Is your output 7.18V peak? Then the peak current in the 10 ohms load is 718mA and the RMS current is 0.508A.

An input higher than about 9.2V peak will result in a very distorted output because then the transistors have no base current near the peaks.
Even if the input is 7V peak then the transistors do not have enough base current.
Then why feed 12.5V into the input?

 

Is your output 7.18V peak? Then the peak current in the 10 ohms load is 718mA and the RMS current is 0.508A.

An input higher than about 9.2V peak will result in a very distorted output because then the transistors have no base current near the peaks.
Even if the input is 7V peak then the transistors do not have enough base current.
Then why feed 12.5V into the input?

Because Jony indicated that the maximum ac output voltage is 7.88 volts peak and the
maximum Ie = .788A. To approach that output you would have to input 12.5V peak,
according to LTS. LTS doesn't show any distortion at that level. I do get the part about over-driving the input.

I don't see the numbers that you are speaking of.

Here is the LTS sim.

 

The simulation shows little low power and low current output transistors that are seriously overloaded so I changed them and I changed RB1 and RB2 to simulate minimum current gain. Then there is horrible clipping because the input level is too high.

 

If I input a 12.5 volt sine into LTS it results in a much different Ie than 788 mA,
like 36 to 38 mA although the ouput is 7.18 volts. I am confused.

I'm confused too. From where did you get this value?

Also if Vo = 7.18V the Ie current must be equal to 7.18V/10Ω = 0.718A

 

I'm confused too. From where did you get this value?

Also if Vo = 7.18V the Ie current must be equal to 7.18V/10Ω = 0.718A

I have attached the sim. Peak output per the transient analysis is 7.18V but
it is 7.53V per the AC analysis.

I don't see the 0.718A or the 0.753A in the sim results.

The total DC current drawn from the supply doesn't make sense.

How can this be?

 

You need to use a transient analysis because the AC analysis does not include power supply limitations. AC analysis is a linearized small signal analysis.

 

Please get rid of the low power, low current transistors and use transistors rated for at least 1A.