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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!
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