0
\$\begingroup\$

here's the circuit - voltage divider with 2 matched thermistors ("matched" means close enough that the measured temperatures when thermistors are right next to each other comes close enough for my needs)

(edited diagram to show series 1K resistor from V2 to DIO1, and V3 measurement)

3.3V--------+------------+
            |            |
            \            \
            T1           T2
            \            \              __
            T1           T2            /V3\
            \            \       +-----\__/------+
            T1           T2      |               |
            |            |       |               |
Vvd --->    o v1      v2 o-------+---/\/\/\/\/\--+--{XBee DIO1 pin}
            |            |              1k ohm
            \            \
            /            /
            \            \
            / 10K        / 10K
            \            \
            /            /
            |            |
GND---------o------------o

In the above circuit exactly as shown, v1 and v2 both measure at 1.31V at common reference T. So far, so good

However, when I connect the o at v1 to XBee pin A and the o at v1 to XBee pin B (XBee shares common 3.3V and GND, power-supply has enough power to power it all) then v1 and v2 are 1.50V and 1.57V respectively - different enough that measurements of T1 and T2 are useless to me.

(the thermistors are right next to each other so that I can validate the temperatures are the still measuring the same when on XBee)

It seems clear to me that connecting the transmitter to v1 and v2 disturbs the system (1.31V becomes 1.50 V and 1.57 V) What's less clear is why it doesn't affect v1 and v2 the same way.

Biggest question I have: is this one of those cases where an op amp should be used to preserve v1 and v2 at their "real" values and prevent the transmitter/instrument from messing with the values?

Edits: V4 measures 5.5mV, so current is about 5.5 microamps.

V across T2 is about 1.8 V and T2 has R about 16K ohm at these conditions so current through T2 is about 0.1 milliamps.

current to DIO1 is about 1/20th of what goes through T2 - doesn't explain the 0.2V difference as far as I can see.what am I missing?

\$\endgroup\$
6
  • \$\begingroup\$ XBee pin A and XBee pin B could be sinking/sourcing current. Or, (I can't tell from your text whether you measured v1 and v2 with a meter or used the ADC) it could be that the ADC is not very accurate. You could try connecting XBee pin A and XBee pin B together, then connecting a controllable Voltage source of some sort to them both, and comparing readings. You can also put a resistor in series with the inputs to check whether they sink/source current (for example, 1k). \$\endgroup\$
    – mkeith
    Nov 29 '14 at 2:05
  • 1
    \$\begingroup\$ Thanks for the ideas. To answer your question, I measured v1 and v2 with an oscilloscope as well as a multimeter \$\endgroup\$
    – user13655
    Nov 29 '14 at 3:15
  • \$\begingroup\$ In that case, it sure seems like the XBee input current is not negligible. I would focus on trying to measure it. Once that is confirmed to be the problem, your op-amp buffer idea will likely fix it. If you can find a specification for the XBee input impedance, that could help us confirm everything, although measuring with a series resistor is very reliable. \$\endgroup\$
    – mkeith
    Nov 29 '14 at 4:17
  • \$\begingroup\$ mkeith - I edited the diagram to show the current measurement I think you mean I should measure. If that is what you mean, I will need to break the present connection and put an ammeter between as shown. \$\endgroup\$
    – user13655
    Nov 29 '14 at 6:23
  • \$\begingroup\$ Location is correct. An ammeter is OK, but I was proposing to just use a 1k resistor. Then you can probe across the resistor with your multi-meter in voltage mode. If the voltage is too small for an accurate reading with your multimeter, you might have to try a bigger resistor. But if you have an ammeter that can display micro-amps, that is fine, too. \$\endgroup\$
    – mkeith
    Nov 29 '14 at 6:51
0
\$\begingroup\$

You say the current flowing into the input is 5.5 uA. The voltage is around 1.5, I think, based on a 1.8V drop across T2 (3.3-1.8=1.5). So that means the input resistance is 1.5 / 5.5 uA = 272 k. It is possible that the input does not behave like a resistance, but more like a current sink. In any event, in the specific conditions above, it makes no difference.

So, in effect, the 10k lower leg is transformed from 10k to 10k in parallel with 272k, which is 9.65k. This changes the input voltage by around 30 mV. It doesn't seem to me that this could fully explain the difference you see between your two inputs (1.57 vs 1.5), but to be sure, you should put a 1k on BOTH inputs at the same time. Maybe the two inputs have different resistances or leakage currents.

I think you should entertain the possibility that the circuit basically works, with a small error due to current into the XBee, and the differences you are seeing are actually caused by differences in thermistor temperature. Could you please work out the implied temperature at 1.31, 1.5 and 1.57V using thermistor properties?

Also, I would like to caution you against comparing the 1.31 V reading obtained before connecting the XBee with the 1.5 and 1.57 V readings obtained after. Readings obtained at different times really can't be compared unless you have very good measuring equipment to record actual thermistor temperature and verify that it is the same at both times. Or, if you have the voltmeter connected and you can sit there and connect/disconnect the XBee and watch the voltage jump up and down by 200 mV. That would also be very convincing that the 200 mV is due to XBee. But if you measure 1.31, then de-power the circuit, mess around with it, and measure again and find it is 1.5, well, I would not draw any conclusions from that. Maybe the NTC just warmed up a little due to handling (which is not uncommon). The calculations I asked for will tell us whether that explanation is plausible.

\$\endgroup\$
2
  • \$\begingroup\$ mkeith - thanks for the help. I'd already disconnected 1 thermistor completely for the same reason you suggest doing the R on both. Sadly I mixed reporting of V - some are reported as voltages, and others as V-drops. The voltage at the divider point varies between 1.51 (no XBee) and 1.57 (Xbee). Voltages are very repeatable (I take 3-5 meas. and average them (heavy scientific training, just not in EE, obviously). Repeated V measurements are within 2-3 mV of each other. Thermistors are untouched, less than 1mm apart, and in a fan air-stream; guaranteed the temps are very close to each other) \$\endgroup\$
    – user13655
    Nov 30 '14 at 6:39
  • \$\begingroup\$ Sounds like you know what you are doing! If you end up buffering the divider with an op-amp, pay attention to the input offset voltage specification (difference between + and - inputs), and the input bias current specification (how much current flows into/out of inputs) for the op-amp. Your application is relatively intolerant of small voltage and current errors that might go unnoticed in some applications. Not every op-amp will be satisfactory, but your specifications are not outlandish, and you will find something that works. \$\endgroup\$
    – mkeith
    Nov 30 '14 at 6:58
0
\$\begingroup\$

To a first approximation, the reason the transmitter doesn't affect V1 and V2 in the same way is because the transmitter's input resistances aren't identical.

In 1, below, the left-hand side of the bridge is shown, and if R2 is precisely 10000 ohms, E1 is precisely 3.3 volts, and E2 is precisely 1.31 volts, then RT1's (the thermistor's) resistance must be ~ 15191 ohms.

In 2, then, if E2 rises to 1.50 ohms and everything else remains the same as in 1, the parallel combination of the thermistor and the transmitter's input resistance must be 12000 ohms.

In 3, if the thermistor's resistance is 15191 ohms and the total resistance of the parallel pair is 12000 ohms, then the effective resistance the transmitter is shunting across the thermistor is 57127 ohms.

The math is the same and the results are similar for the the other half of the bridge, with the transmitter's input resistance on that side being 40122 ohms connected to 3.3 volts.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy