biasing in multiple opamp circuits

[josh777]

Hi,

I've searched for the answer to this question, but haven't found an answer i can understand, so sorry if this does get asked a lot.

I'm trying to understand how to bias op-amps when there is more than one in the circuit, and I'll use the well known tubescreamer circuit as an example.

The first non-inverting input is biased to 4.5V, and the output DC coupled to the second non-inverting input, which is also biased to 4.5v by the virtual ground. As the first non-inverting input is not connected to DC ground, does it therefore not act as a DC amplifier, thus the first DC output will read 0V, which is why the second non-inverting input must be biased again??

Sorry if the wordings not great.

Josh

[R.G.]

Opamps amplify the difference between the noninverting and inverting inputs. In the TS, the first opamp has its NI input biased to 4.5V. The (-) input is prevented from holding any DC voltage other than the same as the output by the capacitor separating it from ground.

The gain of an opamp from the (+) input to its output is 1+Rf/Ri, where Rf is the feedback resistor and Ri is the resistor from the (-) input to the reference voltage. The gain from the inverting input to the output is just Rf/Ri. Where the (-) input has a capacitor to ground, the cap is effectively an infinite resistance at DC, so the DC gain from the  (-) input to output is Rf/infinity, which is 0. The gain from the (+) input to output is 1+Rf/infinity, or 1+0 = 1.

In a circuit with a cap isolating the Ri resistor to the (-) input, the output will always be the same as the bias voltage on the (+) input, +/- the input offset voltage of the opamp.

So:

As the first non-inverting input is not connected to DC ground

,
It's connected to a bias voltage, so the first output will be at essentially the same DC voltage as the bias voltage, as explained above.

does it therefore not act as a DC amplifier, thus the first DC output will read 0V,

It does act as a DC amplifier. It's just that the gain is 1.000000.

which is why the second non-inverting input must be biased again??

The second amplifier has much the same issues, but a different solution. Notice that the DC level on both the (+) and (-) inputs of the second amp are bias voltage. The DC gain there is 1+Rf/Ri for the voltage into the + input, and Rf/Ri for the voltage into the (-) input. The circuit then subtracts the two; the output is 1+Rf/Ri - Rf/Ri, or 1. So the output still has a gain of 1 from the + input to output for DC levels. It has other gains at other frequencies.

[josh777]

thankyou for such a quick reply, opamps are starting to make a lot more sense now.

[josh777]

hold a sec, now there not.

if the first output is 4.5V, and this is connected to the second input, why does another 4.5V have to be added? Surely this would take it up to 9V??

[Sir H C]

It amplifies the difference of the two inputs, so if both are at 4.5 volts then the output is common mode + the gained difference.

[slacker]

hold a sec, now there not.

if the first output is 4.5V, and this is connected to the second input, why does another 4.5V have to be added? Surely this would take it up to 9V??

Voltages don't add up like that. If you take a line at 4.5volts and connect it via a resistor to another 4.5 volt line then you end up with roughly 4.5 volts. You're right though if you take the output of an opamp thats got 4.5 volts DC on it and connect it to the non inverting input of a second opamp then the 4.5 volts biases the second opamp.

[R.G.]

if the first output is 4.5V, and this is connected to the second input, why does another 4.5V have to be added? Surely this would take it up to 9V??

Your question kicks off three replies in my head.

(1) You may be meaning the way that multiple inputs of the same voltage add together at a (-) input. There is a longish discussion to be had there.
(2) As a first order answer, it doesn't have to be added. I'm not sure why they put that second biasing resistor in there.
(3) As a second order answer, maybe it's to swamp out input offset voltage or some other minor effect.

Let's take those in order.
(1) There is a hookup which may be confusing you. If you have an opamp set up with multiple Ri input resistors going to the (-) input, then the inputs do "add"; more properly,  they each subtracts individually from the voltage on the (+) input. If we have a hookup where the (+) input is at 4.5Vdc, there is a Rf of 10K and two Ri's, each of 10K, we can figure out what the DC output will be for several different cases.
(a) if we leave both Ri's open, then the output voltage is 4.5V. This is because there is 0 current coming into the (-) input pin from either Ri, and so no current needs to flow to/from the output pin to balance this input current out. The only way the output can be supplying zero current is if there is 0 volts across it, and the only way that can happen is if there is 0 differential voltage between the (+) and (-) inputs, so the output seeks and finds the exact voltage of the (+) input and sits there. For this circuit, the output must be the same as the (+) input - always throwing in the afterthought that offset errors and noise may make it slightly different, a few millivolts perhaps. Notice that connecting Ri's from the (-) input pin through capacitors to any other DC voltage has the same effect. The capacitors do not let any DC input current flow, so the output pin must balance at the (+) input's voltage.
(b)Both Ri's are connected to 4.5Vdc; in this case, if the output is at any voltage other than 4.5V, a current flows through the Ri's toward the (-) input pin because the (-) input pin is connected to the output by Rf. The (-) input pin is high impedance, and no current can actually go into the opamp there. But the output of the opamp swings the opposite direction from the (-) input. So if the Ri's are above the (+) input pin's 4.5V, the difference is positive and the output swings down. If the Ri's are below the (+) input pin's 4.5V, the output swings up. In either case, the output comes to rest at exactly the voltage that will suck any excess charge off the (-) input and bring its voltage to exactly the DC voltage on the (+) input, because when the (-) input is exactly the same voltage as the (+) input, there is zero difference to be amplified and the output is not pushed either direction by the amplifier circuit. So the (+) pin and (-) pin come to rest at 4.5V, the output is at 4.5V, and the two Ri's have 4.5V on each end, so no current flows. Note that if there is a capacitor in series with an Ri, then since no DC current can flow through Ri, then both ends are at the same DC voltage and so it has no way to make DC current flow into/out of the (-) pin, and so no effect on the output voltage at DC. This is another way to look at (a).
(c ) One or both Ri's are connected to votlages other than 4.5Vdc. In this case, the output does amplify the DC difference. It does it this way: once again, the - input cannot eat current or store charge, so any current that flows into/out of it through the Ri's must be balanced by the current out/in from Rf.  If one Ri is at 4.5V (i.e., not playing in this what-if) and one suddenly changes to 4.0V, here's what happens: the (+) input is at 4.5V still, and the output was at 4.5V. So now there is 0.5V across the 10K Ri. That means that a current of 0.5V/10K flows out of the (-) pin through Ri. This current has to come from somewhere, and the only place it can come from is the output. So as the voltage on the (-) input drops a few micro-volts below the (+) pin's 4.5V, there is now a voltage difference between the (+) and (-) input pins. That voltage difference is amplified by the amp's DC gain, usually upwards of one to several million. The amplifier starts swinging its output voltage upwards because the differential input voltage has the (+) input higher than the (-) input, and as it rises the voltage across Rf increases at the output end. When it gets to a voltage where the same current is flowing through it as is being sucked out of Ri, the (-) pin is brought back to equal the (+) input voltage, and the amplifier is once again balanced, the differential voltage is now 0, and the amp quits moving the output. So the output settles at a voltage that makes the input current at the (-) pin be 0. That has to be the same current as is coming out of the (-) input through Ri, or (Vout-V(-))/Rf = (-0.5V- v(-))/Ri. We can do a little math and find out that the output voltage is just Vout/Vin = Rf/Ri.

Another way to look at this same situation is that for the same situation, (+) at 4.5V, Rf=Ri=10K and 4.0V on the input to Ri, then the voltage at Ri's input and the output of Rf form a voltage divider. The (-) input can't eat any current, as it's quite high impedance, so the voltage at the (-) pin is just whatever voltage happens to be caused by the voltage divider between the input voltage and the amplifier output voltage through Rf and Ri. In the case of the two resistors we have picked, the (-) input rests exactly halfway between the Ri input voltage and the amplifier output voltage. That's the differential voltage that drives the opamp, and so it must come to rest with the (-) input at the same voltage as the (+) input to make the difference in voltage between them be 0 and make the amplifier stop moving the output.

If we put two different input voltages through the two Ri's: One is at 4.0V and one is at 5.0V. The current into the (-) pin is (5.0-4.5)/10K for one of the Ri's and (4.0-4.5V)/10K in the other case. These currents happen to be equal, so what one input resistor pushes in, the other takes out. The Ri's balance themselves and the output remains at 4.5VDc, no amplifier action needed.

If one is at 2.0V and the other is at 5.0V... now what? We know the amplifier is going to make the (-) pin be within millivolts of the (+) pin. So all we have to do is figure out what currents flow. One Ri is supplying 0.5V/10K into the (-) pin, the other is sucking out 4.5V-2.0V/10K from the pin. So one is providing +50uA into the pin, the other is sucking 250uA out. The Rf must be supplying the difference of 200uA into the (-) pin to balance things, so the output voltage must be 200uA*10K or 2V.

Notice that we get the same answer if we take the stock opamp equations of gain = -Rf/Ri for each Ri and add them.

In fact, the gain= -Rf/Ri and gain = 1 +Rf/Ri for the (-) and (+) inputs were derived as I've just explained, based on the current flows.

[Mark Hammer]

You heard the "smart guy" version.  Now hear the "dumb guy" version.  Hopefully, between the two you'll gain some clarity on the matter.

1) Sometimes you want to shape the bandwidth and spectral content of a stage, and that may require a series cap to accomplish (e.g., for trimming the bass).  That series cap blocks any DC from the preceding stage, so you have to re-bias all over again because you've essentially "lost" the bias voltage in the process.

2) Not all op-amps are equally error-free (which is why there are so damn many different kinds of op-amps).  There can be small offset voltages added into the mix, such that you don't want to "share" bias voltages across stages.  I may grasp it wrong, but whenhigh gain is applied in such stages, that error may be magnified.  In those instances, a DC-blocking cap is used and rebiasing done in the next stage just to be sure that the stage has the "true" V+/2 and not (V+/2) plus some offset.

3) Sometimes, stages need to be isolated from each other, such that they are not all drawing from a "common pool" of Vref current.  Perfect case in point is the LFO section of a modulation effect.  These produce a sudden current draw when they generate the large-amplitude square wave that gets transformed into a triangle for sweep purposes.  This is a common source of audible ticking.  Because of this, you will virtually always see such sub-circuits having their very own voltage divider and smoothing/storage cap between Vref and ground to supply them with their own personal Vref that doesn't interfere with the Vref used elsewhere in the circuit.

I suspect the confusion you are experiencing is because, yes, there are many instances where Vref-sharing seems to be just fine, and everything is tied to the same Vref point, and there are also many instances where a Vref is inserted early in the path and "preserved/reused" for the remainder of the circuit.  (One illustrative example here is the bias voltage that all bucket-brigade delay chips need to be able to pass signal.  Often you see this supplied at the input pin of the BBD, but very often it will be inserted much earlier in the circuit (e.g., prior to the lowpass anti-alias filtering before the BBD), as long as there is no DC-blocking cap along the way.)  At the same time as such "recycling" of Vref or bias voltages takes place, there are also instances where the slate appears to be wiped clean and rebiasing is undertaken.  Obviously, the guidelines for what happens under what conditions are not always clear to people.  To paraphrase that old John Houseman commercial for an investment firm: "Your confusion was gotten the old-fashioned way: You earned it!" 

The general principle, though, is that if it is possible to re-use a bias voltage, and advisable to do so, then it gets re-used.  If it is possible but not advisable, then a new and distinct Vref/bias is obtained.  If it is not contra-indicated, but impossible to do (i.e., DC-blocking along the way), then you *have* to fetch yourself a new Vref all over again.

[josh777]

Thankyou very much for the replies, it's helped greatly. I understand why the op-amp needs to be biased, and the 'extra' bias point after the first stage is just to help stabilise the bias voltage that would be provided by the first stages output anyway, as stated by slacker et al.

Thanks again.

Josh

[petemoore]

  You heard the "smart guy" version.  Now hear the "dumb guy" version.  Hopefully, between the two you'll gain some clarity on the matter. 
  Yes, and again I feel like we're mighty darn welcome,
.. thanks again Mark and RG !!!