The design of a standard inverting integrator is simple when bipolar supplies are available, but it's cumbersome with a unipolar supply. To allow adequate headroom for the output, the circuit must be biased away from ground, often at half the supply voltage. Because neither the input nor output is referenced to ground, the design becomes awkward and often impractical.
One solution is a Howland current source with a capacitive load, also known as a "Deboo" integrator (Fig. 1). Powered by bipolar or unipolar supplies, this noninverting circuit has both a ground-referenced input and output. An intuitive understanding can be gained by considering the circuit in three parts: the input resistor, R; the integrating capacitor, C; and the right side of the circuit, which is equivalent to a negative resistor connected to ground.
The right side of the circuit (Fig. 2a) is easily simplified. Its defining equations are:
and
resulting in
Note that the input current equals the input voltage divided by a negative resistor. The negative resistor merely changes the direction of the input current. The corresponding equivalent circuit is inferred by inspection (Fig. 2b).
After simplification, the circuit (Fig. 3) reveals a "T" configuration, in which VIN drives the capacitor through a positive resistor, R, and ground drives the capacitor through a negative resistor, R1R3/R2. The total current into the capacitor is:
If
then
This important result shows that if R = (R1R3)/R2, then current into the capacitor depends only on the input voltage VIN and R, not on the capacitor voltage. In other words, the capacitor is driven by a current source whose value is VIN/R. If VIN is a function of time, the voltage across the capacitor is:
VC is then amplified by (1 + R2/R1), so VOUT is
The circuit shown in Figure 4 is a practical Deboo integrator with two inputs and a reset. The input R is simply the parallel combination of the two 20-kΩ resistors. The circuit can sum any number of inputs as long as the parallel combination of the input resistors equals the required R, which in turn equals R1R3/R2.
A single-pole/double-throw analog switch was chosen to perform the reset function. This configuration (versus a single switch across the capacitor) eliminates any offset error due to voltage division between the input R and on-resistance in the switch. To prevent the switch from going into its short-circuit-protection mode during reset (which prolongs the reset), the voltage across the MAX4624 is limited to less than 0.6 V by adding 10 Ω in series. (For capacitors of 0.1 µF or less, this voltage-limiting resistor is unnecessary).
Many parameters should be considered when designing an integrator. Fast integrators require wideband op amps with low bias-current. Very slow integrators require closely matched resistors, op amps with extremely low bias-current, capacitors with low leakage-current, and a physically clean board layout. Polypropylene, polystyrene, or teflon are the dielectrics of choice for capacitors with low leakage and low dielectric absorption.
To implement a single-input integrator, the four resistors can have the same value. Quad equal-value resistor packs with ratio tolerances down to ±0.01% are available from Caddock Electronics. Less demanding requirements can allow the use of four discrete ±1% resistors. The reset switch should be chosen carefully, since its off-leakage is integrated by the capacitor. The switch on-resistance must be low enough to allow the capacitor to be reset as quickly as necessary. Finally, the charge injected by the switch during turn-off must be sufficiently low to avoid an objectionably large offset on the capacitor.
The Deboo integrator is an easily implemented and very useful circuit, particularly for single-supply applications.
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:)
If you want to know more about the philosophy of these legendary circuits, visit my "Circuit stories on the whiteboard" located on www.circuit-fantasia.com and the Wikipedia page about negative resistance... or just email me.
--- the end ---
email: cyril@circuit-fantasia.com
Cyril Mechkov -May 01, 2007
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).
--- to be continued ---
Cyril Mechkov -May 01, 2007 (Article Rating: )
Great article, just the kind of hint I needed for a university project. Cheers.
Steve -September 14, 2005 (Article Rating: )
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