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In the circuit given below, I'm working on an audio amplifer. It is driving a 4-Ohm load (speaker.)

  • Input : 1.1V peak-peak sine wave of frequency 1Khz. Input dc = 12v

  • Output : 4-Ohm load. Gain = (8 peak-peak) / ( 1.1 peak-peak) = 7.2

Circuit Description :

This circuit is meant to amplify audio input from a typical smartphone audio jack.

Input stage of amplifer is driven by 2N3904 low power NPN. At output stage I have used TIP147 and TIP142 to amplify the signal. Input potentiometer is used to control biasing current of driving transistor at 34% of its value(1k potentiometer), it gives me optimal results, after that it starts to clip.

100uF(C4) bootstrap is also used to feed-back at R9-R2 junction increase the voltage gain. Rest are just biasing resistors and filtering caps.

enter image description here

Yellow = Output & Blue = Input

enter image description here Question:

Q) What can I do to increase the gain of this circuit and consequently increase the output power (I don't want to increase supply voltage?)

Keeping this circuit in mind, also suggest me some better design options too.

Give me a heads up if other circuit specific information is needed to entertain my query.

Please bear in mind that this circuit is for my learning experience only. I know there are other more efficient ways to do what I am doing, but they are a bit advanced for me at this stage.

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  • \$\begingroup\$ I don't think you have room for more gain. You have a 12V power supply. Signal 1.1Vpp, gain of 7.2, output of 8Vpp. You lose a couple of volts from the high and low voltage ends because of the transistors. 8Vpp is really about all you can expect for this design and power supply. \$\endgroup\$
    – JRE
    Oct 27, 2019 at 23:16
  • \$\begingroup\$ If this design clips on the +12V DC supply, you might consider using two amplifiers in anti-phase. Look into "bridge amplifier". A very good design will give 4X power than you have now, with the same supply. However, that supply will have to deliver twice the current, and the power transistors work harder. \$\endgroup\$
    – glen_geek
    Oct 27, 2019 at 23:19
  • \$\begingroup\$ What's \$R_9\$ doing for you? (I usually expect to see the speaker in that position when someone says "bootstrapped" with regard to the output stage.) \$\endgroup\$
    – jonk
    Oct 28, 2019 at 6:40

1 Answer 1

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You want to increased output power without increased power supply voltage? Are you willing to have the lowered load resistance; i.e. multiple parallel connected speakers at output?

If not there is no point of any sort in discussing modifying the circuit.

If you are willing to have the lowered load resistance, your circuit need to be modified to enable it handling high currents at it's output stage. To do this, add base resistance of 100ohms to each output transistors and repeat its counts to 4 pairs.

That will give you 0.5ohm capable output stage. That's 100 watts with 12 volt power supply.

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  • \$\begingroup\$ I think the OP is only asking for \$2\:\text{W}=\frac{\left(4\:\text{V}\right)^2}{2\cdot \,4\:\Omega}\$. Maybe a couple of orders of magnitude below \$100\:\text{W}\$. Separately, to drive \$100\:\text{W}\$ into \$4\:\Omega\$ you need to swing almost \$\pm 30\:\text{V}\approx \sqrt{2\cdot 4\:\Omega\cdot 100\:\text{W}}\$. That's not happening with a single rail \$12\:\text{V}\$ supply. \$\endgroup\$
    – jonk
    Oct 28, 2019 at 6:28

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