The sound of two transistors switching

[Nelson Pass]

Prior to that he designed the Quintessence stuff.

 

[millwood]

Millwood - I don't fully understand this idea, perhaps we should both do some research, simulation or thinking or some of each.
Mike

Mike, it looks like I was wrong. I did some simulations and it turns out loading the output with a current source will indead put one transistor into class a: conducting 360 degrees, under one important condition: there got to be a emitter resistor.

I used a modified Citation 12 (T-driver stage, complementary output stage) loaded with a 1amp CCS to the negative rail. The base for the upper driver / output device went up by a little bit, just enough to generate the needed current on the emitter resistor to satisfy the CCS. As such, there will not be any DC voltage offset: the increase in base voltage will be fully absorbed by the added voltage drop on the emitter resistor.

so two lessons:
1) there has got to be an emitter resistor;
2) the CCS should be bigger than the peak current generated on the load to the opamp. For example, if you are looking for 1v peak swings on a 1k load, the CCS should be at least 1ma. That way, you will completely avoid any switching and only one of the two output transistors will be conducting.

 

[Tube_Dude]

so two lessons:
1) there has got to be an emitter resistor;
2) the CCS should be bigger than the peak current generated on the load to the opamp. For example, if you are looking for 1v peak swings on a 1k load, the CCS should be at least 1ma. That way, you will completely avoid any switching and only one of the two output transistors will be conducting.

Glad to see that you have find this results!

About the lessons:

1) No...if there is no emitter resistor, the result , will be the some ,as the CCS have forced one of the output transistor to supply more current ( in your case 1mA) and he only left class A for more than 1 mA peak output current..
2)Yes...you got it!!

 

[mikelm]

I did some simulations and it turns out loading the output with a current source will indeed put one transistor into class a: conducting 360 degrees, under one important condition: there got to be a emitter resistor.

Thanks for this research millwood, I must play with this in simulation as well to fully get my head round what is happening.

I think that the topology of the o/p stage and how it is biased will govern wether the lower o/p tr will actually turn off.

Some o/p stages are especially designed so that the unused tr will always have a small bias current flowing even when the other one is conducting to the load so it would be wrong to make blanket assumtions but I think in most cases the unused tr will turn off if enough bias current is introduced externally.

After all don't we expect one tr to turn off when the other one is conducting in AB amps ? I think so - This is why I prefer class A

I think this is what tube dude was saying.

Hey Mr Tube Dude or any knowledgable Dude ..... can you confirm this for us ...

cheers

mike

 

[Tube_Dude]

I think this is what tube dude was saying.

Hey Mr Tube Dude or any knowledgable Dude ..... can you confirm this for us ...

cheers

mike

Hi Mike

When a ouput transistor in a push pull output stage is forced by a CCS to suplly a current say 10 mA DC ...only for a current output of more than 10 mA (AC now!!!) the other transistor come in...

I hope have been clear...but English Language is not my speciality!!! I prefer the solder smoke!!!

PS . See also the some discussion at :
http://www.diyaudio.com/forums/showthread.php?s=&threadid=28042

Cheers

PS(2): And please don't call me Mister...call me Jorge because here we are a group of friends sharing the same passion!!

 

[mikelm]

OK thanks Jorge

 

[millwood]

Some o/p stages are especially designed so that the unused tr will always have a small bias current flowing even when the other one is conducting to the load so it would be wrong to make blanket assumtions but I think in most cases the unused tr will turn off if enough bias current is introduced externally.

mike

intuitively, the unused transistor will turn off, unless the original bias is so huge.

one way I thought about is to tie a small resistor to the output before the feedback pick up. like the schematic below. R1 is selected so that its voltage drop (I1*R1) is about 50% of the peak voltage swing of the opamp.

 

[jan.didden]

Millwood, what do you want to accomplish with this? I don't understand the reason.

Jan Didden

 

[Tube_Dude]

one way I thought about is to tie a small resistor to the output before the feedback pick up. like the schematic below. R1 is selected so that its voltage drop (I1*R1) is about 50% of the peak voltage swing of the opamp.

With the resistor you only loose output voltage swing...

 

[millwood]

Millwood, what do you want to accomplish with this? I don't understand the reason.

Jan Didden

the resistor is there so that even if the opamp doesn't have emitter resistors, you will still not get a DC offset at the output.

With the resistor you only loose output voltage swing...

it will not, unless the feedback pick up is before the resistor.

 

[Tube_Dude]

it will not, unless the feedback pick up is before the resistor.

The resistor here are only increasing the open loop output impedance...but it don't change nothing in the issue of the CCS forcing a enriched bias...

 

[jan.didden]

the resistor is there so that even if the opamp doesn't have emitter resistors, you will still not get a DC offset at the output.



it will not, unless the feedback pick up is before the resistor.

Sorry Milwood, I don't follow you on the first point. I cannot see that this does anything on the offset, except make it worse (the feedback being less effective). How do you calculate that the resistor should exactly drop half the output voltage?

On the second point, Jorge is right. The feedback will try to get more output voltage, but with say 12v supply, and 6 v across the resistor, that only leaves 6 v on the load. In fact, the feedback will try so hard that you probably end up with hard clipping.

Jan Didden

 

[millwood]

On the second point, Jorge is right. The feedback will try to get more output voltage, but with say 12v supply, and 6 v across the resistor, that only leaves 6 v on the load. In fact, the feedback will try so hard that you probably end up with hard clipping.

Jan Didden

essentially, you raise the "true opamp output" (the output pin on the opamp, just before the resistor) to half of its full output. After the bias current going through that added resistor, the output after the resistor is exactly zero, as it should be.

Say that you have an opamp that goes from -12v to +12v, and CCS at 1ma. You drop in a resistor of 6k in there. The voltage drop on the resistor will be 6v, and the output of the opamp will be at 6v, and the feedback will see 6v - 6v = 0v.

The maximum swing this particular setup is able to achieve is +6v and -6v (from +12v to 0v at the opamp output pin).

By comparison, say you used a 9k resistor. The opamp output pin will idle at 9v. Its maximum swing is from 3v (12v-9v) to -9v (0v-9v). or +/- 3v assuming music is symmetric on average.

as you can see, as far as the feedback is concerned, it is dealing exactly with the same amp. Albeit the amp's swing is reduced because you want it to work in class A, thus only one of the two transistors will be conducting.

 

[Tube_Dude]

By comparison, say you used a 9k resistor. The opamp output pin will idle at 9v. Its maximum swing is from 3v (12v-9v) to -9v (0v-9v). or +/- 3v assuming music is symmetric on average.

as you can see, as far as the feedback is concerned, it is dealing exactly with the same amp.

In this point you are correct!!

Albeit the amp's swing is reduced because you want it to work in class A, thus only one of the two transistors will be conducting.

In this second case the resistor don't do any change!!

PS : I must leave for a while...i go to hear my tweked digital amp in a friend house that have some Rehdeko speakers....3 speakers full range with 104dB W /m....
Anybody know this speakers??

 

[dmfraser]

Biasing Op AMps

Any op-amp based on the LM324/LM358 architecture (quite a few are - its the most common op-amp in the world) have a Class-B output stage. It is a common trick to put a 6.8K resistor from the output to the negative rail to bias these into Class-A operation. If you look at the schematics of cheap Asian auto equalizers, you will usually see this resistor on each op-amp. It is apparantly cheaper for them to do this than to use a decent op-amp.

When this resistor is added, the LM324, which was never meant for audio use, experiences an improvement of 6db to 20db in THD products. However, the mediocre noise performance does not improve. I have the feeling that the LM series op-amps in this person's CD player are based on the LM324 design and it is indeed possible to get a sonic improvement by doing this.

However, a good op-amp like the LM833 or OP275 will see essentially no improvement and may actually get worse.

Note - LM324 is a quad device. All the others mentioned are dual op-amps in an 8 pin package and are pin for pin interchangeable.

Our research shows that for a dual op-amp for audio it is hard to beat the OP275 from Analog Devices with the LM833 not too far behind. If you need a quad device, we recommend the LM837.