Folded cascode headphone amp

[nelsonvandal]

I've been toying around with a folded cascode amp, inspired by AD829, AD797 and LT1469. The aim is as usual portable use. I've come up with this simplified circuit. It simulates very well, seems robust (changing resistor values doesn't affect it that much), swings to within 1 V from rails, simulates well down to 3 V supply. R22/23 and R24 could be pots for trimming DC-offset and bias. 100 Ohms and 47 uF in series with the input simulates my iRiver IHP120.

Please comment.

[AndrewT]

The three front end currents seem very low.
Is this to suit battery power?
You would probably be better using a FET input opamp for the front end if battery life is a concern.
How low can you take the output stage bias currents and still retain good crossover distortion suppression?

The 6p nested feedback cap looks unusual (can't read the number).
What about leaving alternative pads/traces/routes for other compensation strategies?

What Vbe is being predicted for the two folded cascodes?

[darkfenriz]

Looks nice to me.
100uF caps to ground seems a little overkill to me though

[nelsonvandal]

Battery life is a concern, but not a major one. A cople of mA's more or less doesn't matter. What simulations should I run to find "proper" currents to the front stages? Increasing the currents doesn't alter the square wave response, phase response, voltage swing or distortion.

I don't really know how to detect crossover distortion but with a 30 Ohm load and 2.2 V output, I can see a slightly broken sine wave when the bias on the output transistors is 650 uA. With a 300 Ohm load, I can see no crossover distortion at any level. Please help me to learn how to measure or detect crossover distortion with LTSpice!

I'm just not clever enough to predict Vbe. I only know how to simulate. I usually try with different resistors, LED's, voltage supply etc to see to that there are some margins.

The two small (5 mA) capacitors are remnants from fiddling with the AD797 topology. They improve the sqaure wave, eliminating peaking without slowing it down like when the feedback capacitance is raised.

I'm done with monolithic opamps. There are some OK ones like AD797, AD829, AD8599, AD825, but even with my low level of electronic knowledge, I've made an all discrete amp that I find better.

[jam]

You might try replacing the Widlar current mirror with a Wilson mirror.

Jam

[nelsonvandal]

Originally posted by darkfenriz
Looks nice to me.
100uF caps to ground seems a little overkill to me though

Together with the 10 Ohm resistors, they reduce rail noise by a magnitude on the input stage. I doubt that it's audible though. I guess I could do it with a larger resistor and smaller capacitor, but I don't want to loose too much voltage swing.

[nelsonvandal]

Originally posted by jam
You might try replacing the Widlar current mirror with a Wilson mirror.

Jam

Like this, AD829

or like this


What improvents are expected? They simulate rather the same. If it's better, should I use it on the output stage as well?

I was hoping to keep the number of parts down since I inted it to fit inside a pocket sized box.

[AndrewT]

The red LED should have between 1.5 & 1.7V across it when lit.
You have only 0.3mA passing and I suspect this reduces the LED voltage and places it on a steep part of the IvsV curve.
What voltage is dropped across the LTP collector (load) resistors. I can't tell if they are 220r or 820r.

[jam]

Nelsonvandal,

Both circuits would work the first circuit which is the improved Wilson mirror is slightly better but might be overkill.

Q14 in the bottom circuit keeps the junction of Q14 from heating and unbalancing the mirror. I find it improves the performance of the mirror sonically ia an amplifier circuit I designed in another forum.

It might be worth a try on the output stage but I suspect it won't be as dramatic.

Good reading

http://users.ece.gatech.edu/~mleach/...jtmirrsu06.pdf


If you are not a fan of feedback don't use this circuit.

Jam

[nelsonvandal]

Originally posted by AndrewT
The red LED should have between 1.5 & 1.7V across it when lit.
You have only 0.3mA passing and I suspect this reduces the LED voltage and places it on a steep part of the IvsV curve.
What voltage is dropped across the LTP collector (load) resistors. I can't tell if they are 220r or 820r.

The drop should be almost exactly 1 V. The resistors are 820 Ohm.

The current through the LED is 0.86 mA. Sorry about the crappy picture quality.

I'm glad you're willing to help. Where on the Scottish borders do you live? I've spent vacation in Coldstream a couple of years ago.