-1
\$\begingroup\$

I am making an audio amplifier circuit in LTspice. This is the circuit that I've made:

enter image description here

Notice that the input is 20 mV. The output plot is as follows:

enter image description here

That is around the gain I want to obtain. The speaker that I will be using is an 8 Ω, 0.5 W speaker that can sustain a maximum 10 V peak-to-peak input.

The problem is that as soon as I model this in LTspice, the voltage just drops too much and the shape distorts.

enter image description here

Why is this happening and how do I fix it? Does it have anything to do with impedances?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ You write: "The speaker that I will be using is an 8 Ω, 0.5 W speaker that can sustain a maximum 10 V peak-to-peak input." \$0.5\:\text{W}\$ means \$2.8\:{\text{V}_\text{PK}}\$ or \$5.6\:{\text{V}_\text{PP}}\$. Not \$10\:{\text{V}_\text{PP}}\$. Also, you show a picture with \$4\:{\text{V}_\text{PK}}\$ and say "That is around the gain I want to obtain." But that is \$1\:\text{W}\$ and more than the speaker handles. Step #1 is to get these numbers squared away. If you can't get it right for this part, none of the rest of the design is going to go well when the numbers really start to matter. \$\endgroup\$ Commented Jan 23, 2024 at 5:09
  • \$\begingroup\$ Your circuit includes places that might allow global NFB for both DC bias and AC gain management. But this cargo cult approach to design leaves that (and much else) out. \$\endgroup\$ Commented Jan 23, 2024 at 5:19
  • \$\begingroup\$ I don't understand why you're using opamps to buffer the signal between discreet transistor stages. What do the 1st & 2nd transistor stages give you that you couldn't get by just combining them together into a single opamp stage, using the opamp to give you the required gain? \$\endgroup\$ Commented Jan 23, 2024 at 14:31

2 Answers 2

Reset to default
2
\$\begingroup\$

There are other, better topologies for an audio amp, but what you have can be improved without starting over.

  1. As mentioned by JS, R10 and R11 are too large. Starting with your 4 V peak output voltage and the subsequent peak output current, take the transistor gain at that current (from the datasheet), divide that in half for operating margin, and use that to calculate the required base current. 100% of the base current comes from R10 and R11.

  2. You have two opamps plus a 1-transistor gain stage all for a total gain of 6. You can keep C1, eliminate U1 and Q1 (and its parts), and configure U3 for a gain of 6.

  3. There is no DC reference at the input to U1. Now that U1 is gone, add a 100K resistor from the U3 non-inverting input to GND.

  4. Delete R9, C2, and C4. I don't know why they are there, but they make things worse, The output stage needs a low driving point impedance, such as the direct output of an opamp.

The way the circuit actually works, U3 does not provide any base current to the output transistors. It shunts current from R10 and R11 away from the output transistors. For example, in a positive half-cycle, R10 pulls up the Q2 base and sources its base current. for any particular instantaneous voltage, R10 pulls up the base until D1 starts to conduct, which is when the base is 0.7 V greater than the U3 output voltage. To pull the base down, the U3 output goes down, pulling down D1, which pulls down R10 and the Q2 base.

Note that a 741 might not be able to provide enough output current to drive the output stage to the level you want. Still, you should be able to get things closer your desired output.

\$\endgroup\$
3
\$\begingroup\$

Your output transistors have finite current gain, likely below 100. If you drive them via 100 kΩ resistors, their output impedance will be around 1 kΩ, which is so high that they can't drive your load.

You need to build a proper Class-AB output stage instead. For such a simple amp, you'd typically use a Darlington configuration with a Vbe multiplier for biasing. Don't forget to add emitter resistors, otherwise your amp might go into thermal runaway. Collector fuses are probably a good idea as well.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.