maybe a new idea on lm1875

[krizis]

hello everyone here,

this is a new amp based on lm1875,two lm1875 work parallel to provide 50W at 8 ohms load,which could provide lower source impedance to drive your speakers easily.

And I designed a new but simple preamp with single-end to drive your lm1875.
Also please pay attention to the power supply for preamp,this circuit will make you unit get a good signal-noise-ratio.

Attached is the topology,please give some advises.

Thanks very much and have a good day.block diagram.pdf

[krizis]

sorry,I have deleted the sub out finally,so this is the new topology,Please see the attachement.

Thanks
block diagram new.pdf

[Bone]

Two 1875 in parallel will not give any more power in 8 ohms than one will. You need to drop the impedance to 4 ohms to get more power.

[pacificblue]

Exactly. If you want more power into 8 Ohm than a single LM1875 can deliver, you have to bridge two of them. In theory that will give you four times the output power from the same rail voltage into the same load. It will also give you four times the heat dissipation to deal with.

[geraldfryjr]

I had explained this in great detail here Would you use a 4780 in a parallel or bridged configuration with 6 ohm speakers? and starting here Would you use a 4780 in a parallel or bridged configuration with 6 ohm speakers?
with regaurds .jer

[krizis]

hello everyone,
yes,50w at 8ohms is not correct,because if the rail voltage is 25v,and if it is rail-to-rail(this is only a hypothesis)it will provide only 25/1.414*25/1.414/8=39w.

[krizis]

I wirted a simple pdf file to explain my understaning to this,
please go to the attachement.
If any problems,please refer out,
Thanks
amp model.pdf

[krizis]

For the amp model pdf file,I have some other ideas:
1:why class a sounds better than class ab?
for example a 1a idal current class a and a 10ma ideal current class ab
Ro=(vcc-i*RL)/i
so class a Ro=(25-1*8)/1=17ohms
but class ab Ro=(25-0.01*8)/0.01=2.5k
the difference is quite big,so class a will be much more powerful to drive the Rl than class ab.because speaker is not a simple pure resistor,it mix inductors and capcitors,so a low impedance source is very important.
by the way if you can not bear the heat cause by class a,you could also make the output stage working in parallel,this will also lower the source impedance,this is why I choose this lm1875 working like this.

Thanks everyone.

[geraldfryjr]

"hello everyone,
yes,50w at 8ohms is not correct,because if the rail voltage is 25v,and if it is rail-to-rail(this is only a hypothesis)it will provide only 25/1.414*25/1.414/8=39w. "

yes, this is correct and into 4ohms you would get 78 wats rms. jer

[pacificblue]

Originally Posted by krizis

I wirted a simple pdf file to explain my understaning to this,
please go to the attachement.
If any problems,please refer out,
Thanks
Attachment 171877

Your Pmax equations are wrong. Pmax for a single amp will be V²/(Ro+Rl)/2 and for parallel amps V²/((Ro/2)+Rl)/2. How big would Ro have to be for a significant effect?

Use realistic values for your Ro and take into account that Ro is non-linear.

Originally Posted by krizis

Ro=(vcc-i*RL)/i
so class a Ro=(25-1*8)/1=17ohms
but class ab Ro=(25-0.01*8)/0.01=2.5k

Your i is the current through the transistors, not through the speaker. Class A amplifiers are used with low output power, because their inefficency leads to enormous heat dissipation. If you build exactly the same amplifier twice and bias one of them into class A, but the other not, you will still get the same maximum output power from both of them. The sonic advantages are mostly due to the fact that the transistors in class A don't switch during zero-crossing, thus avoiding the corresponding distortions.