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Old 2nd September 2009, 06:32 PM   #1
Mark245 is offline Mark245  United States
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Default Two Stage Amplifier Question

Hello,

I am building a simple two stage amp that consists of a CE stage followed by a CC state. When the load is sufficiently high i.e. 1k, the output it a clean sine wave. However, for a 10 ohm load or a speaker load of 8 ohms, the output gets clipped off at the bottom. I have attached a schematic and plot of the simulation. The simulation matches what is actually happening on my breadboard testing.

Notice in the simulation plot, the bottom of the sine wave gets clippped for a 10 ohm load. However, at the input to the CC stage, the wave looks clean. So the CC stage is distorting the output. Anyone know why this is happening and any suggestion of how to fix it?

Thanks,

Mark
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File Type: jpg schematic.jpg (52.8 KB, 212 views)
File Type: jpg plot_Rl=10.jpg (62.2 KB, 167 views)
File Type: jpg plot_Rl=1k.jpg (52.1 KB, 142 views)
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Old 2nd September 2009, 07:36 PM   #2
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It is happening because a common collector stage that is called an emitter follower may be viewed as a diode linearized by feedback. For the upper half-period your transistor delivers current to the load; but for the lower half-period of the sine the resistor in it's emitter delivers the current to the load.

Now, what do you need to change to get higher negative current delivered to the load?
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Old 2nd September 2009, 08:18 PM   #3
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Originally Posted by Wavebourn View Post
It is happening because a common collector stage that is called an emitter follower may be viewed as a diode linearized by feedback. For the upper half-period your transistor delivers current to the load; but for the lower half-period of the sine the resistor in it's emitter delivers the current to the load.

Now, what do you need to change to get higher negative current delivered to the load?
The output of your amp is grossly mismatched to the load.
You need to lower the output impedance of the final stage.
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Old 2nd September 2009, 08:21 PM   #4
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Quote:
Originally Posted by nigelwright7557 View Post
The output of your amp is grossly mismatched to the load.
You need to lower the output impedance of the final stage.
It is the answer with common words.

I've asked, what needs to be done to get it right?
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Old 2nd September 2009, 08:46 PM   #5
Mark245 is offline Mark245  United States
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Ok so when the current through the 10 ohm resistor into the emitter equals the dc bias current, the transistor cuts off right, thereby clipping the negative half cycle?

Is this why people go with a push-pull output stage instead of the single-ended design?
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Old 2nd September 2009, 08:48 PM   #6
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You need to put in a transistor to drive the negative half cycle.
You would then need to bias the output stage in class b if your not fussy about the quality or class AB if you want a decent sound.

Look up class ab output style. There are literally hundreds of circuits out there on this forum and elsewhere.
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Old 2nd September 2009, 09:33 PM   #7
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Quote:
Originally Posted by nigelwright7557 View Post
The output of your amp is grossly mismatched to the load.
You need to lower the output impedance of the final stage.
Clearly said Anatoliy.
How about this: this output stage (emitter follower) is a class A stage. Therefore, it can only deliver non-clipped signal at 2*quiescent current, which I think is about 3.9V/390ohms *2 is about 20mA into 10 ohms which in turn is about .2V peak-peak. So any signal above that causes clipping because even if the transistor completely closes up, the emitter resistor cannot sink more current.

Solution: increase the standing current by decreasing Re. But watch the heat in that emitter follower!

Or go to a push-pull stage in class AB/B as mentioned. Then you have two transistors that can take turns in sourcing and sinking to arbitrary amounts of current (within reason).

jd
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Old 2nd September 2009, 10:11 PM   #8
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Originally Posted by janneman View Post
Solution: increase the standing current by decreasing Re. But watch the heat in that emitter follower!
Right Janneman;

It is one half of the answer;

another half is, a standing voltage on emitter must be 2/3 of the B+, instead of a half of it.
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Old 3rd September 2009, 01:34 AM   #9
Mark245 is offline Mark245  United States
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Thanks for all the replies.

Wavebourn, why is it neccessary to have 2/3 of B+ on the emitter? It seems like you could have just whatever as long as you are drawing enough current through the emitter resistor to allow for the full voltage swing. Can you further explain this? Thanks

Mark
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Old 3rd September 2009, 04:34 AM   #10
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Quote:
Originally Posted by Mark245 View Post
Thanks for all the replies.

Wavebourn, why is it neccessary to have 2/3 of B+ on the emitter? It seems like you could have just whatever as long as you are drawing enough current through the emitter resistor to allow for the full voltage swing. Can you further explain this? Thanks
Mark;
what do you mean enough? Do you have an unlimited supply?

Suppose, you still want your amp to be efficient, i.e. to deliver max power on the load per dissipated watt.
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