Simple Class A Amplifier

Thread Starter

LemoNSK

Joined Dec 23, 2019
8
Hello, I was recently working ona Class A amplifier, using one BJT NPN 2N3904 transistor and NPN Darlington pair transistor. The main problem I face is the current flowing from the collector of the first transistor to the base of the darlington pair. I want to get rid of this current, but when I add coupling capacitor between, it changes overall capacitance. The C3 acts as a low-pass filter and if I try to somehow change it, it will affect my signal very badly. Is there any way to get rid of the DC current from collector of the first transistor, or any improvements to my circuit and proper explanation?

Here's the schematic of the amplifier:
Bez názvu.png
If you need more details, I'll be glad to provide you if you ask for anything.
Thank you in advance!
 

crutschow

Joined Mar 14, 2008
24,299
the current flowing from the collector of the first transistor to the base of the darlington pair. I want to get rid of this current
Why?
That base current is very small current (<1mA) and can be readily compensated by proper biasing of the first stage.

And remove R5 and R6.
They serve no useful purpose in that circuit.
Bias at that point should be determined by the first stage resistors.

Note that Class A amps are very inefficient for driving an 8 ohm load.
 

Thread Starter

LemoNSK

Joined Dec 23, 2019
8
Well, the first stage Q point is biased in the middle of the load line (Uce = 4.3 volts). I'm getting like 40 mW from an output.. So I think I will not hear anything when I have 0.5W 8 ohm speaker?
 

Jony130

Joined Feb 17, 2009
5,145
I want to get rid of this current, but when I add coupling capacitor between, it changes overall capacitance.
Why?

I add coupling capacitor between, it changes overall capacitance.
You add a coupling capacitor and still left the C3 capacitor?
In this case, what you have done is a Capacitive voltage divider.
The C3 acts as a low-pass filter and if I try to somehow change it, it will affect my signal very badly.
Why did you add C3 into the circuit?
 

Thread Starter

LemoNSK

Joined Dec 23, 2019
8
Remove C3.
Add a coupling capacitor from the collector of Q1 to the base of Q2.
Then my DC current will be blocked from the collector of the first Q1 and now I have to bias the second one.. Should I choose the reactance value according to the frequency which is somewhere in the middle of 20 Hz to 7 kHz? For example, for 3,5 kHz and reactance approximately 0,5 Ohms.. that gives me 90,99 uF capacitor.
 

Thread Starter

LemoNSK

Joined Dec 23, 2019
8
I don't understand that.
Q2 is already biased by R5 and R6.
Yes, but I was told to delete it.. so I added it back, the same values as before. But the problem is that somehow collector-emitter voltage on Q2 is 6.8V and emitter resistor voltage is 2.1V. I wanted it to be 1/2 and 1/2 (4.5V and 4.5V). And my signal looks like this:
asdas.png
 

Jony130

Joined Feb 17, 2009
5,145
Yes, but I was told to delete it.. so I added it back, the same values as before. But the problem is that somehow collector-emitter voltage on Q2 is 6.8V and emitter resistor voltage is 2.1V. I wanted it to be 1/2 and 1/2 (4.5V and 4.5V). And my signal looks like this:
In this case, try to reduce R5 and R6 value by a factor of 10.
 

Thread Starter

LemoNSK

Joined Dec 23, 2019
8
Well, I have 8 Ohm, 0.5W speaker allowing 7.000 Hz at most. I'd like to have at least 100 or 200 mW to hear at least something.
 

crutschow

Joined Mar 14, 2008
24,299
The 30 ohm emitter resistor Re2 has to provide the minus going current for the 8 ohm load, which means the negative excursion can be no more than about 4.5V * 8/(8+30) = 947mV peak.

That's why a Class B (or AB) stage is normally used to drive a speaker from a low supply voltage.
That will allow the peak voltages to be close to ±4V for a 9V supply.
 

Jony130

Joined Feb 17, 2009
5,145
I'd like to have at least 100 or 200 mW to hear at least something.
So you need a peak voltage across the load equal to:

VL = √(2 * 0.1W * 8Ω) ≈ 1.3Vpeak and a current IL ≈ 162.5mA

or

VL = √(2 * 0.2W * 8Ω) ≈ 1.8Vpeak and IL ≈ 225mA


In the emitter follower, the maximum negative voltage swing across the load is limited by RE resistance value and DC voltage at the emitter.



The RE and RL form a voltage divider so that the maximum negative voltage across is equal to:

VL_max = Veq * RL/(RL + RE) = 4.5V * 8Ω/(8Ω + 30Ω) = 0.94V prak (at best assuming emitter resistor voltage is 4.5V).

So you would need RE around 10Ω and Ie1 = 0.45A.

As you can see this type of amplifier is very inefficient.
 
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