你知道JHL1969的静态电流到底应该取多少吗?

WANGCLZL

又认真看了一遍,支持楼主30楼的观点.应该加分.好帖!

911xp

支持楼主的研究精神

我只知道厂家标推挽的甲类功率是P＝I2 X  2R

但如果单端甲类效率达到50%，那就值得商榷

关于zen（就是1969的单电压版），我曾请教过locky_z 兄，他用的方法计算效率是25％，我认为是对的，因为这个是单端。


楼主敢于出来捅娄子，勇气可嘉，理不辨不明，我们这旁观者学到了很多，而这之前，很多人不懂装懂。

楼主牺牲一人，抛砖引玉，探索真理，应该支援

hy1635

我看在玩1969的人不少,有些朋友问的问题比较简单,比如电源要多高,静态电流电流应该多大,牛的功率多大才够用,输出功率有多少,总说1969功率太小,有20W就好了等等,我想集中一下高手们的观点,给新手们一个指点.

millwood

原帖由 911xp 于 2009-2-24 01:35 发表
他用的方法计算效率是25％，我认为是对的，因为这个是单端。

jlh1969 is a classic push-pull. with perfect transistors (Vce(sat)=0), you get a 50% efficiency - the excel spreadsheet I posted will confirm that.

in reality, you get about 33% efficiency out of BJTs and 25 - 30% out of MOSFETs.

ZHU110

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认真学习了

xitonghan

原帖由 millwood 于 2009-2-18 08:27 发表


xlf0602 is right about this.  and the theory behind it is fairly elementary: in a push-pull Class A design, the current output from the positive half or the negative half swings from 0 to 2x  ...

翻译原文如下:
xlf0602 is right about this.  and the theory behind it is fairly elementary: in a push-pull Class A design, the current output from the positive half or the negative half swings from 0 to 2x of its idle current.

关于这个,xlf0602的观点是正确的,但是分析还是基本清楚的:在推挽甲类设计中,当信号正半周到负半周,输出电流是从0到二倍于静态电流之间波动.
when the positive half swings to its maximum of 2x idle current output, the negative half is at its minimum of 0 amp. so the total output current on the load is 2x of the idle current.
当正半周达到二倍静态电流时,负半周接近最小值0,因此整体输出到负载的电流就是二倍于静态电流,
in reality, due to non-linearity, the swing is considerably less. For example, 1.2x - 1.3x of idle current is fairly common for JLH1969.
但是由于非线性的原因,这种波动相当低,以JLH1969为例,输出电流通常为1.2被到3被静态电流.
now, the above discussion is on the output capability. how much current an output stage CAN output depends on a lot other factors and typically differs from how much current that output stage WILL output - which is your confusion.
现在,以上讨论焦点是输出能力上,多大的电流可以有多大的输出能力,还依赖于其他因素.典型的不同就是多大的电流提供多大的输出能力这也正是你所搞错的地方.

xitonghan

supply rail voltage is usually determined by your transformer and is fixed.

供电电压通常依赖于你的变压器且是固定的.
what you can change is the idle current. so the right question should have been, given my transformer, what is the right idle current?
你所能改变的是静态电流的大小,因此正确解决问题的办法设定多大的静态电流?
here is a rough calculation.
以下是略算:
let's say that you have a 27vac transformer. the DC output is about 27vac*1.3=35v. assume that you use bjt output devices and the saturation voltage is 1v, the maximum voltage swing on the load is then 35/2-1=16.5v.
例如你有一台27v 交流的变压器,那么直流输出电压是27Vx1.3=35V,假设你使用半导体三极管,且压降为1V,那么在负载上的最大输出电压是35/2-1=16.5V
further, we are designing for a 6ohm load (fairly typical for an 8ohm nominal impedance speaker), the peak output current is 16.5v/6ohm=2.5amp.
当我们使用6欧姆负载的时候,(8欧姆扬声器的平均值),那么峰值输出电流是16.5V/6=2.5A
that means the idle current is 2.5amp/1.3x=2amp, assuming a push-pull output stage with a 1.3x multiplier.
这也就意味着,静态电流是2.5A/1.3X=2A,(假设推挽输出电流是静态电流x1.3倍关系).
each device will dissipate about 2amp*17.5v=35w. That's about the limit for a to247/to3p device rated at 100w dissipation (at 25c).
每个功率管要耗散掉2Ax17.5v=35w的功耗,这就限定to247/to3p封装设计耗散功率是100w的(25摄氏度)原因
the whole amp dissipates about 35wx2=70w, and you for all likelihood will need a 100w transformer (27vac*2.5amp).
双声道耗散功率就是35Wx2=70W,为了考虑各种可能性,变压器功率需要100w
the output power is 2.5amp*2.5amp*6/2=20w rms. or total efficiency of 25w rms / 70w rms = 30%.
输出功率是2.5Ax2.5Ax6/2=20W整体效率是25W/70W=30%

xitonghan

a big assumption in the above calculation is that we are using bjt transistors, and as a result run the amp in Class A.
以上假设计算的前提是,使用双极型晶体管,目的是使用在甲类放大电路中.
if you use mosfets output devices, you have the option to run the output in Class B, and idle the output devices at much lower current levels (typically 100ma).
如果使用场效应管做输出,你可以使用乙类放大电路,且静态电流可以设置在比较低的水平上(典型100ma)
however, mosfets have much higher saturation voltages (4-5volt typically), so you will likely need to run the amp at higher supply voltage.
可是场效应管有着更高的饱和压降,(电芯那个4-5v),因此你需要让这个电路工作在更好的电压上.
here is a quick calculation, assuming that we generate the same output voltage on the same load, but with mosfets.
这有一个快速计算办法.假设,我们在相同负载输出相同的电压,但是使用场管
to output 16.5v on a 6ohm load, the mosfets should sustain 16.5v+5v saturation voltage = 21.5v, or 43vdc supply rail. That means 43vdc/1.3=33vac from yourpower transformer.
在6欧姆负载上输出16.5v.场管需要16.5V+5V=21.5v供电来维持,也就是需要21.5x2=43v的供电电压,这也就意味着,变压器需要33V交流输出
assume that we idle the mosfets at 100ma, each device will dissipate about 21.5v*100ma=2.2w, or 5w per amp. far less than a typically to220 device can take. So an IRF540 class mosfet will work perfectly here.
假设我们设置其静态电流为100ma,每个管子需要耗散21.5x0.1=2.2w,也就是双声道需要5W,远低于TO220封装的耗散功率,因此IRF540一类场管可以用于此
as you can see, with mosfets you dissipate less than 20% of the heat generated by bjts.

由此看来,场管低于双极型晶体管20%的发热量

xitonghan

原帖由 millwood 于 2009-2-18 19:59 发表


if you really need someone to show that to you with a chart, you shouldn't be talking about jlh1969's idle current, .

here is essentially the jlh1969's output stage: the driver is loaded ...

这段貌似没必要翻译,直接查看放大电路技术书籍即可

xitonghan

原帖由 millwood 于 2009-2-20 04:11 发表
here is a quick excel spreadsheet for the calculation. you just need to change the cells highlighted in yellow.

good luck.

JHL功率设计计算器中文版下载:

weiweizhu649

文字是认识 就是不理解

mzsrz

分析得很有道理

xlf0602

原帖由 xitonghan 于 2009-3-23 23:25 发表
翻译原文如下:
关于这个,xlf0602的观点是错误的……

你翻错了吧？

mzsrz

原帖由 xlf0602 于 2009-3-24 12:30 发表

你翻错了吧？

呵呵！是xitonghan翻译错了！ 你的观点才是对的。

xitonghan

原帖由 xlf0602 于 2009-3-24 12:30 发表

你翻错了吧？

哈哈 走眼 更正

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huimeihj

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hy1635

甲类发热量不可小看,区区10W输出的小甲,当你听到美妙声音的时候会犯愁了,这么大的发热量咋办?所以,不要盲目加大静态电流.也不要用电压太高的牛.当然,如果发热量你吃得消,电压高些,静态电流调大点,输出会更大,不再是10W小甲了,也许是15W甚至20W.

hzx999

学习了，虽然看得不太懂，但是这个气氛非常好，楼主的作业总结对我们非常有用。