# Need Help with Analog Circuit

Discussion in 'General Electronics Chat' started by Ryanj, Dec 3, 2012.

1. ### Ryanj Thread Starter New Member

Dec 3, 2012
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How do I calculate the peak to peak voltage required across the 8 ohm resistor (R9) to get a 3 W output? I've attached the circuit and would appreciate any help with this circuit.

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Apr 6, 2009
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3. ### bountyhunter Well-Known Member

Sep 7, 2009
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Is the 3W continuous power or peak power?

Power is equal to v(squared)/R, so you can calculate it directly.

Peak voltage is equal to 1.414 x RMS voltage for AC sine waves.

4. ### mbohuntr Active Member

Apr 6, 2009
413
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I'm not sure your asking the right question, 3W might be the limit of the power this circuit can handle. I'm not even sure what it does, kinda looks like an amplifier or oscillator to my limited abilities.

5. ### Ryanj Thread Starter New Member

Dec 3, 2012
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Ok your explanation is making sense thanks for the help

6. ### Audioguru New Member

Dec 20, 2007
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The RMS output voltage for 3W into 8 ohms is the root of (3 x 8)= 4.9V. The peak-to-peak voltage is (4.9V x 2.828)= 13.9V. The peak current into the 8 ohms load is (4.9V x 1.414)/8= 869mA.

The 2N3904 and 2N3906 have an absolute maximum allowed current of only 200mA.
If the peak current is 200mA then the RMS current is 141.4mA then the power output is only (141.4mA squared x 8=) 0.2W. Use power transistors instead.

The emitter resistors for the output transistors have a very high value of 4.7 ohms so they throw away more than half of the output power. Most amplifiers use 0.22 ohms.

Your amplifier does not include the output transistors in the negative feedback loop so the distortion might be fairly high.

The amplifier has a dual-polarity supply so the 1000uF output coupling capacitor is not needed.

7. ### MKCheruvu Member

Nov 20, 2012
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As the Amplifier is with dual Supply , if the load happens to be resistive 8 ohms output 1000 micro farad capacitor is not required..

However if the load happens to be Speaker of 8 Ohms impedance , its resistance value is likely to be very low. In such case the Amplifier's output DC offset Voltage can damage the Speaker with high DC Current.

Also 1000 micro farads capaitor normally likely to be Electroylitic.The Amplifier ac output voltage being Bi-polar (due to Dual supply) will impress both polarity voltages across across the output Electrolytic capacitor, which can lead to degradation failure.

8. ### Audioguru New Member

Dec 20, 2007
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If the negative feedback is connected directly to the output like most amplifiers then there will be no DC at the output.

EDIT: I forgot about the extremely high gain of 751 that amplifies the high noise level from the lousy old 741 opamp.
Also, with such high gain then its high frequencies drop above only about 1kHz.

Last edited: Dec 4, 2012
9. ### Ryanj Thread Starter New Member

Dec 3, 2012
4
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How do I determine how much voltage the power supplies should have if I want an 14V peak to peak output for the op amp?

10. ### Audioguru New Member

Dec 20, 2007
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895
Did you reduce the value of the 4.7 ohm emitter resistors?
If you use 0.22 ohms then they reduce the output level by only 0.4V peak-to-peak.

The emitter followers need an input signal that is about 2V peak-to-peak higher than at their emitters.

The peak output current is 869mA and TIP31 and TIP32 power transistors have a minimum hFE of 25 so the output current from the opamp must be 35mA which is not available from an old 741 opamp.

Then you need darlington output transistors or use a Sziklai pair for each transistor.

Darlington output transistors need an input that is about 3.2V peak-to-peak higher than their output.

An old 741 opamp has an output that could be 6V peak-to-peak less than the power supply voltages.
The voltage losses are 0.4V + 3.2V + 6V= 9.4V. You want an output that is 13.9V peak-to-peak so your power supply must be 13.9V + 9.4V= 23.3V.