Electronic Design

# What's All This Tempco Stuff, Anyhow?

In the recent picoammeter circuit I wrote about in the column entitled "What's All This Thermistor Stuff, Anyhow?"(Electronic Design, Jan. 6 1997, p. 171-172), I showed a circuit for compensating for the tempco of some high-megohm resistors with a fat tempco such as −250 ppm/°C, or larger.

But what if the UNCERTAINTY of that tempco was sloppy, too? What if its tempco was −250 ±150? What if other RN55D resistors added enough slop to the tempco, that it was not performing nearly as well as desired? Well, here's a semi-classical solution-an improvement that needs 2 new trim pots, and a new trim procedure.

The basic circuit (Fig. 1) is very similar to the original, except that we have added new components P1, P2, and two new 10k resistors (*). Formerly, Rf was just tied to the juncture of the two resistors marked ** that were connected to the thermistor network.

Why did we add these extra resistors? It's very simple. If we rotate P1 so that the wiper is connected to the left end, the tempco will be overcompensated. If we slide it to the RH end, it will be under-compensated. SOMEWHERE IN THE MIDDLE, IT WILL BE JUST RIGHT. So here is the Calibration Procedure:

Start at room temperature.

Put in I = 0 and trim P3 for zero output (well below 10 mV).

Put in I = -10 pA, and turn the pot P1 to the LEFT end. Trim P4 to get 1.00 V.

Keep I = -10 pA and turn P1 to the RIGHT end. Trim P2 to get 1.00 V also.

Now, repeat the calibration sequence, as there may be minor interactions. When you have it right, turning P1 from end to end will have no effect. (This is true because the impedance of the pot P1 is more than 1000 X lower than Rf.)

Now, put the whole thing in an oven, and allow it to warm up to some suitable temperature in its working range, such as +35 or 40°C.

Put in +10, 0.00, and -10 pA. Before, if you moved the P1 pot from one end to the other, the output stayed at 1.0 V. Now, as you turn the pot, the output voltage goes both above and below 1.0 V. So ,just reach into the warm oven and trim P1 to get 1.0 V to match the room-temp value. (Better yet, if Iin alternates between +10 and -10 pA, trim P1 to get the output voltage to be 2.0 V p-p. This helps to simplify the rejecting of the op amp's Vos.)

And how do we generate an accurate, stable +/- 10 pA current?? You should refer to Fig. 2. A3 is an integrator and A4 is a detector that turns the integrator around when it hits +/- 4 V. This is a basic triangle-wave generator that I have been building over the last 30 years. When A4's output is LOW (-1.3 V) A3's output is ramping up at +0.5 V/s. When the integrator gets to +4 V, the 1 µA through RJ just balances the 1 µA coming through the diode bridge, and A4 will switch its output to + 1.3 V. The triangle wave changes direction about every 16 seconds. This gives everything time to settle.

The ±1 µA coming through the diode bridge into A3 gets integrated nice and slow; it is effectively attenuated by a factor of 100,000-by the ratio of the 2-µF to the 20-pF capacitor-down to ±10 pA. Therefore, as that voltage ramps up and down at ±0.5 V/s, the current through the 20 pF cap is ±10 pA.

It's not hard to measure and confirm the peak + and - voltages at Vj. It's easy to measure the time it takes for each half cycle. As a result, you can compute the ramp rates easily. The accuracy of this calibration depends on the fact that you can measure a 20-pF capacitance at 1 MHz and its capacitance at 30 milliHertz will be the same. Many capacitors will not do that, but silver-mica is pretty good, and Teflon, NP0 or C0G ceramics are very good: The capacitance does not change much over this wide range of frequencies. Of course, to measure a 20-pF capacitor with 1% accuracy, you will need to mount it securely and stably in a hole in a metal guard wall, and use a meter such as the Boonton Bridge (Electronic Design, "What's All This Picofarad Stuff, Anyhow?" Jan 22, 1996, p. 99.) I inserted a 22- resistor in the path, to make it easy to disconnect the capacitor from the amplifier for measuring....

Note, almost every 20-pF ceramic capacitor is NP0. You can check that by heating it 100 degrees and watching the capacitance change less than 1/2%. Good stuff!

A3 and A4 can be any op amp with less than 1 nA of input current. Just be sure to use low-leakage diodes, such as the emitter of a transistor with C and B tied together. Avoid those leaky 1N914s or 1N4148s, as in this kind of circuit, their 20-nA leakage could cause over a percent of error compared to 1 µA.

The actual current fed out of the calibrator to the picoammeter will be:

I=Cx(V/t).

The temperature coefficient of the polypropylene integrating capacitor is down near -200 ppm/°C. The tempco of the NP0 ceramic is ±30 ppm/°C, max. The tempcos of the 4-M and 5.4-M resistances will all track if they are made out of a lot of 1-M 1% resistors. The tempco of the 1 mA currents will be about + 300 ppm/°C, due to the tempco of the diodes, which adds to the -200 ppm of the Poly to cause the dV/dt to increase at about +500 ppm/°C. Thus, the calibrator should be used in an air-conditioned room, unless you wanted to use a trick circuit for the 1-µA current sources.

I showed this improved temp-compensated picoammeter to the guy in South Africa who was originally trying to get good tempco despite imperfect parts. He thought about it. He tried it. He decided it was worth doing. The added complexity of adding 2 pots and a trim cycle in a warm oven was worth it, compared to the uncertainty of the tempco of your whole instrument if you didn't.

I also showed this to Ed Walker at DAC/I: http://www.cyberspy/daci. He had suspected my original circuit could lose accuracy if the conspiracy of all the resistors? tempcos ganged up to hurt you. I pointed out to him that this trim scheme would compensate out the tempco of every resistor in the gain path. It didn't really matter if some of the gain resistors were + or -50 ppm/°C. Ed did some analysis and indicated that if all the resistors were at ±100 ppm/°C, in the worst possible directions, you might run out of trim range. I agreed that it was possible-but unlikely. So, while my original circuit did make some good nominal improvements, and could also cancel out any consistent tempco trends of the main feedback R, this improved circuit could fix almost any tempco, even if the tempcos moved around considerably. Even if the tempco of some resistors has a little curvature, this does a pretty good job fixing it. Ya gotta admit, a "tempco adjust knob" that has no effect on room-temp gain or offset is an attractive concept!

All for now. / Comments invited!
RAP / Robert A. Pease / Engineer

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