I'm trying to understand how speakers behave when connected to a single ended class A amp circuit. Let's say we've got one output transistor biased to the full amplitude of the input signal. This transistor is connected between 0 and let's say 25 volts. Let's assume that the dc offset is 0 volts. In my understanding, if a positive voltage is connected to a speaker, the cone moves outwards, and with a negative voltage, inwards compared to it's resting position.
What I see here with the class A design, that the speaker will receive a variable voltage but always in the positive range, which makes me think that the cone will be displaced outwards (and vibrate in that range) and never inwards relative to it's resting state. Also because of this, when the signal amplitude declines, it's only the speaker construction (spider and edges) that are going to help the cone to return to a lower position, which means it will be controlled less and might not have the time to move without distortion. I hope someone can confirm this, or contradict it to give me some clarification.
What I see here with the class A design, that the speaker will receive a variable voltage but always in the positive range, which makes me think that the cone will be displaced outwards (and vibrate in that range) and never inwards relative to it's resting state. Also because of this, when the signal amplitude declines, it's only the speaker construction (spider and edges) that are going to help the cone to return to a lower position, which means it will be controlled less and might not have the time to move without distortion. I hope someone can confirm this, or contradict it to give me some clarification.
Thanks for the hints I think I'm starting to understand it now. I did a little experiment, connecting a capacitor charged up to 6 volts in series between my power supply and multi meter. When the power supply is set to 6 volts, the multi meter shows 0 volts. When I increase the voltage over 6, I see increasing voltage, but when I lower it below 6 volts, I see negative voltage in the multi meter. So I guess the direction of the current changes when I go below the charge of the cap and it starts to discharge. Is that what you were referring to?
Also interesting what capacitance we would need to deliver many amps into a speaker.
Also interesting what capacitance we would need to deliver many amps into a speaker.
The situation you're describing cannot exist without damage to the speaker. Even with 0 VDC offset on the input.
Look up single ended class A amp topologies and you'll see that all of them employ DC-blocking which means that the speaker only gets the AC making it work from its normal resting state.
Edit: someone beat me to it.
Yes, you varying the power supply voltage is equivalent to an AC signal (which audio is). The constant part (DC) is blocked, hence the function of the capacitor in this application which is quite different from its function when used in a DC power supply as a filter cap.
You could repeat the experiment without any charge on the cap and you will get the same result from 0 V.
When you stop varying the power supply, you will see that the cap charges up and the current stops flowing, so your multimeter will start to go to 0 VDC, even if the voltage on the cap would be higher that that.
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When the single-ended amp is turned on, the output capacitor will charge up and during that time period, there will be voltage at the output end. Effectively, this is a low frequency AC output. Capacitor coupled outputs should have a resistor, say 200 ohms, across the output to provide a current path when no speaker is present. If a speaker is connected, you may hear a thump.