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. View attachment block diagram.pdf

 

[krizis]

this is new version,only deleted the subout

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

Thanks
View attachment 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 http://www.diyaudio.com/forums/chip...configuration-6-ohm-speakers.html#post2187960 and starting here http://www.diyaudio.com/forums/chip...configuration-6-ohm-speakers.html#post2185797
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]

amp works in parallel

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

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

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.

 

[krizis]

geraldfryjr thanks

"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

hi geraldfryjr,
Thanks,
for the 78watts in bridge mode,maybe we have to judge if the current is enough for the power.

 

[krizis]

thanks pacificblue

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.




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.

hi pacificblue,
Thanks,my equation is for the rms power,and it was only indicate a transient status in clipping.
I did not get what you meaned,could you please explain why you "/2" in "V²/(Ro+Rl)/2" and why you "Ro+Rl"?

"Your i is the current through the transistors, not through the speaker"
Sorry,why not through the speaker?
Generally,there is no ideal current through the speaker,that is because the curent into the speaker equals the current(this comes from the Complementary PNP in the half) out from the speaker,so there is "no current through the speaker".

Thanks very much,and nice to contact with you .

 

[pacificblue]

could you please explain why you "/2" in "V²/(Ro+Rl)/2"

1/1,41 x 1/1,41 is the same as 1/2

and why you "Ro+Rl"?

You are right, the output power in your example is (V/(Rl/(Rl+Ro)))²/(2*Rl). So, how big does Ro have to be to get significantly more power from (V/(Rl/(Rl+(Ro/2))))²/(2*Rl)?

Sorry,why not through the speaker?

The idle current flows when no signal is present directly from the positive rail through the transistors to the negative rail. At the speaker terminal there will be 0 V in that situation, which means no current at all will flow through the speaker.

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.

When 0,01 A flow through the load both amplifiers will be equally powerful with 800 µW into 8 Ohm. When 1 A flows through the load both amplifiers will be equally powerful with 8 W into 8 Ohm. The idle current does not change anything there.
When 2 A flow through the load the class AB amplifier will be more powerful. It will be the only one to drive the load with 32 W into 8 Ohm, because the class A amplifier is limited to 1 A.

 

[krizis]

hi pacificblue

1/1,41 x 1/1,41 is the same as 1/2


You are right, the output power in your example is (V/(Rl/(Rl+Ro)))²/(2*Rl). So, how big does Ro have to be to get significantly more power from (V/(Rl/(Rl+(Ro/2))))²/(2*Rl)?


The idle current flows when no signal is present directly from the positive rail through the transistors to the negative rail. At the speaker terminal there will be 0 V in that situation, which means no current at all will flow through the speaker.



When 0,01 A flow through the load both amplifiers will be equally powerful with 800 µW into 8 Ohm. When 1 A flows through the load both amplifiers will be equally powerful with 8 W into 8 Ohm. The idle current does not change anything there.
When 2 A flow through the load the class AB amplifier will be more powerful. It will be the only one to drive the load with 32 W into 8 Ohm, because the class A amplifier is limited to 1 A.

hi pacificblue

Thank for the "/2",it is very convenient the figure out the rms power.

For the current through speaker and source impedance please see the attached pdf file View attachment typical output stage.pdf

Thanks ,have a good day

 

[pacificblue]

View attachment 172116

You are mixing several things.
- The idle current or better bias current flows through the transistors only. No bias current flows through the speakers.
- The DC current through the speakers when no signal is present results from the offset voltage.
- The current that flows through the speakers due to the audio signal is not connected to the bias current.
- In a single ended output stage there is no DC current through the speakers, because it is blocked by the output capacitor.
- The output impedance does not depend on the bias current. It depends on the static impedance of the output stage, and on the dynamic impedance determined by the open loop gain and the actual gain setting.
- When you bias a push-pull output stage into class A the output impedance remains the same. In the situation you are looking at only the idle current flows. The output voltage is 0 which means the output impedance is infinite for both amplifiers at that moment, whether the idle current is 10 mA, 1 A or even 1 kA. The difference is that the higher bias keeps the transistors switched on up to a higher output current. Therefore switching distortion will set in earlier in a class B output stage than in a class A biased output stage. In a single ended output stage there is no switching distortion. The lack of that distortion is what objectively improves the sonic performance of class A amplifiers over that of class B amplifiers. Subjectively some people even like class B amplifiers better, because they have gotten used to switching distortion and come to like it. These people usually say that class A amplifiers sound dull and lifeless.

 

[krizis]

hi pacificblue

You are mixing several things.
- The idle current or better bias current flows through the transistors only. No bias current flows through the speakers.
- The DC current through the speakers when no signal is present results from the offset voltage.
- The current that flows through the speakers due to the audio signal is not connected to the bias current.
- In a single ended output stage there is no DC current through the speakers, because it is blocked by the output capacitor.
- The output impedance does not depend on the bias current. It depends on the static impedance of the output stage, and on the dynamic impedance determined by the open loop gain and the actual gain setting.
- When you bias a push-pull output stage into class A the output impedance remains the same. In the situation you are looking at only the idle current flows. The output voltage is 0 which means the output impedance is infinite for both amplifiers at that moment, whether the idle current is 10 mA, 1 A or even 1 kA. The difference is that the higher bias keeps the transistors switched on up to a higher output current. Therefore switching distortion will set in earlier in a class B output stage than in a class A biased output stage. In a single ended output stage there is no switching distortion. The lack of that distortion is what objectively improves the sonic performance of class A amplifiers over that of class B amplifiers. Subjectively some people even like class B amplifiers better, because they have gotten used to switching distortion and come to like it. These people usually say that class A amplifiers sound dull and lifeless.

hi pacificblue,
I think it is no need to go on to stare at the current through the speaker.
It seems you did not get what I wanted to say,oh sorry,this must be caued by my poor English.I just wanted to arouse everybody to think about the BJT-CE impedance,which is very important in our amplifiers,and the expressions is just for the model I drawed,it is not all-purpose.if somebody want to use and will find there is no current through the loads.

what is the relation between the bias current and offset voltage?

the output impedance,the model I showed was open loop,so there is no relation with the loop gain at this time.and the impedance did not relate to the dynamic,I just talk about the static.

switching distortion,class b amplifier have this problem,but class ab and class a don't have this distortion now.

Best regards.

 

[pacificblue]

It seems you did not get what I wanted to say

Let's check. I believe you wanted to say that a BJT's impedance changes with the current that flows through it. You swapped cause and effect here. A transistor is basically an adjustable impedance and the current through it depends on how high you gate it.
Then you concluded that a class A amplifier has a lower output impedance, because the idle or bias current is higher than in a class B amplifier. That is not so. Only the transistor has a lower impedance at that moment to let a higher current pass. The amplifier as a complete circuit has not.
The third thing you seemed to want to say is that an amplifier with a lower output impedance sounds more powerful. It does not. The advantage of low output impedance aka higher damping factor is that the voltage at the speaker terminals remains more constant across the frequency when a real speaker is connected. That does not correspond to more powerful sound.

what is the relation between the bias current and offset voltage?

Except for an inevitable interaction in real circuits, they are individual concepts.

the impedance did not relate to the dynamic,I just talk about the static.

On the contrary. The static impedance of a transistor is what you get, when the transistor is fully open. You described a situation where only the idle current flows. The transistor is not fully open then, but only as far open as the bias setting determines. And when you talk about good sound, the static impedance is of no avail, because it only comes into play, when the amplifier is clipping. A clipping amplifier is not a good sounding one, at least not objectively.

 

[krizis]

Let's check. I believe you wanted to say that a BJT's impedance changes with the current that flows through it. You swapped cause and effect here. A transistor is basically an adjustable impedance and the current through it depends on how high you gate it.
Then you concluded that a class A amplifier has a lower output impedance, because the idle or bias current is higher than in a class B amplifier. That is not so. Only the transistor has a lower impedance at that moment to let a higher current pass. The amplifier as a complete circuit has not.
The third thing you seemed to want to say is that an amplifier with a lower output impedance sounds more powerful. It does not. The advantage of low output impedance aka higher damping factor is that the voltage at the speaker terminals remains more constant across the frequency when a real speaker is connected. That does not correspond to more powerful sound.


Except for an inevitable interaction in real circuits, they are individual concepts.


On the contrary. The static impedance of a transistor is what you get, when the transistor is fully open. You described a situation where only the idle current flows. The transistor is not fully open then, but only as far open as the bias setting determines. And when you talk about good sound, the static impedance is of no avail, because it only comes into play, when the amplifier is clipping. A clipping amplifier is not a good sounding one, at least not objectively.

Hi pacificblue,
yes,you did not get what I wanted to say.
"You swapped cause and effect here",
the expressions was just for figuring out the impedance at that moment,it was not telling you the impedance caused by the current through it.

"Only the transistor has a lower impedance at that moment to let a higher current pass"
this is right,but also what I said was right.in a push-pull class a,the current is higher than class ab,(from the words you said,I think you have understand the class a bjt-ce impedance is lower than class ab)but it is not for getting more current to drive the speaker,because just as you said,the idle current was through the two transistors,not through the speaker.high current is a result when we want a low impedance,but this high current is not what we
want.(just in idle static).I think the situation you talked about is not idle static.

"damping factor",in a loop system the damping factor not same as a open loop system,what you talked about was not the same thing I talked.


"Except for an inevitable interaction in real circuits, they are individual concepts." aha,I dont know what you want to say.

"And when you talk about good sound, the static impedance is of no avail,"
just for your words,an ideal amplifier should have a low output impedance,this is not only for getting more power,also for better performance

Thanks

 

[pacificblue]

you did not get what I wanted to say.

Maybe you see my dilemma when you re-read your posts:

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.

in a push-pull class a,the current is higher than class ab,(from the words you said,I think you have understand the class a bjt-ce impedance is lower than class ab)but it is not for getting more current to drive the speaker,because just as you said,the idle current was through the two transistors,not through the speaker.high current is a result when we want a low impedance,but this high current is not what we
want.(just in idle static).I think the situation you talked about is not idle static.

Do you want low output impedance to get a better sound (post #8)? Or do you want low transistor impedance in idle state (post #18)?
How does low transistor impedance in idle state affect the sound? There is no sound at all in idle state. And the output impedance in idle state is always infinite per definition.

Look at real-world output impedances. You will be hard pushed to find class B or AB amplifiers that even go as high as 0,5 Ohms. Good amps remain below 0,1 Ohm. Compare that to the 2,5 kOhms from your post #8 and explain the significance of that number.

 

[AndrewT]

I cannot keep quiet any longer.

Pacific,
bow out gracefully.