you need more knowledge in electronics , to understand Papa's answer ;
but - funny - in case that you have more knowledge , your question will not be asked .
every project is sum of compromises ....... and I'm sure that final result can be declared as compromise , too .
even if Papa is certainly keen to call it different - best solution .
The concept is loosely similar to the following:
Car A is a luxury car and has a 0-60 mph of 10 sec and has a top speed of 130 mph
Car B is a sports car and has a 0-60 mph of 5 sec and has a top speed of 130 mph
Car B is faster but both of them have the same top speed.
Now subsistute 0-60 with db and top speed with Watt.
Both are nice cars but are just optimized in different ways.
Hope it helps instead of creating more confusion...
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What was the question? I only saw you ask why the discrepancy, but I'm not seeing it as a discrepancy.
An amplifier's power output rating and its gain are not inextricably linked.
A given output stage will only be able to swing a given amount of voltage and deliver a given amount of current into a given load. This is essentially what establishes the amplifier's output power rating, whether you put a 5dB gain stage in front of it or a 20dB gain stage in front of it.
Consider the F4 for example. It's just a power buffer and has no voltage gain (0dB). It's capable of delivering 25 watts into 8 ohms before it starts clipping.
Stick a 20dB gain stage in front of it and it will still only deliver 25 watts into 8 ohms. The only difference is how much signal you have to feed it in order for it to output 25 watts.
So I don't see what the discrepancy is that you speak of.
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Let's make it very simple (or simplisitic).
I put 1 volt into a device and 2 volts come out. My gain is 6dB. If this is driving a 8 Ohm transducer, then the power is V times I . In this case the Power would be 16 watts.
In my mind, if the input voltage is the same (for the two devices) and they are driving the same load impedance, then the gain and power should be the same for the two devices.
I must be missing something very basic ( but soon I will have the "ah ha" experience).
I put 1 volt into a device and 2 volts come out. My gain is 6dB. If this is driving a 8 Ohm transducer, then the power is V times I . In this case the Power would be 16 watts.
In my mind, if the input voltage is the same (for the two devices) and they are driving the same load impedance, then the gain and power should be the same for the two devices.
I must be missing something very basic ( but soon I will have the "ah ha" experience).
The I is not the 8 ohms, but the current that 2 volts puts
through 8 ohms, which is 2/8 = 0.25
Then V*I is 2 * 0.25 = 0.5 watts.
Commonly we express it as Power = V * V / R since I = V / R
😎
through 8 ohms, which is 2/8 = 0.25
Then V*I is 2 * 0.25 = 0.5 watts.
Commonly we express it as Power = V * V / R since I = V / R
😎
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Hi WithTarragon
Gain of an amplifier is determined by the gain of the devices being used and the amount of negative feedback implemented.
The Power is determined or limited by the power supply specs and the amps ability to dissipate heat (ie heatsinking).
Hope that makes more sense
Gain of an amplifier is determined by the gain of the devices being used and the amount of negative feedback implemented.
The Power is determined or limited by the power supply specs and the amps ability to dissipate heat (ie heatsinking).
Hope that makes more sense
After a good night's sleep my brain has re-engaged.
My thinking was cloudy because I was not remembering that at some point the amp is starved for voltage (essentially "depressing" the gain number) or it is starved for current (essentially "depressing" power number).
Sorry for going off-topic, it's just that I thought that maybe everything I knew was wrong. A most disturbing thought.
Thanks for you collective indulgence
I'll confess that this morning when I re-read the thread, I did have a chuckle that an experienced engineer was having to remind me of Ohm's law (and that I needed to be reminded).
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