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I've been reading an article discussing single-supply op-amp design. Regarding Sallen-Key low-pass filter circuits, the author states that:

A few filter topologies, such as the Sallen-Key low pass (see section 1), require a resistive divider to DC bias the input. Addition of this network negates the requirement of a Virtual Ground.

By virtual ground, the author is referring to ICs such as the TLE2426 virtual ground chip. Can someone explain why Sallen-Key requires a resistor divider? Why won't biasing the AC-coupled signal (i.e. put a capacitor in series with the input) to Vcc/2 (provided by e.g. a TLE2426) work?

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Let's look at a Sallen-Key low-pass filter:

enter image description here

One thing you will notice: the filter does not introduce any additional DC path to ground. C2 is connected to "ground", but since there is no DC path to it, it doesn't actually matter where it's connected, as long as it's a fixed voltage. We could just as well connect it to \$V_{CC}\$, or any other power rail. It doesn't matter, except for power-on transients.

How about a high-pass filter?

enter image description here

Here, we have a path to ground through R2, but R2 is 10kΩ. The point of a virtual ground IC is to provide a low impedance virtual ground, but here we need a 10kΩ ground. We don't need an IC for that, we just need a voltage divider made of two 20kΩ resistors. Sure, you could use a virtual ground IC and follow it with a 10kΩ resistor, but what's the point? A pair of 20kΩ resistors is a lot simpler.

Look at the Sallen-Key topology in general:

enter image description here

In this topology, there is always some impedance (\$Z_4\$) between the filter and ground. Since the point of a virtual ground IC is to make a low impedance ground, but we would never need that, the Sallen-Key "negates the requirement of a Virtual Ground". In other words, it isn't that you couldn't use a virtual ground IC: it's that you'd never need to use one.

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  • \$\begingroup\$ In your first diagram, the DC path to ground is through the output impedance of whatever is providing Vin, and if it weren't there, input bias currents would saturate the amp. \$\endgroup\$ – Scott Seidman Jan 5 '15 at 17:35
  • \$\begingroup\$ @ScottSeidman Sure you can think of it that way, but that's just another way of saying that the input voltage must be somewhere between the supply rails. But I can be a bit more specific it my wording... \$\endgroup\$ – Phil Frost Jan 5 '15 at 18:03
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Actually the resistor divider is providing a virtual ground. The difference is that unlike other filter topologies that need a stiff ground to absorb current through network or feedback resistors, the Sallen-Key low pass filter only needs a high impedance bias voltage (its only load is the opamp's non-inverting input, which should draw negligible bias current).

At unity gain there should be no problem DC coupling subsequent stages, but higher gain will magnify any offset voltage present. If the opamps have unacceptably high offset voltage at high gain then AC coupling may be required. Of course any AC coupling actually turns the 'low pass' filter into a bandpass filter!

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  • \$\begingroup\$ This is my point: the resistor divider also gives Vcc/2, just like the TLE2426. So why can't 2426 be used? \$\endgroup\$ – seertaak Jan 5 '15 at 11:53
  • \$\begingroup\$ Here the "resistor divider" does not provide a virtual ground; it just provides a "soft" biasing voltage (1/2Vcc) that can be easily affected by the input voltage. This is a usual biasing technique for single-supplied op-amp circuits. \$\endgroup\$ – Circuit fantasist Jan 5 '15 at 11:53
  • \$\begingroup\$ @Circuitfantasist: well, one of my favorite quotes from an electronics book is: The one thing that should be remembered whenever the term "ground" is used, is that "Ground is a place where potatoes and carrots thrive"! Almost every time we use "ground" it's a convention for some return current through some more or less negligible resistor. \$\endgroup\$ – Fizz Jan 5 '15 at 13:10
  • \$\begingroup\$ Seertaak, you are asking "So why can't 2426 be used"? My question: At which location (node) of the Sallen-Key filter you would place the 2462? Do you find the answer to your question by yourself? \$\endgroup\$ – LvW Jan 5 '15 at 13:47
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    \$\begingroup\$ You shorted your signal to (the virtual) ground! Two 100k resistors in a voltage divider present the same input impedance as one 50k resistor going to Vcc/2. So connect a 50k resistor from the TLE2426 to the output of the capacitor, and the signal will be able to develop an AC voltage across it. \$\endgroup\$ – Bruce Abbott Jan 5 '15 at 17:44
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Each opamp based circuit that is intended for ac signals (like filter circuits) must have the capability that its output can swing below and above the quiescent operating point. Hence, it is logical that this Q point is in the middle between both supply voltages (for dual supply: 0 V).

For single-supply operation this Q point is at half of the supply voltage. Therefore, we use a voltage divider which produces this bias that is connected to the non-inverting input. In case of 100% feedback (unity gain) this bias voltage also appears at the output (as desired). As a consequence, the output now can swing around this quiescent operating point.

As mentioned at the end of the referenced article, the voltage divider, of course, has an influence on the filter circuitry (time constant); therefore, it can be wise to use an additional opamp (as shown in the last figure) for decoupling purposes. On the other hand, it is, of course, possible to take this divider into account during calculation of the filter elements.

However, note that this works only for DC unity gain. Therefore, this scheme is applicable for the Sallen-Key unity-gain version only. For other circuits (e.g. Sallen-Key with gain) we are required to discriminate between ac and dc feedback (using a series capacitor).

Remark (edit): The referenced document also shows the GIC based Fliege filter. In this case, no voltage divider is necessary to produce the Q point (50% of the supply voltage) because the non-inv. input of the lower opamp is not connected to the input signal. However, a large coupling capacitor is necessary (not shown in the figure).

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Any op-amp circuit (filter or amplifier) needs to have input signals that are constrained to within the rails of the op-amp(s). Normally this is achieved with a midpoint at 0V (as per the case of an op-amp with +V and -V supply rails) but, if the input signal p-p amplitude is only a volt then there is no reason why the "midpoint" should not be offset from 0V - the only proviso is that if the op-amp provides gain then the output (which may be 10Vp-p for example) is also constrained to within the supply rails.

On a +/- 15V rail, having a signal output of 10Vp-p, the "ground" dc voltage could be easily 6V above 0V. There are some circuits that might require this and there is nothing wrong with this conceptually - the 10Vp-p output will be actually peaking at +11V and +1V. This is entirely supported by most op-amps. Op-amps don't care where the "midpoint" is.

Single rail op-amp circuits are no different - the internal "midpoint" can be at +6V and from a single supply of +15V, the output would still be at +11V (max peak) and +1V minimum peak. A more natural mid-point would be 7.5 volts but the op-amp doesn't care about this at all providing the output doesn't get too close to positive rail and 0V (negative rail) - if it does then use a rail-to-rail output op-amp.

By the way, there is no such thing as a single rail op-amp - all op-amps are capable of single rail operation.

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@Respawned Fluff, will you say that the voltage divider R1-R2 in the transistor circuit below creates a virtual ground?

enter image description here

There is a lot of philosophy in this topic...

To make a current flow between two points inside a circuit, we must create a potential difference (voltage) between them. For this purpose, we keep the voltage of the one point (the ground) constant, and change the (input) voltage of the other point. So the ground is the other point, not the input point.

"Ground" makes an association with something stable, solid, stiff... that does not "move" when we change the input voltage from the "other side" (or source/sink a curent to/from the ground). So, this point may have any voltage (to be a "shifted ground") but obligatory it must be low resistive enough... just to call it "ground".

So, here the voltage divider output does not serve as any ground... it produces another input (bias) voltage... it is another input voltage source. Thus we have two voltage sources connected in parallel to the same point that strive to set its voltage.

The one of them (the voltage divider supplied by Vcc) is permanently connected (no matter of the frequency). It is a "bad voltage source" (having significant output resistance)... or more precisely, it is an "intentionally worsened voltage source". So, if you still want to think of it as of a kind of virtual ground, you can name it "intentionally worsened virtual ground".

The other (input) source is perfect... but, at low frequency, it is "disconnected" (by the capacitor) from the common point... and it does not affect its voltage that is set to 1/2Vcc by the imperfect voltage source (the voltage divider). At high frequency, the capacitor "connects" the perfect input source in parallel to the imperfect one... and forces its voltage upon the common point.

This is the biasing trick here - at high frequency, to connect a "strong" (AC+DC) voltage source to a "weak" (DC) voltage source.

Another unconventional viewpoint is to think of this arrangement as a voltage summer with weighted inputs (applying the superposition principle). The input from the side of the voltage divider has relatively low but constant weight, while the input from the side of the input source has variable weight (big - at high frequency, and small - at low frequency).

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