quad 405: error correction / current dumping amplifiers

ary

原帖由 millwood 于 2007-12-16 05:10 发表
here is a more modern version of the quad 405.

who care about whether a amplifier circuit is modern version or not,

May you want to say "here is a more modified version of the quad 405." ?

ary

欢迎讨论Quad405

millwood

when we talk about error correction amps, we cannot do so without mentioning bob cordell.

Bob is a lengendary figure in audio and together with Otala (Curl to a lesser extent) started the discussion of tim which basically lead to today's thinking of low negative feedback.

anyway, Bob designed a mosfet amp that is capable of 0.1% thd at 20khz OPEN LOOP, with just 5% resistors.

the complete design is avaialble at his website (http://www.cordellaudio.com/poweramp/mosfet.shtml) but here is the output stage.

wongkm33

Ary兄，如果图中R11删除的话，R9 220是不是也应该短路？若果还保留它，Op正、负供给电压岂不是会不等？

millwood

a few interesting things about the cordell ops (output power stage):

a) r50/r51 gate stopper: at just 47ohm, they are very very small for a power mosfet in a voltage-feedback amplifier. typical values are 110ohm or 220ohm.
b) c8/r52 and c9/r53: those are dampening networks. part of the reason he can use small gate stoppers r50/r51 is because of those dampening networks. also notice that the dampening networks are grounded, not to the sources of the mosfets.
c) c10: I never figured out why it is there.
d) this is a typical tripple darlington design (actually if you count the mosfets, it is a quardrople darlington design), very similar to the Leach amp (which by itself is a great amp too).

ary

原帖由 wongkm33 于 2007-12-16 20:28 发表
303475

Ary兄，如果图中R11删除的话，R9 220是不是也应该短路？若果还保留它，Op正、负供给电压岂不是会不等？

对，应该短路的，不过不短路关系也不是很大，这里R9好像是为了Quad的一款喇叭的保护而用的。

保留它的话，如果用在其它喇叭驱动上的话，正电压的峰值会受点影响（即原作的保护功能吧应该），不过一般用不到那么大的峰值的。

可以短路的。

millwood

you may also see another popular amp being dubed "error correcton" by Graham Maynard (who is a jerk, if you ask me).

here is the linnk to Maynard's site: http://www.zen22142.zen.co.uk/Circuits/Audio/thegem.htm.

the GEM (as Maynard would call it) is a bastardization of the jlh1969.

Rather than the fast curret feedback input stage, Maynard opted to use a differential input stage making it an voltage feedback amp.

it is essentially two amps in one: a class A jlh1969 and a Class AB amp.

there are two ways to view this amp:

a) a class C amp with a class A amp for better cross-over distortion: if you cross out the parallel output deivces, the part circulated in red is a class a amp. and the rest looks very much like a typical class B amp.

b) alternatively, a JLH1969 + a negative current sink. if you take out the part circled in green, you have almost a jlh1969 (the difference is the input stage). the part circled in green is essentially the negative current dumper / current sink.

either way, it is interesting to look at but I wouldn't build it: as is, it is not stable.

流泪的背影

http://fanyi.cn.yahoo.com/translate_txt

millwood

JLH also designed a class AB amp (see the attached schematic).

the GEM is essentially the marriage of the Class AB JLH + JLH1969 + differential input stage.

millwood

ary:

the error nulling in that circuit is actually quite difficult that's why R6 is a variable resistor:  you adjust it until you reach the lowest THD. in simulation, you will see that by giving different values of R6 you can get non-sine output from sine input.

Anyway,

Z1=r2;
Z2=r6;
Z3=impedance on L1 = sqrt(Rl1^2+(2*pai*f*L)^2). where Rl1 is the serial resistance of L1. if Rl1<<2*pai*f*l, Z3=impedance of 2*pai*f*L.
Z4=impedance of (R4+C4//R5), which is roughly 1/(2*pai*f*C4) if R4 is sufficiently small and R5 is sufficiently big.

the error nulling is reached if Z1*Z2=Z3*Z4.

or r2*r6=L1/C4.

in this  analysis, we omitted a lot of non-ideal situations, especially the output impedance of the opamp.

the gain is r2/r1 (about 5x), and the input impedance is r1 (very small).

you can reconfigure the input so that it is non-inverting.

the current load on the opamp depends on your output pairs and your load. One issue with this particular design is that the opamp used isn't powerful enough for anything greater than 10ma output current so you will need to use a beefier opamp (lm3875 or tda2030) if you want to drive a big load.

the original 405 buffered the open collector output of lm301 with one transistor and then followed it up with a triple darlington. so the current load on the opamp (lm301) is minimal.

millwood

原帖由 audioboy 于 2007-12-16 15:37 发表
hi.
I think in the original Quad405,LM301 is not a fast OP.

for the error correction to work, you actually don't need an opamp those the high gain of the opamp makes the analysis a lo easier.

the later versons of 405 (606 for example) abandoned the opamp idea and went fully discrete, which I think is the better approach.

lm301 was a fast opamp by its days.

ary

原帖由 millwood 于 2007-12-16 21:06 发表
ary:

the error nulling in that circuit is actually quite difficult that's why R6 is a variable resistor:  you adjust it until you reach the lowest THD. in simulation, you will see that by giving different values of R6 you can get non-sine output from sine input.

Anyway,

Z1=r2;
Z2=r6;
Z3=impedance on L1 = sqrt(Rl1^2+(2*pai*f*L)^2). where Rl1 is the serial resistance of L1. if Rl1<<2*pai*f*l, Z3=impedance of 2*pai*f*L.
Z4=impedance of (R4+C4//R5), which is roughly 1/(2*pai*f*C4) if R4 is sufficiently small and R5 is sufficiently big.

the error nulling is reached if Z1*Z2=Z3*Z4.

or r2*r6=L1/C4.

in this  analysis, we omitted a lot of non-ideal situations, especially the output impedance of the opamp.

the gain is r2/r1 (about 5x), and the input impedance is r1 (very small).

you can reconfigure the input so that it is non-inverting.

the current load on the opamp depends on your output pairs and your load. One issue with this particular design is that the opamp used isn't powerful enough for anything greater than 10ma output current so you will need to use a beefier opamp (lm3875 or tda2030) if you want to drive a big load.

the original 405 buffered the open collector output of lm301 with one transistor and then followed it up with a triple darlington. so the current load on the opamp (lm301) is minimal.

你好，感谢回复。
我先就这个帖子继续讨论（非常欢迎这样的讨论氛围，呵呵）

1.Z3/Z4的算法我知道，所以我才得出Z3*Z4=L3/C4这个结论（另外，你算Z3的时候不应该加一个sqrt，没有这个的开根号的，不过没关系，我们都知道你不是故意写成这样的）

我是说按这个“桥”式算法：Z1*Z2与Z3*Z4好像是不相等的，你算一下看看？

2.增益用R2/R1来替代可以的，我本以为你还会考虑到15uH的电感，不过没关系，其阻抗的确是很小的。可以忽略。

3.OP电流的问题，我想你理解不对的。任何功放都有电压放大和电流放大，所以你这里谈OP的电流是没有意义的。它是电压放大级，你对它要求电流放大是完全没有必要的。如果要如你所说用LM3875的话，根本没必要用405了，我想你如果理解了405的工作原理你至少能够得出一个基本的结论：405本身设计的初衷就是所谓的“电流放大零失真”。所以我反对谈OP的电流要求，不知有没有道理。

4.欢迎讨论405，请就我其它问题的疑问展开讨论，没关系，如果我有错误的理解请尽管指出来，
I would very appreciate if you 讨论 with me, pls do not hold any hesitation!

ary

qpzmg

看得懂的就看，看不懂的……也看一下

linaux

原帖由 ary 于 2007-12-16 20:40 发表


对，应该短路的，不过不短路关系也不是很大，这里R9好像是为了Quad的一款喇叭的保护而用的。

保留它的话，如果用在其它喇叭驱动上的话，正电压的峰值会受点影响（即原作的保护功能吧应该），不过一般用不 ...

保留的话运放供电实测就是不平衡的，所产生的零飘难道是靠负反馈强行抑制的。如果是这样的话还是短路的好啊

millwood

1.Z3/Z4的算法我知道，所以我才得出Z3*Z4=L3/C4这个结论（另外，你算Z3的时候不应该加一个sqrt，没有这个的开根号的，不过没关系，我们都知道你不是故意写成这样的）

我是说按这个“桥”式算法：Z1*Z2与Z3*Z4好像是不相等的，你算一下看看？

because nothing is ideal.

the Vanderkooy / Lipshitz paper ("Feedforward error correction in power amplifiers", 63rd AES) expanded almost the exact same circuit (picture 1) and the error nulling conditions to be (picture 2).

the calculation got you roughly there under highly simplified assumptions.

2.增益用R2/R1来替代可以的，我本以为你还会考虑到15uH的电感，不过没关系，其阻抗的确是很小的。可以忽略。

15uh at 20khz has an impedance of <2ohm, or 0.2% of the 1kohm resistor. When was last time you needed gain to be within 0.2%? or you have used 0.1% resistors?

3.OP电流的问题，我想你理解不对的。任何功放都有电压放大和电流放大，所以你这里谈OP的电流是没有意义的。它是电压放大级，你对它要求电流放大是完全没有必要的。如果要如你所说用LM3875的话，根本没必要用405了，我想你如果理解了405的工作原理你至少能够得出一个基本的结论：405本身设计的初衷就是所谓的“电流放大零失真”。所以我反对谈OP的电流要求，不知有没有道理。

the whole error correction amp thing comes from actually feedforward. the Vanderkooy / Lipshitz paper showed that the quad405 is just a form of it.

The essence of error correction is actually in error nulling. The notion is that if there is any error signal (due to imperfect amps, or distortion or group delay), it will show up on the virtual ground of the opamp.

and if you pick up that error signal and null it against the output, you will get a distortion free output signal, as shown in the 3rd picture.

the ancillary amp A2 is picking up the error signal at the inverting end of the main amp. the summation is done by Z3 (for the main amp) and Z4 (for the ancillary amp).

so in  theory, you do not need any current output capabilities from the main amp (thus the use of ne5534 in their demo circuit and lt1022 in my demo circuit). All the main amp does is to reproduce the error signal on its virtual ground.

[4.欢迎讨论405，请就我其它问题的疑问展开讨论，没关系，如果我有错误的理解请尽管指出来，
I would very appreciate if you 讨论 with me, pls do not hold any hesitation![/quote]

you should really get hold of the AES paper and read through it. it has some good analysis.

ary

原帖由 millwood 于 2007-12-16 22:01 发表


because nothing is ideal.

the Vanderkooy / Lipshitz paper ("Feedforward error correction in power amplifiers", 63rd AES) expanded almost the exact same circuit (picture 1) and the error nulling conditions to be (picture 2).

the calculation got you roughly there under highly simplified assumptions.

你提到这篇文章，但为何回避下面这些内容呢？下面的内容很不合理？有错误？

ary

原帖由 millwood 于 2007-12-16 22:01 发表


The essence of error correction is actually in error nulling. The notion is that if there is any error signal (due to imperfect amps, or distortion or group delay), it will show up on the virtual ground of the opamp.

事实上，如果你看过Walker（405的发明者）关于405的原理，你应该知道，它并不是为了弥补OP的失真，完全不是。OP的失真是电压级的失真，玩过电路的人都知道，失真度是极小的，稍微留意一下都可以控制在0。01%以下。

我这里再替Walker转达一下，他发明这个电路是为了弥补电流放大级的失真，乙类的交越失真等等，对吧！

你在仿真图里面的运放后面的失真正是由于这个“桥”电路本身带来的，没有桥的话也不会失真的，所以你这样谈运放后面输出失真是没有意义的。

millwood

that's the same circuit. the one I posted is 6(a).

What the authors showed is that you can expand the "feedward" concept to the quad405.

and if you flip two more pages to page 12 you will see their mention of the output impedance for the main amp.

Equation 11 you had quoted is the nulling condition if a lot of assumptions went right.

unfortunately, none of us live in the ideal world where capacitors are nothing but capacitors and inductors are nothing but inductors.

so your nulling conditions will get a little bit more complicated than what high-school math can reproduce.

ary

原帖由 millwood 于 2007-12-16 22:01 发表

so in  theory, you do not need any current output capabilities from the main amp (thus the use of ne5534 in their demo circuit and lt1022 in my demo circuit). All the main amp does is to reproduce the error signal on its virtual ground.

你的意思是不需要电流放大？ 开玩笑？

millwood

原帖由 ary 于 2007-12-16 22:22 发表


事实上，如果你看过Walker（405的发明者）关于405的原理，

I am sure that Walker would really appreciate your conveying his points.