常能搜索到的一个简单耳放的电路分析［学了晶体管电路设计后练练］

hylpro

这个最简单,所以分析这个了.

如果到处乱逛,难免看到下面的这款耳放. http://www.redcircuits.com/Page31.htm


(原文件名:1.GIF)


(原文件名:2.gif)





R1_____________10K 1/4W Resistor                  R2____________100K 1/4W Resistor
R3_____________68K 1/4W Resistor (see notes)      R4______________1K5 1/4W Resistor
R5______________3K3 1/4W Resistor                 R6____________330R 1/4W Resistor
R7______________4K7 1/4W Resistor                 R8______________2R2 1/4W Resistor

C1______________1µF 63V Polyester Capacitor       C2____________100µF 25V Electrolytic Capacitor
C3____________470µF 25V Electrolytic Capacitor   

Q1____________BC239C 25V 100mA NPN High-gain Low-noise Transistor
Q2____________BC337 45V 800mA NPN Transistor
Q3____________BC327 45V 800mA PNP Transistor

J1____________Stereo 3mm. Jack socket     SW1___________SPST Switch  B1____________3V Battery (two 1.5V AA or C cells in series, but see Notes)

这个东西的直流工作点也依赖于负反馈, 所以计算分析需要考虑所有涉及电路. 先来模拟看看, 这里没有截到图,probe1处是1.19V 6.53ma.

(原文件名:dcbsxfpf_881ff8gdqgg_b.jpg)


尝试计算分析:
Q2,Q3组成一个达林顿连接的变形. 具体分析就是:
1. R7上的电压是0.6V左右, Q2 基本是个恒流源, 电流大约120uA.
2. Probe1处电流取决于4点电压, 而4点电压又取决于流过Q1 的集电极和发射极电流, Ie~=Ic.
3. Ie又取决于Probe3的电压.
4. Probe3的电压又是对Probe1的电压通过反馈电阻R2得到的. 点6的电压是对Probe1处的分压值,因为R2+R3远远大于R8+R9(负载), 并且可以忽略Q1的基极电流.

根据这个环路对Q1 Re Rc列个简单的方程就能知道Porbe1处的电压和电流,见下图:

(原文件名:1.jpg)
R4压降: 0.617V-0.6
R5+R6压降: 因为Ie~=Ic, 所以是: (0.617V-0.6)x(R5+R6)/R4  ==> (0.617v-0.6)*4.8K/330
4点电压: V+0.6

从4点往上到R5接Vcc即3V,所以  (0.617V-0.6)*4.8K/330+ (V+0.6) = 3   ==>  8.9V-8.72+V+0.6 = 3  =>  9.9V=11.12
得到V=1.12V, 模拟结果是1.19, 可以接受的计算误差. 反过来设计的时候也要遵守可省略原则,否则影响静态工作点,就不能准确设计了.就是:
1.默认都工作再放大状态,
2. Vbe~=0.6
3. Ib 需要可以忽略
4. Q1基极电流还要比反馈电流小10倍.

最后这个电路就是一个共射极放大加上一个变形达林顿链接的射随. 负反馈不仅有直接从输出到输入的,还有通过C2给第一极放大的负反馈. 如果去掉C2, 则失真在1%以上(模拟结果), 总的模拟结果失真率和下面的参数值基本一样.

Notes:

    * Can be directly connected to CD players, tuners and tape recorders.
    * Tested with several headphone models of different impedance: 32, 100, 245, 300, 600 & 2000 Ohm.
    * Schematic shows left channel only.
    * B1, SW1, J1 & C3 are common to both channels.
    * R3 value was calculated for headphone impedance up to 300 Ohm. Using 600 Ohm loads or higher, change R3 value to 100K.
    * An interesting upgrade for this circuit was suggested by Mike Baum, NY USA. This involves the use of a Lithium-Ion Prismatic Rechargeable 340948 Battery, featuring a nominal voltage of 3.7V and a current rating of 1200mAH. These weight 38 grams and are 34.2mm x 8.5mm x 48mm. Correct value of R3 when using a 3.7V supply and 32 Ohm impedance headphones will be 100K: under these conditions, the amplifier will deliver about 3V peak-to-peak undistorted output. This means that the output power on 32 Ohm load will be almost doubled in respect to 3V supply. Current drawing will raise to 40mA. When powering a stereo version of this amplifier, the battery will last about 15 hours.

Technical data:
Current drain:
        35mA per channel with 32 Ohm impedance headphones.Much less with higher impedance loads

Output voltage:

         Above 2V peak-to-peak on all loads

Sensitivity:

        90mV RMS input for 2V peak-to-peak output

Frequency response:

        Flat from 30Hz to 20KHz
Total harmonic distortion
    @ 1KHz & 10KHz:Below 0.05% into 32 to 600 Ohm loads and up to 1.5V peak-to-peak output.
    Below 0.1% at maximum outputUnconditionally stable on capacitive loads

stdio

我也要算算。。

gzhuli

这玩意输出有直流成分，能用么？

hylpro

回复【2楼】gzhuli  咕唧霖
这玩意输出有直流成分，能用么？
-----------------------------------------------------------------------

找个大个的,非常便宜的,杂用也不心疼..          

hylpro

后来又看到多个直接作为直流一部分的设计, 为了简单? 凑合也用了. 不过这样,即便是30-20khz平坦, 失真低,  好像喇叭在通电就被预紧了下, 然后以那个位置为中心震动, 应该是不怎么科学的用法了

xslff

模拟，怕怕的东西！

millwood0

your analysis is basically wrong.

for maximum output swing, Q2's emitter is should be at roughly Vc/2, or slightly lower. from that, you get the voltage on Q1's base (and Q3's idle current).

Once you have the voltage on Q1's base, you know its Ic by going through the Vbe+R4 route.

if you know the voltage on Q2's emitter, you know the voltage on Q2's base / Q1's collector.

since you know Q1's Ic and the voltage on its collector (thus the voltage drop over R5/R6), you know the resistance required for R5/R6.

you are done.

the topology isn't the base one. for a better design, you can take the JLH1969 and stripe out the output stage - the resulting design is usually ultra-linear design.

millwood0

Q3 isn't needed, unless you need tons of output current - like in a headphone amp.

hylpro

回复【6楼】millwood0  
-----------------------------------------------------------------------


>>for maximum output swing, Q2's emitter is should be at roughly Vc/2, or slightly lower.
正是1.2V左右, 3/2略低.
>>from that, you get the voltage on Q1's base (and Q3's idle current).
>>Once you have the voltage on Q1's base, you know its Ic by going through the Vbe+R4 route.
我是看电路计算的, 你的思路是设计思路.

总之,我看是basically right.

>>Q3 isn't needed, unless you need tons of output current - like in a headphone amp.
这个倒是没有注意,功率没有计算...

hylpro

millwood0 老大, 这里有另一幅电路,也是那个日本人写的<高低频电路设计>,  原图2点是OP放大器输入, 右边的push pull也是接上的.
书中是用这个电路做电压放大的. 可是,分析来分析去,感觉他不太能工作阿, Q1,Q2的集电极电流总要相等, 暂时困在这里不知道Q1,Q2的集电极电位是如何
取出放大的电压的. 模拟显示,有1Vpp的输入输出就是方波了...

莫非真想要的是方波阿, 书中不想阿,<高低频电路设计> ,p77. 大家指教下..


(原文件名:1.jpg)

astudent

学习

millwood0

it is a follower.

when the voltage on point 2, V2, goes up, it pushes up the voltage on g; this raises the voltage on Q2's base, thus Q2's emitter / Q4's base, assuming Q4 is connected.

a rise on Q4's base pulls up Q4's emitter, thus the output swings with it.

the amp has an open-loop gain of slightly less than 1x.

if Q4 / Q3 aren't connected, and you are taking the output off Q2/Q1's collector, the amp will have an open loop gain, depending on Q2/Q1's hFE, typically quite high, causing the appearance of square wave output.

hylpro

>>when the voltage on point 2, V2, goes up, it pushes up the voltage on g; this raises the voltage on Q2's base, thus >>Q2's emitter / Q4's base, assuming Q4 is connected.
问题是, Q4不是接到e的是接到c的(书中). 所以不是follower..., 还是很疑惑.

>>a rise on Q4's base pulls up Q4's emitter, thus the output swings with it.
后极是folloer,可以理解, 但是Q1,Q2接到一起的电路很操蛋, 电流无法增加:输入电位高的时候Q1倾向于提高IC,而Q2倾向于减小电流, 不知道主要矛盾在哪里.


模拟用的开环, 原电路有负反馈..

stdio

找不到9楼的那本 高低频 的书，楼主能不能整个书中的原图来看看啊？

hylpro

我也没有pdf的,我的是纸的...
搜到一个封面.

(原文件名:9198467-1_o.jpg)

stdio

后极是folloer,可以理解, 但是Q1,Q2接到一起的电路很操蛋, 电流无法增加:输入电位高的时候Q1倾向于提高IC,而Q2倾向于减小电流, 不知道主要矛盾在哪里.
------------------------------

我的理解，请指正。

这是恒流负载的一个变形应用。IC1,IC2相反变化的时候，导致Q1,Q2的集电集电位发生变化。三极管的集电集并不是理想的恒流源，而是有一定内阻的恒流源，钥匙就在这个内阻上。Q1,Q2的集电集电位变化，通过其集电集的内阻，影响IC1和IC2，使二者相等而达到平衡。大的集电集内阻，使得电压放大倍数非常大。

litteworm

mark

hylpro

>>这是恒流负载的一个变形应用。IC1,IC2相反变化的时候，导致Q1,Q2的集电集电位发生变化。三极管的集电集并不是理想的恒流源，而是有一定内阻的恒流源，钥匙就在这个内阻上。Q1,Q2的集电集电位变化，通过其集电集的内阻，影响IC1和IC2，使二者相等而达到平衡。大的集电集内阻，使得电压放大倍数非常大。

用“大内阻的恒流源“来理解,感觉上是这样子的. 只是很纠结: Q1,Q2的E极变化肯定是主要的矛盾(否则就不会是放大状态),然后,说是恒流源吧,不太像阿, 明明是反相变化的Q1,和Q2阿. 感觉不存在一个"形像的“理解... 继续纠结吧.

lions

要是大家能发《高低频电路设计与制作》电子书上来多好啊。

hylpro

感觉这本没有晶体管电路设计经典了, 不过还好,介绍了许多电路的原理和作用

hylpro

回复【15楼】stdio  
-----------------------------------------------------------------------
>>。IC1,IC2相反变化的时候，导致Q1,Q2的集电集电位发生变化
现在就是想不清楚这个..., 方波是因为没有负反馈, 放大倍数过大..

millwood0

"问题是, Q4不是接到e的是接到c的(书中). 所以不是follower..., 还是很疑惑. "

for that to work, you have to close the feedback loop.

hylpro

回复【21楼】millwood0  
"问题是, Q4不是接到e的是接到c的(书中). 所以不是follower..., 还是很疑惑. "
for that to work, you have to close the feedback loop.

-----------------------------------------------------------------------
关闭反馈就是方波了.

gzhuli

close the feedback loop是闭合反馈环路，不是关闭反馈。

hylpro

>>close the feedback loop是闭合反馈环路
汗误解了,         Closed feedback loop认识, 动词了就不认识了