BIAS Formula?

[Yoshi]

Hello... I´ve a question... is there any formula that can define the BIAS of an op-amp ?

[Rob Strand]

Hello... I´ve a question... is there any formula that can define the BIAS of an op-amp ?

You need something more concrete.   Most effects pedals are very simple:

       Vout = Vref

If you look at page 6 from this datasheet you can see it has an R4 resistor which has a DC path from the opamp -ve input to ground.  That can be use to bias the output of an opamp to a specific DC voltage.

The specific problem in this case is internal Vref for the chip is 1.8V but you want the output of the opamp to bias at Vcc/2 for best output swing.   In the examples Vcc=6V for this chip so you want the output to bias at 3V.

https://www.onsemi.com/pdf/datasheet/sa571-d.pdf

See figures 6 and 7 and read the paragraphs and equations in between those figures.  The resistors internal to the chip are R4=30k, R3=20k.  The RDC resistors are external to the chip.

[Yoshi]

oooooo cool!!!

for example i´ve this buffer... and in pin 8 is going +18v and in pin 4 is going -9v, The Vref is 4.5v ( the idea of a klon centaur), But I dont know how to set a correct Bias Resistor, plz can you help me?

 

[Rob Strand]

for example i´ve this buffer... and in pin 8 is going +18v and in pin 4 is going -9v, The Vref is 4.5v ( the idea of a klon centaur), But I dont know how to set a correct Bias Resistor, plz can you help me?

The simple answer is since  Vout = Vref, for Vcc=18V you want Vout=18/2 = 9V so you need to connect the 1M resistor to 9V not 4.5V.

If you connect the 1M to 4.5V the circuit will work just that the opamp will swing positively 18V-4.5V=13.5V and swing negatively 4.5V  (that's rough because I ignored 1V to 2V swing loss due to the opamp).  However you are the losing headroom gained by running the opamp at 18V.

The thing is, in this case you can't play with the resistor value to *change* the bias voltage.  That's why you need a Vref (for that opamp) which is 9V not 4.5V.

Depending on the rest of the circuit your Vref might actually be 9V when running from 18V and not 4.5V like you think.

[Yoshi]

oooooooooooooooo that´s great.... But... if V- is conected to -9volts... With the +18v and the -9v ... this dosen´t create a 27 volt range?

[Rob Strand]

oooooooooooooooo that´s great.... But... if V- is conected to -9volts... With the +18v and the -9v ... this dosen´t create a 27 volt range?

Yes you can do that.   To maximize headroom Vref for that opamp should be biased at (18V + (-4.5V))/2 = +6.75V but like before you can bias at anything with some loss of headroom.  For convenience your existing(?) +4.5V Vref isn't too far off +6.75V.

Obviously the output of U1A goes somewhere, like the two paths on your schematic.   When these circuits power up the output of the opamp output could glitch anywhere between -4.5V to +18V.    If the following circuits can't handle -4.5V to +18V it could blow the next stage up!!!.

[antonis]

To maximize headroom Vref for that opamp should be biased at (18V + (-4.5V))/2 = +6.75V

Can't get that (-4.5V), Rob..

For +18V/-9V supply, +4.5V Vref should be ideal..
(considering rail-to-rail op-amp, it could equally swing 13.5V to both rails..)

P.S.
I just follow your promting to check what you write while being wakeful..

[m4268588]

I may not understand this issue.

+18V -----+
          |
         10k
          |
          +-----+-> Vref
          |     |
         10k   Cap
          |     |
-9V  -----+    Gnd

[R.G.]

The issue is that an opamp with a V+ of +18V and a V- of -9V can swing its output voltage (mostly) anywhere between -9V and +18V. Different opamp chips will vary in how close their output pin can swing to the power supply voltages, but in theory, the full power supply range.

It is common practice to bias the output pin of an opamp at half the available power supply voltage total. This allows the biggest possible voltage swing on the output pin before the output clips from not-enough-voltage in one direction or the other. In your diagram, the bias voltage is 27V/2 = 13.5V lower than the +18V supply, meaning it's +4.5V compared to the "ground" on the capacitor. So far, so good.

Standard opamps have no real idea what "ground" is. All they know is the voltage on their two input pins. They work just fine with any DC bias voltage connected to their + input that happens to be within their power supply voltage range. It is only the circuits before and after the opamp that may care what voltage is "ground" compared to the opamp power supplies. In the circuit as shown, the opamp could swing to (nearly) +18V and down to (nearly) -9V. But the circuits after the opamp output might not work OK with an input voltage that swings to +18V. Or, if the output of the opamp is capacitor coupled, the circuits after the opamp might be presented with +-13.5V. If their power supply is not that big, they could be driven into ugly clipping behavior, or in some rare cases even damaged.

[Rob Strand]

Can't get that (-4.5V), Rob..

For +18V/-9V supply, +4.5V Vref should be ideal..
(considering rail-to-rail op-amp, it could equally swing 13.5V to both rails..

Thanks for catching it.  I somehow misread the -ve power connection to the opamp as -4.5V when it should be -9V.

[Yoshi]

The issue is that an opamp with a V+ of +18V and a V- of -9V can swing its output voltage (mostly) anywhere between -9V and +18V. Different opamp chips will vary in how close their output pin can swing to the power supply voltages, but in theory, the full power supply range.

It is common practice to bias the output pin of an opamp at half the available power supply voltage total. This allows the biggest possible voltage swing on the output pin before the output clips from not-enough-voltage in one direction or the other. In your diagram, the bias voltage is 27V/2 = 13.5V lower than the +18V supply, meaning it's +4.5V compared to the "ground" on the capacitor. So far, so good.

Standard opamps have no real idea what "ground" is. All they know is the voltage on their two input pins. They work just fine with any DC bias voltage connected to their + input that happens to be within their power supply voltage range. It is only the circuits before and after the opamp that may care what voltage is "ground" compared to the opamp power supplies. In the circuit as shown, the opamp could swing to (nearly) +18V and down to (nearly) -9V. But the circuits after the opamp output might not work OK with an input voltage that swings to +18V. Or, if the output of the opamp is capacitor coupled, the circuits after the opamp might be presented with +-13.5V. If their power supply is not that big, they could be driven into ugly clipping behavior, or in some rare cases even damaged.

bro ... how can I prevent that "ugly" clipping?

[amptramp]

Check the spec sheet for whatever op amp you are using.  There will be a spec for common mode input voltage range.  Your bias supply should center itself in that range, not the center of the power supply voltages.

For example, if you are using an LM358, the input common mode range is V- to V+ -2 volts.  Therefore, if you are operating from a 9 VDC supply, your range is 0 to 7 volts and you should bias your circuitry to the middle of that range or 3.5 volts.

If you are running a TL072, the worst case bias is V- +4 volts and V+ -4 volts but typically V- +3 volts and V+.  If you are running typical values with a 9 VDC supply, this means the common mode input range is 3 to 9 volts and the center of that should be +6 volts, not the +4.5 volts that a lot of people use.  If you use worst-case values, the range is 4 to 5 volts and the center would be 4.5 volts, but a lot of circuits that run more than 1 volt peak-to-peak seem to work without problems, so the worst case doesn't seem to happen all that much.

Biasing the circuit to the halfway point between the high and low input common mode range should allow the most signal to pass through with the least clipping.  It will not necessarily be the Vcc/2 voltage most people use.  Note that you may have a signal that has much more excursion in one direction or the other so a pulse signal that is AC-coupled into an op amp stage will be close to the average signal, so it would be possible to run into clipping of pulse peaks.  If you have peaks always going in the same direction then you need to adjust the bias to get all of the signal through.

[antonis]

I think OP's initial query was about optimum bias point for a given power supply..
(neither about particular device maximum supply range nor about succeeding stage supply compliance..)

Like something V_CC / 2 for a grounded CE amp or (V_CC +V_E) /2 for an Emitter degenerated CE amp with bypassed Emitter resistor..