BJT amp to drive an 8 ohm load

Thread Starter

automagp68

Joined Nov 13, 2011
81
Hey guys


So i have build many a BJT amps but I've never tried to make one that supports an 8 ohm load with a 6 volt swing

Any ideas on this?

Right now i just have a few CE ganged up to get my gain correct.

If i use a CC as an output stage with the coupling cap i loose all my voltage swing because I'm obviously blocking DC

Any one got any ideas on how to accomplish this in a simply fashion with out a lot of fancy stuff

This is just a for fun project. I have seen many designs but they all include very small swings, say 10mv or so. I need 6 volts across




Here is what i got so far. Please excuse the diff amp on the left its from another project that i had built so I'm just using it.
Im getting the gain from my CE stage and trying to get current from the CC stage but as i said before when i put the cap in across the 8 ohm load i loose all my swing and if i take the cap out the DC current in the CC shoots through the roof of course.

And ideas? a basic direction perhaps? As i said above I'm not really familiar with power amplifier design at all
 
Last edited:

Veracohr

Joined Jan 3, 2011
695
You can use a CC output stage but it needs higher quiescent current than what your schematic shows. Reduce R7 a lot. Also increase the coupling cap to at least 1000uF.

Or you can look up class AB output stages for another approach.
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
Thanks for the replies.

I updated the picture above.

So i i reduced it to say 300 ohms giving me an increased IE.

I also upgraded the cap value

Any ideas on how to get 6 volts across the speaker?
I am not understanding how I'm gonna see 6 volt swing with the coupling cap still there
 
Last edited:

dl324

Joined Mar 30, 2015
9,620
Q4 needs to provide your 6V signal swing. Q5 is an emitter follower (common collector to some) that provides a low impedance to drive the speaker. C2 decouples any DC bias from Q5 and passes your 6V AC signal.
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
ok i think i understand what your saying.

So the Q4 stage is where my swing comes from. Now my AC is no longer small signal its large signal but still acc thus allowing it to pass through c2

ok thanks that was helpful to understand. now how to get it to work lol
 

crutschow

Joined Mar 14, 2008
24,122
It's difficult to get that kind of output into 8Ω using just an emitter follower unless the follower has a very high bias current.
You should go to a push-pull output stage, which is a common-way to drive a speaker.
 

Thread Starter

automagp68

Joined Nov 13, 2011
81
It's difficult to get that kind of output into 8Ω using just an emitter follower unless the follower has a very high bias current.
You should go to a push-pull output stage, which is a common-way to drive a speaker.

Thanks

I do understand that is typically the way its done but i don't have that experience and don't have the time to learn it right this second.
This was just a small experiment simulation only.

6V peak ? And do you know that V_max_negative_output_voltage = Ieq * Re||RL
6v peak swing yes
Amplitude at this time is unimportant just as long as it swings 6 volts
 

kyka

Joined Jun 7, 2015
24
Ok, here's the analytical approach.

If by "voltage swing" you mean peak-to-peak value, then at some point the output transistor will have to supply 3/8=375 mA. Let's assume that the output transistor has enough gain and proper biasing to do that (it's not difficult at all). The problem is the negative cycle. At that point the coupling capacitor must get discharged in a controllable manner in order to follow the input signal. This can't be done, because its peak current will be 375 mA which will force the output transistor to get cutoff and the discharging process will be abrupt.

I ran a simulation on LTSpice and here's what came out. I biased the transistor to the same DC emitter current.



The only way to fix this problem, without changing the topology, is by reducing the emitter resistance. See the next picture,

The only problem now is that the transistor conducts almost an amp of DC current and you must make sure that your transistor can handle it.
 

crutschow

Joined Mar 14, 2008
24,122
..................
The only problem now is that the transistor conducts almost an amp of DC current and you must make sure that your transistor can handle it.
Also it will dissipate about 8W at idle so the transistor will need to be a power type (such as a 2N3055) mounted on a good heat sink.
 
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