your source? Most don't. If you know none of your sources leak DC (or subsonics), then omit all coupling caps. Myself, I sometimes have to plug-up strange gear, and sometimes with huge subsonics that will flutter headphone diaphragms to no good effect, so I like a 17Hz cutoff. (I don't try to judge "balance" on 20Hz-40Hz tones in headphones on location, so the small response droop is not a problem.)

> And don't forget pentodes, I hear our bottlehead brethren saying

Q1 and IDC can be replaced with hollow-state devices, obviously.

Because vacuum is a much worse conductor than a semiconductor, you'll have to run much higher rail voltages. If the tubes are fat enough to pump headphone impedances well, the op-amp supply rail can stay about the same, though for a low-Mu tube you might shift to +10V -20V to allow for negative-grid operation. (Positive grid operation should be practical, with the high current available from a chip. This might get double the current from a given tube, but few small tubes have grid-current ratings so you might melt a grid.)

This can give a tube amp without an output capacitor. NOTE: the servo loop won't work right when the tubes are half-hot. Let the tubes hot-up 10-20 seconds before you connect the phones.

Op-amp pushing hollow-state:


This is more a joke than a serious plan.

Thanks(?) to the op-amp, THD below clipping is point-oh stuff. Any "tube sound" will be suppressed by the op-amp.

The 6c33c tube is needed to get "serious" power in 32Ω; lesser tubes won't do near as well. The 6c33c is a refugee from USSR heavy electronics: a regulator pass-tube with 40W of heater demand, and we need four for stereo.

This simulates as ~250mA plate current, though I do not trust either the simulator or the russian tube factory too far. Some adjustment will be needed. It is very sensitive to supply voltages: lower will clip, but too high runs hot and is hard to drive.

This runs a little bit into grid-current at high level in low impedance. I really doubt a 5532 could melt a 6c33c's grids, and for speech/music signals I'm sure it is not an issue.

I played with zero-bias operation. Make R5=zero. Perfomance is good, but really better with R5 in there to jack-up TU5's plate resistance. If plate-power efficiency is critical, zero-bias may be the way to go. (But who cares about 20W in the plates when we need 80W/channel to hot-up the cathodes?) In zero-bias you get several mA grid current at several volt swing. Probably not a problem for a low-Mu tube with 60W plates, especially with plates idling around 10W.

Thanks to both you guys for the Sunday school lesson :wink:
No. I can see a million of em coming, but I'll pass. :twisted:
Did you hear the one about the Priest, the Rabbi and the Engineer?


I'll have to work on that one.

Thanks again.
Kelly

Good Grief---4 6C33's. But they would look so cool.

BTW by an interesting coincidence the latest issue of AudioXpress has an article by Borbely on a hybrid headphone amp, using a tube first stage and MOSFET outputs.