hi guys,
with my limited knowledge of electronics i haven't really been able to find out whether it's possible to bias an opamp into class a without using a buffer.
If it's not possible: o well, i'll let'er switch. If it is; can anybody give me an example?
I'm building a opa2132 / 627 / 637 based linestage (without a buffer, obviously).
if i get it to work nicely i might mod an old arcam dac...
with my limited knowledge of electronics i haven't really been able to find out whether it's possible to bias an opamp into class a without using a buffer.
If it's not possible: o well, i'll let'er switch. If it is; can anybody give me an example?
I'm building a opa2132 / 627 / 637 based linestage (without a buffer, obviously).
if i get it to work nicely i might mod an old arcam dac...
hum... of course.
Looking at a inverting topology, having the input on end of the feedback resistors and the non inverting input is usualy to ground..
Consider that the opamp is trying to keep a difference of potential of 0 volts across its inputs, that means the out equal :
Out = -((((input - 0v)*R2)/R1)- 0V)
So, what happens to the formula if you apply a voltage to the Non inverting input?
5v for eg;
Out = -((((input - 5v)*R2)/R1)-5V)
The signal is inverted and 10v above 0v.
Heres a drawing.
Looking at a inverting topology, having the input on end of the feedback resistors and the non inverting input is usualy to ground..
Consider that the opamp is trying to keep a difference of potential of 0 volts across its inputs, that means the out equal :
Out = -((((input - 0v)*R2)/R1)- 0V)
So, what happens to the formula if you apply a voltage to the Non inverting input?
5v for eg;
Out = -((((input - 5v)*R2)/R1)-5V)
The signal is inverted and 10v above 0v.
Heres a drawing.
Attachments
Ilianh said:hum... of course.
Looking at a inverting topology, having the input on end of the feedback resistors and the non inverting input is usualy to ground..
Consider that the opamp is trying to keep a difference of potential of 0 volts across its inputs, that means the out equal :
Out = -((((input - 0v)*R2)/R1)- 0V)
So, what happens to the formula if you apply a voltage to the Non inverting input?
5v for eg;
Out = -((((input - 5v)*R2)/R1)-5V)
The signal is inverted and 10v above 0v.
Heres a drawing.
I think he has in mind adding a resistor between Out and -V, to bias the output stage of the opamp in Class A.
If you do bias one of the inputs, you'll get large DC at the output (if the gain is valid for DC it'll even saturate and nothing's going to come out of that opamp) and I'm not sure that's what you want (DC at the output doesn't mean that the output stage works inclass A).
This sort of stuff?
http://tangentsoft.net/audio/opamp-bias.html
I'm biasing the buffer into class A too. Works fine.
JF
http://tangentsoft.net/audio/opamp-bias.html
I'm biasing the buffer into class A too. Works fine.
JF
... yes, but you can't get more than 1-3 mA or so as max output current. This technique is best if you use it together WITH a buffer. Without this OPA627 has 0.00003% dist!matjans said:hi guys,
with my limited knowledge of electronics i haven't really been able to find out whether it's possible to bias an opamp into class a without using a buffer.
This means that the load can't take more than 0.35 mA or so.matjans said:i think i'll just give it a try and bias it at 0.4 ma or something. we'll see.
carlosfm: what do you mean double the pleasure ?
Using an active constant current source from the op amp output to the V- supply rail provides a lot better biasing that a simple resistor. Using a simple resistor to supply this current will result in the current changing with the output signal. This is because the current through a resistor is voltage dependent and is non discriminate between AC signals and the DC supply voltage.
An active current source maintains a constant current in the face of changing Voltage both AC and DC. The only requirement is that the Minimum dropout voltage and the maximum breakdown voltage of the active current source are maintained. See my post on Head-fi that got tangent and others interested in the concept. http://www4.head-fi.org/forums/showthread.php?threadid=3254&highlight=Cascode+current+source
John how much current are you using to bias the Buffer?
An active current source maintains a constant current in the face of changing Voltage both AC and DC. The only requirement is that the Minimum dropout voltage and the maximum breakdown voltage of the active current source are maintained. See my post on Head-fi that got tangent and others interested in the concept. http://www4.head-fi.org/forums/showthread.php?threadid=3254&highlight=Cascode+current+source
John how much current are you using to bias the Buffer?
ppl said:John how much current are you using to bias the Buffer?
Hi PPL!
I just prototyped something this weekend. So I haven't done this a lot. I just used 8 mA. I'm thinking 40mA maximum (for 300 ohm loads). However, as you've mentioned before, it doesn't take much voltage to generate volume in headphones. My AC voltmeter showed 25mV and it was loud. Not rock loud, but for Debussy it was loud. In that case, 8mA is fine.
I learned other things. I've got to do something about turn on/off pop. Also, the OPA637 generates a large DC offset with the attenuator set to minumum or the maximum (doesn't do this with the '627). This is with long jumper wires into a breadboard and minumum bypass capacitance, so it's not the ultimate test.
Good to hear from you. Borbely, Gilmore, and PPA are all my reference circuits now. I'm not sure it's possible to improve on any of them. Are you working with your discrete buffer yet?
Below you can see my sketch of the current source in the upper left corner. If I stick with transistors, I will use a cascode circuit.
JF
Attachments
matjans said:
carlosfm: what do you mean double the pleasure ?
I mean the LISTENING pleasure.
Soud quality.
Dynamics.
Speed.
Detail.
What were you thinking of?
Ilianh said:heres an idea I just had....
I imagine many ways could be used to do the same.
here, the resistors should be matched to perfection, and the opamps too, otherwise, some may not like the DC offset at the output
That won't work.
You can't use an op-amp with single voltage as you use with +/- rails.
I think he has in mind adding a resistor between Out and -V, to bias the output stage of the opamp in Class A.
If you do bias one of the inputs, you'll get large DC at the output (if the gain is valid for DC it'll even saturate and nothing's going to come out of that opamp) and I'm not sure that's what you want (DC at the output doesn't mean that the output stage works inclass A).
By inserting DC in the signal, don't we get off the switching in the B stage.
And didnt I insert the dc Before the B stage?
Of course, I could be wrong...
That won't work.
You can't use an op-amp with single voltage as you use with +/- rails.
well... As I see it, the only thing that it changes is that evrything that goes in the negative area will be clipped.
Why don't you explain why?
Atleast I based myself on the National lm3875 datasheet.
Ok then...
This is from the LM3875 datasheet.
Single supply.
Notice the two 100k RA resistors.
They are dividing the voltage for the input.
And check the 4700f/50v cap on the output.
Essential if you use single supply (or you'll have huge DC), but very bad for the sound.
This is from the LM3875 datasheet.
Single supply.
Notice the two 100k RA resistors.
They are dividing the voltage for the input.
And check the 4700f/50v cap on the output.
Essential if you use single supply (or you'll have huge DC), but very bad for the sound.
Attachments
Notice the two 100k RA resistors.
thats the bias for the input signal.
And check the 4700f/50v cap on the output.
Thats to get the dc off the signal before getting into the speaker, analyse the circuit I provided to see why there is no offset either.
Essential if you use single supply (or you'll have huge DC), but very bad for the sound.
humm? ok....
anyway, At first I wasnt refering to that, but to the supply specs of the lm3875.
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