Post by solderdude on Jun 27, 2013 7:29:33 GMT
Biasing op-amps in class-A is a complicated subject as there is no cut and clear 'value' for the needed 'bias' current as this will depend on the current drawn from the output PLUS that of the feedback resistor.
Highly debatable as well....
Let me elaborate... WARNING nerdy technical stuff ahead.
It also is circuit dependent, in this case when using a BUF364 buffer behind the OPA637/627 the buffer (BUF634) is the actual output device and for an amplifier as a whole to be in class-A the output buffer device (BUF634) should be the one set in 'class-A'.
In this aspect THIS tangentsoft article is perhaps a bit misleading if not interpreted correctly.
The intro is correct but ONLY the voltage gain stage is ´biased´ here and NOT the output stage which would be the idea.
And as it happens to be the case THAT output part (which actually provides the output current) is still class-AB and thus the amplifier design is class-AB with the ONLY difference being one 'half' of the output stage of the OPA637 is not 'in use' any more (at least they assume it is not).
Forcing the OPA's output stage to class-A is not even needed at all as all op-amps are operating in class-AB.
It does appear to be tempting to get rid of the 'B' but the 'A' in the AB part means for low currents the output stage runs in class-A mode sloping to classB mode when currents drawn become higher than those of the idle current of the op-amps output stage.
The output current of an op-amp is determined by the value of the feedback resistor, input current of the op-amp (in practice irrelevant as they are way too small to make a difference), the load resistance, the load capacitance (when higher frequencies are involved) and of course the output voltage.
a LOT of variables.
In case of the tangensoft 'explanation' and considering the BUF634 is used as an output buffer the 'class-A' offset is complete nonsense and not needed as the OPA637 already IS in class A when used without that extra resistor/current source.
This is why...
The output of the opamp is loaded ONLY by the input of the BUF634 and is not loaded by a (local) feedback resistor.
As the BUF634 is basically nothing more than a class-AB emitter follower circuit the input current of the BUF634 is dependent on the output current of this buffer that is drawn by the headphone.
The BUF634 can be set to a higher BandWidth via the BW-pin within a certain range.
In the highest BW setting more current is needed in the output stage and thus also more in the input stage.
Since we want to determine the maximum current the BUF634 input requires we should use the worse case scenario numbers.
Max BW in this case.
The datasheet has given values for a 100 Ohm load in both low and high bandwidth modes.
BUT headphones can also be lower in impedance than 100 Ohm.
Most headphones are around 32 Ohm so a factor 3 lower in value than the specified 100 Ohm.
Since it is an emitter follower, with some 'control and protection' circuitry surrounding it, the input current will thus be 3x higher.
So again we calculate worse case scenario which is a factor 3 above the values in mentioned in the BUF634 datasheet.
Also there is the matter of input capacitance. This is only 8pF, so does not appear to be that high but the higher the frequency is the more current a capacitor will draw.
Let's assume 20kHz... 8pF this means the 'impedance' of the capacitor will be down to 1Mohm at this point.
Now in practice we will NEVER see a 20kHz 0dB output voltage except when we use a tone-generator and drive the amp full power.
But again we assume worse case scenario, so 1MOhm it is.
The load the input of the BUF634 imposes on the OPA637 thus consists of 3x the load a 100 Ohm resistor would have.
With a 100 Ohm load in wide BW mode the datasheet states: 8 MOhm//8pF (// means in parallel to).
That 8 MOhm will thus be 3x smaller so 2.6 MOhm (when driving a 32 Ohm load) in parallel with the calculated 1 MOhm load the 8pF capacitance creates at 20kHz.
This makes a grand total 730 kOhm.
Current = Volts/Resistance. Let's assume the opamps work on their max supply voltage, in this case (OPA637) +/- 15V.
In reality the +/- 15V output voltage can never be reached as there as the OPA637 is not a so-called Rail-to-Rail output device.
Still... let's assume the theoretical maximum of 15V is possible as this is worse case.
The max. current that will flow into the BUF634 (is the output current of the OPA637) and assuming the amplifier design has a minimal gain of 5x, and thus is not secretly oscillating, the MAX current the BUF634 will draw from the output is 15/730000 = 20uA.
So 20uA is the absolute max current that can possibly be drawn and in practice it will be much lower.
The OPA637 is a high-speed op-amp and thus it needs to have a relatively high idle current in it's output stage.
The current consumption of the OPA637 = 7mA (at 15V). Part of this current is drawn by the amplification parts inside but at least 1mA is running in the class-AB output section (is probably even higher !).
For this reason the output stage will ALWAYS be in class-A when the drawn output current is LOWER than 1mA.
20uA is 50x lower than that idle current and in reality the idle current will be higher and the current load the BUF634 draws will be much lower.
Conclusion the OPA637 (OPA627) when loaded with a BUF634 will ALWAYS be in class-A no matter how 'severe' the load on the BUF634 is and will NEVER be close to operating in class-B
Therefore the Tangentsoft article is complete and utter... errm ... sales pitch talk when it comes to the idea that an op-amp driving a buffer needs to be forced into class-A because it already IS in class-A.
I say ... remove the resistor and IF you still want ONE output device in the OPA637 being switched off the extra load resistor would need to draw >7mA to be sure.
When a resistor is used (which will become a NON linear current load b.t.w.) the voltage across that resistor varies and is why they recommend to use a current source it would thus have to be 2k2 (when 15V power supply voltage is used) and lower when lower supply voltages are used.
so when operating on +/- 7V for instance the resistor should be 1kOhm.
The in forums often 'proposed' 10k is too high in value and may even set the OP637 on the threshold of going from class-A to class-B which is the exact opposite of the intended goal.
IF the amplifier would have to be real class-A the BUF634 should be loaded with a current source/resistor and draw a current that will be determined by the load impedance and maximum output voltage.
The BUF634 would get quite hot and thus one needs to use a BUF634T (the TO-220/5 case) with a large heatsink attached to it.
I can do the math if you want...
As input I will need:
power supply voltage and impedance(s) of the load(s) being headphones in this case.
Highly debatable as well....
Let me elaborate... WARNING nerdy technical stuff ahead.
It also is circuit dependent, in this case when using a BUF364 buffer behind the OPA637/627 the buffer (BUF634) is the actual output device and for an amplifier as a whole to be in class-A the output buffer device (BUF634) should be the one set in 'class-A'.
In this aspect THIS tangentsoft article is perhaps a bit misleading if not interpreted correctly.
The intro is correct but ONLY the voltage gain stage is ´biased´ here and NOT the output stage which would be the idea.
And as it happens to be the case THAT output part (which actually provides the output current) is still class-AB and thus the amplifier design is class-AB with the ONLY difference being one 'half' of the output stage of the OPA637 is not 'in use' any more (at least they assume it is not).
Forcing the OPA's output stage to class-A is not even needed at all as all op-amps are operating in class-AB.
It does appear to be tempting to get rid of the 'B' but the 'A' in the AB part means for low currents the output stage runs in class-A mode sloping to classB mode when currents drawn become higher than those of the idle current of the op-amps output stage.
The output current of an op-amp is determined by the value of the feedback resistor, input current of the op-amp (in practice irrelevant as they are way too small to make a difference), the load resistance, the load capacitance (when higher frequencies are involved) and of course the output voltage.
a LOT of variables.
In case of the tangensoft 'explanation' and considering the BUF634 is used as an output buffer the 'class-A' offset is complete nonsense and not needed as the OPA637 already IS in class A when used without that extra resistor/current source.
This is why...
The output of the opamp is loaded ONLY by the input of the BUF634 and is not loaded by a (local) feedback resistor.
As the BUF634 is basically nothing more than a class-AB emitter follower circuit the input current of the BUF634 is dependent on the output current of this buffer that is drawn by the headphone.
The BUF634 can be set to a higher BandWidth via the BW-pin within a certain range.
In the highest BW setting more current is needed in the output stage and thus also more in the input stage.
Since we want to determine the maximum current the BUF634 input requires we should use the worse case scenario numbers.
Max BW in this case.
The datasheet has given values for a 100 Ohm load in both low and high bandwidth modes.
BUT headphones can also be lower in impedance than 100 Ohm.
Most headphones are around 32 Ohm so a factor 3 lower in value than the specified 100 Ohm.
Since it is an emitter follower, with some 'control and protection' circuitry surrounding it, the input current will thus be 3x higher.
So again we calculate worse case scenario which is a factor 3 above the values in mentioned in the BUF634 datasheet.
Also there is the matter of input capacitance. This is only 8pF, so does not appear to be that high but the higher the frequency is the more current a capacitor will draw.
Let's assume 20kHz... 8pF this means the 'impedance' of the capacitor will be down to 1Mohm at this point.
Now in practice we will NEVER see a 20kHz 0dB output voltage except when we use a tone-generator and drive the amp full power.
But again we assume worse case scenario, so 1MOhm it is.
The load the input of the BUF634 imposes on the OPA637 thus consists of 3x the load a 100 Ohm resistor would have.
With a 100 Ohm load in wide BW mode the datasheet states: 8 MOhm//8pF (// means in parallel to).
That 8 MOhm will thus be 3x smaller so 2.6 MOhm (when driving a 32 Ohm load) in parallel with the calculated 1 MOhm load the 8pF capacitance creates at 20kHz.
This makes a grand total 730 kOhm.
Current = Volts/Resistance. Let's assume the opamps work on their max supply voltage, in this case (OPA637) +/- 15V.
In reality the +/- 15V output voltage can never be reached as there as the OPA637 is not a so-called Rail-to-Rail output device.
Still... let's assume the theoretical maximum of 15V is possible as this is worse case.
The max. current that will flow into the BUF634 (is the output current of the OPA637) and assuming the amplifier design has a minimal gain of 5x, and thus is not secretly oscillating, the MAX current the BUF634 will draw from the output is 15/730000 = 20uA.
So 20uA is the absolute max current that can possibly be drawn and in practice it will be much lower.
The OPA637 is a high-speed op-amp and thus it needs to have a relatively high idle current in it's output stage.
The current consumption of the OPA637 = 7mA (at 15V). Part of this current is drawn by the amplification parts inside but at least 1mA is running in the class-AB output section (is probably even higher !).
For this reason the output stage will ALWAYS be in class-A when the drawn output current is LOWER than 1mA.
20uA is 50x lower than that idle current and in reality the idle current will be higher and the current load the BUF634 draws will be much lower.
Conclusion the OPA637 (OPA627) when loaded with a BUF634 will ALWAYS be in class-A no matter how 'severe' the load on the BUF634 is and will NEVER be close to operating in class-B
Therefore the Tangentsoft article is complete and utter... errm ... sales pitch talk when it comes to the idea that an op-amp driving a buffer needs to be forced into class-A because it already IS in class-A.
I say ... remove the resistor and IF you still want ONE output device in the OPA637 being switched off the extra load resistor would need to draw >7mA to be sure.
When a resistor is used (which will become a NON linear current load b.t.w.) the voltage across that resistor varies and is why they recommend to use a current source it would thus have to be 2k2 (when 15V power supply voltage is used) and lower when lower supply voltages are used.
so when operating on +/- 7V for instance the resistor should be 1kOhm.
The in forums often 'proposed' 10k is too high in value and may even set the OP637 on the threshold of going from class-A to class-B which is the exact opposite of the intended goal.
IF the amplifier would have to be real class-A the BUF634 should be loaded with a current source/resistor and draw a current that will be determined by the load impedance and maximum output voltage.
The BUF634 would get quite hot and thus one needs to use a BUF634T (the TO-220/5 case) with a large heatsink attached to it.
I can do the math if you want...
As input I will need:
power supply voltage and impedance(s) of the load(s) being headphones in this case.