Power requirement for 8ohm speaker

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Nagamai Srinivasan

New Member
Hai Everyone,
Greetings of the day, We designed a Cascade amplifier with output power of 81mV. Will this be sufficient for 8ohm speaker? Or else what could be the suitable speaker used for this output power? If not how could we increase this power and use it for 8ohm?

Pommie

Well-Known Member
That's a power output of less than 1mW. I doubt you'll be able to hear it. Post more details for any help.

Mike.

crutschow

Well-Known Member
You could bump it to about 0.5V-1Vrms with another amp stage and then connect that to an audio amp with a power out of a few watts such as one of these.

MikeMl

Well-Known Member
..We designed a Cascade amplifier with output power of 81mV. Will this be sufficient for 8ohm speaker?
First, 81mV is voltage, not power. What do you mean? Is it 81mVpeak-to-peak or 81mVrms?

If you apply 81mVrms to an 8Ω load, then the power P = E^2/R = 0.081*0.081/8 = 0.0008W or 0.8mW, which is not enough. Good volume would require at least 5Vpp into an 8Ω speaker, which would be 5/2.8 = 1.77Vrms which would 1.77^2/8 ≈ 400mW.

Nagamai Srinivasan

New Member
You could bump it to about 0.5V-1Vrms with another amp stage and then connect that to an audio amp with a power out of a few watts such as one of these.
First, 81mV is voltage, not power. What do you mean? Is it 81mVpeak-to-peak or 81mVrms?

If you apply 81mVrms to an 8Ω load, then the power P = E^2/R = 0.081*0.081/8 = 0.0008W or 0.8mW, which is not enough. Good volume would require at least 5Vpp into an 8Ω speaker, which would be 5/2.8 = 1.77Vrms which would 1.77^2/8 ≈ 400mW.

Nagamai Srinivasan

New Member
Hai Sir,
Good Morning. Now I have changed my Cascade stage such that it's output is 8V peak to peak and my output power is 1W. but what could be my impedance value of speaker used in this case like 4ohm or 8ohm or 16ohm,or anything like that?

audioguru

Well-Known Member
8V p-p is 2.83V RMS. If the output power is 1W then its RMS current is 1W/2.83V= 353mA (0.5A peak) and the speaker impedance will be 2.83V/353mA= 8 ohms.
Does your Cascade stage have an extremely low output impedance using lots of negative feedback to damp the resonances of the speaker?

Nagamai Srinivasan

New Member
8V p-p is 2.83V RMS. If the output power is 1W then its RMS current is 1W/2.83V= 353mA (0.5A peak) and the speaker impedance will be 2.83V/353mA= 8 ohms.
Does your Cascade stage have an extremely low output impedance using lots of negative feedback to damp the resonances of the speaker?

Nagamai Srinivasan

New Member
Hai Sir,
I actually wanted to design an audio amplifier and I have uploaded it's simulated diagram below, will this work for an input from microphone and load resistance as 8ohm speaker?

Nagamai Srinivasan

New Member
Hai sir,
I actually wanted to design an audio amplifier. I have attached the circuit diagram below. Will this work for an input from the microphone and load resistor as a 8ohm speaker?

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Nigel Goodwin

Super Moderator
Hai sir,
I actually wanted to design an audio amplifier. I have attached the circuit diagram below. Will this work for an input from the microphone and load resistor as a 8ohm speaker?
No, nothing like - it's not even a power amplifier.

audioguru

Well-Known Member
There are many things wrong with your "amplifier":
1) Most microphones today are an electret type and must be powered. Yours is not powered.
2) The very old BC107A transistor has a minimum current gain of about 80 then its maximum base current in this circuit is about 63μA. But the resistor values for R1 and R2 are so high that they supply only 50μA. They must supply 630μA or a transistor with a higher current gain must be used. Why are the resistor values so high??
3) The second transistor also has its biasing resistor values much too high.
4) Both transistors are biased wrong.
5) The emitter resistor values are much too low.
6) C5 is 100mF which is 0.1F which is 100,000μF. It will take a few days to charge.
7) The output power of your circuit is extremely low.
8) With no negative feedback the distortion will be very high.
You need to learn the basics about transistors.

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