Talk:Circuit Idea/Deboo Integrator

I would like to start the discussion about the famous Deboo integrator by inserting my comment to the EDN's article Consider The "Deboo" Integrator For Unipolar Noninverting Designs. In this story that I wrote in May 2007, I showed how we can convert, in two different ways (Miller's and Deboo's), the passive RC integrator into an active op-amp one. To my much surprise, six months later I received an email from the very inventor of this clever circuit - Gordon Deboo (see the two emails below). In my reply I asked him to tell how he has invented this clever circuit. Circuit-fantasist (talk) 15:35, 24 December 2007 (UTC)

Reader's comment to a related EDN's article
Consider The "Deboo" Integrator For Unipolar Noninverting Designs

The secret of an active integrator is simple: it is just an improved passive integrator. Then, how do we convert the passive integrator into an active one? Here is a possible "scenario".

RC-INTEGRATOR. The bare capacitor is the simplest integrator. It is perfect, if only we drive it by a constant current. Most frequently, we need to drive the capacitor by a voltage. For this purpose, we connect before a resistor R acting as a voltage-to-current converter. Only, a problem appears - the voltage Vc across the capacitor disturbs the input voltage Vin. As a result, the input current decreases: Ic = (Vin - Vc)/R = Vin/R - Vc/R. What do we do to compensate the losses in the capacitor? We may borrow the remedy from the human routine: connect an additional power source, which "helps" the input source. From the Kirchoff's times, we can do this connection in two ways - series or parallel.

INVERTING (MILLER) INTEGRATOR. Here, we connect in series with the capacitor an additional following voltage source (op-amp voltage "inverter"). It produces a compensating "mirror" voltage -Vc equal and inverse to the voltage drop Vc across the capacitor. In this way, the op-amp adds as much voltage to the input voltage source as it loses across the capacitor; thus, it "helps by voltage" the input source to create the right current Ic = (Vin - Vc + Vc)/R = Vin/R. In this arrangement, the capacitor becomes "flying" but the compensating voltage source is grounded. That is why, we use its "mirror" voltage -Vc as an output that gives two advantages: first, the load is connected to the common ground; second, it consumes energy from the "helping" voltage source (the power supply) instead from the input source (so, the load may be low-resistive enough). A disadvantage is the need of a bipolar power supply. All the circuits with parallel negative feedback (i.e., all the inverting op-amp circuits) exploit this great idea.

NON-INVERTING (DEBOO) INTEGRATOR. Here, we connect in parallel to the capacitor an additional "helping" current source. It produces a compensating current Vc/R, which is proportional to the voltage drop Vc across the capacitor. In this way, the source adds the "missing" current Vc/R to the insufficient input current Iin = Vin/R - Vc/R; thus, it "helps by current" the input source to create the right current Ic = Vin/R - Vc/R + Vc/R = Vin/R. Thus we obtain the famous Howland circuit, which acts as a perfect voltage-to-current converter (if we include the input voltage source, we name it Howland current source).

A negative resistor -R can serve as the "proportional helping current source" needed. It does this magic by "pushing" a current I = Vc/R through the capacitor instead by "sucking" the same current (if it was an ordinary "positive" resistor R). In order to create a negative resistor -R, we may convert a positive resistance R into a negative one -R ; thus we will create the famous negative impedance converter (NIC). For this purpose, we connect in series a "positive" resistor R and a proportional voltage source, which produces a compensating voltage 2Vc (i.e, an amplifier with K = 2). Half the voltage (Vc) vanishes in the capacitor; the rest half (Vc) remains applied across the resistor R, which acts as a voltage-to-current converter. As a result, the whole combination produces the "helping" current needed I = (2Vc - Vc)/R = Vc/R. We may implement this idea into a practical NIC circuit by using a resistor R (R3 on Fig. 1) and a classical op-amp non-inverting amplifier with K = 2 (the op-amp and the resistors R1, R2 on Fig. 1).

From this negative resistance viewpoint, we may present the Howland current source as two parallel connected resistors - a "positive" R and a negative -R. The result of this connection is that the negative resistor has "neutralized" (absorbed) the positive one and the effective differential resistance (the internal source's resistance) is infinite. In the circuit of a Deboo integrator, both the capacitor and the compensating source are grounded; the power supply is unipolar. The voltage drop across the capacitor can serve as an "original" output; only, this is not a good idea because the load will affect the output (we have to isolate the load). Then, where to take a perfect output from? As above, we may use the compensating voltage 2Vc (which creates the compensating current) as a buffered output. In order to enlarge the active region, we may scale the negative resistance by the ratio R1/R2 > 1 (see the article). Note that all the resistances and CMRR affect the accuracy (Miller integrator does not have these problems). For the first time, I encountered the negative resistance phenomenon and its implementations - the exotic circuits of negative impedance converter, Howland current source and Deboo integrator - in the French book "L’amplificateur opérationnel et ses applications Masson, Paris, 1971". Then I was a student and was trying without success to grasp the ideas behind these famous circuits. I was asking the simple questions, "What did the op-amp actually do in these circuits? What problem did it solve? Why was it connected there?"; only, they were not giving me reasonable answers. Until many years later, I managed finally to penetrate into the philosophy of negative resistance circuits. Up to now, I continue revealing the mystery of the negative resistance phenomenon to my students following the "scenario" above.

I have decided to expose my viewpoint about these legendary circuits because, looking at this article (and the entire web), I see that the idea behind them is still hidden. Although the article is useful from a practical point of view, it does not tell us anything new about root of the matter. It only shows another possible application of MAX4250 op-amp; obviously, it is its purpose.

The sentence "...ground drives the capacitor through a negative resistor..." does not sound well as a ground is just a piece of wire; it is not a source. The negative resistor is the source here. So, more correctly is just to say "...a negative resistor drives the capacitor..." Of course, I see that the author would like to show a fine symmetrical "T" configuration on Fig. 3. Really, the configuration is symmetrical - two voltage sources, Vin,Vout, drive the common capacitor through corresponding "positive" resistors,R,R3. The entire problem is that a "positive" resistor contains only a resistor while a negative resistor contains a resistor and a voltage source. Actually, a negative resistor is not a resistor, it is a varying source. IMO, it deserves a special "source" name; for example, "1-port voltage-controlled current source" or "2-terminal voltage-to-current active converter" can serve as full descriptive names:) At worst, we may name it "antiresistor" or "reverse resistor"; "negative resistor" is the most inappropriate word:)

Cyril Mechkov

email: cyril@circuit-fantasia.com

A reply from Gordon Deboo
From: Gordon Deboo To: cyril@circuit-fantasia.com Sent: Thursday, October 04, 2007 8:27 PM Subject: Deboo Integrator

I notice from http://electronicdesign.com/analog/consider-deboo-integrator-unipolar-noninverting-designs   that you expressed an interest in one of the circuits I designed in my old NASA days, viz. the Deboo Integrator. There is another more complicated circuit, which serves a different purpose, that includes a Deboo Integrator along with other components. It can be found in one of my now obsolete patents:

1. Patent # 3493901 - A GYRATOR TYPE CIRCUIT 1968

A "gyrator" makes an effective inductor using an capacitor, which is important because capacitors are easy to make with integrated circuits, while inductors are not. Therefore a gyrator enables large value inductors to be made in integrated without using physically large coils of wire, because the coil can be replaced with a capacitor.

I have been retired from NASA for many years now and I am interested and surprised when some of my old circuits are mentioned in current literature.

It's great to see younger people still working on and improving on what we old fogies did. Keep up the good work!

Gordon Deboo California

Another reply from Gordon Deboo
From: Gordon Deboo To: "Cyril"  Sent: Tuesday, October 09, 2007 3:34 PM Subject: Re: Deboo Integrator

...As you can see from the dates of my publications, I did most of my creative work in the 1960s and early 1970s working for NASA (The  USA’s National Aeronautics and Space Administration). After that I got into management, which wasn’t nearly as much fun! So I have been away from circuit design for 30 years and most of what I did has been improved upon by the many younger people who have followed me. Also I haven’t kept up with developments in the field. A brief look at what has happened tells me that the part of the work I did that is  still somewhat applicable is in circuits using operation amplifiers.

I have looked at what you have on the Internet and am impressed with your ability to explain many complicated circuits. I think your students are lucky to have a teacher with your approach.

I am of course now retired and health and other issues do not permit me to give detailed answers to the many interesting questions you raise. However, let me respond to some of your points...

''...It is interesting to know, if you had known about Howland idea when you invented Deboo integrator; if you applied directly the Howland circuit just adding a capacitor to it or you walked the whole way, etc... Circuit-fantasist''

... I was not aware of the Howland current source when I designed my integrator. I conceived my integrator a couple of years before it was published, so I don’t which came first...

...I’m sorry I’m not able to give you a more detailed reply, but already this is fairly lengthy. I can’t imagine you would want to use anything from this note in your various publications but, if you should wish to, feel free to do so...

Best wishes,

Gordon

A copy from the Wikipedia talk page about negative resistance
I have written the text below in Howland circuit section of the Wikipedia talk page about negative resistance. Circuit-fantasist (talk) 15:42, 3 March 2009 (UTC)

The second version of Howland current source (see Fig. 1 of MAXIM's AN1155) consists of an input voltage source VIN, a positive resistor R, a load and a negative impedance converter INIC (R1 = R2 = R3 = R and the op-amp). The input voltage source and the resistor R constitute an imperfect current source with positive internal resistance R that is connected in parallel to the negative resistor INIC and the load. The role of the negative resistance -R is to neutralize the positive internal resistance R of the source. As a result, the load is driven again by a constant current source with infinite internal resistance. Well, let's include some algebra to be more cogent.

The imperfect current source produces a current IL = (VIN - VL)/R = VIN/R - VL/R. As you can see, it differs from the ideal result IL = VIN/R by the term VL/R. It is more than obvious that we may compensate this error by adding the same term; then, IL = VIN/R - VL/R + VL/R = VIN/R. That means to inject an additional "helping" current IH = VL/R that is proportional to the load voltage. What is this circuit that can produce such a current IH? Of course, it is a voltage driven negative resistor with resistance -R; it is a negative impedance converter with current inversion (INIC) that is connected in parallel to the load.

As a conclusion, Howland voltage-to-current converter consists of two connected in parallel resistors with equivalent but opposite (positive and negative) resistances; Howland voltage-to-current converter = INIC + resistor. The "neutralization" between them produces an infinite resistance and, as a result, a constant current. The circuit is stable since the positive load resistance remains after the "neutralization" (another wisdom: the positive resistance has to dominate over the negative resistance to have stability).

Deboo integrator (see again Fig. 1 of MAXIM's AN1155) is just a Howland voltage-to-current converter driving a capacitive load; Deboo integrator = Howland voltage-to-current converter + capacitor. It has two advantages over the dual inverting integrator: it is a non-inverting circuit and the capacitor is grounded. But it has two disadvantages as well: it requires an op-amp with differential input and it has a bigger error depending on the resistor tolerances (here, the op-amp does not monitor the result of the compensation as it monitors the virtual ground at the inverting circuit).

It is interesting that although there was nothing new in a Deboo integrator (actually, it was just a Howland pump driving a capacitor) Deboo was not aware of the Howland current source when he invented his integrator. I found out this fact in a personal correspondence (see above) between me and Deboo a year ago (Deboo was impressed by my comments to an EDN article about his famous circuit and emailed me). By the way, then I also exchanged interesting thoughts with Deboo about the unique feature of negative feedback to reverse the causality in circuits (to swap circuit inputs and outputs). For example, the negative inductance circuit is based on this idea.

Today, I found an extremely interesting source about the invention of the Howland circuit. Although it is nameless, it seems it is written by the very Howland. Circuit-fantasist (talk) 15:42, 3 March 2009 (UTC)