The sound of two transistors switching

[Nelson Pass]

Prior to that he designed the Quintessence stuff.

[millwood]

Originally posted by mikelm
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]

Originally posted by millwood


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]

Originally posted by millwood

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]

Originally posted by mikelm


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 :
Op amp questions for Mr. Wurcer

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]

Originally posted by mikelm
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.

[janneman]

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

Jan Didden

[Tube_Dude]

Originally posted by millwood


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]

Originally posted by janneman
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.

Originally posted by Tube_Dude
With the resistor you only loose output voltage swing...

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