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From: Winfield Hill
Subject: Re: Current Source Multiplexer
Date: 23 Nov 2002 13:09:00 -0800
Organization: Rowland Institute
X-Newsreader: Direct Read News 4.11
Winfield Hill wrote...
>>>> ... current sources ranging between 1 nA and 10 mA...
> You're still left with the problem of a huge 1nA to 10mA dynamic
> range. Our 1M resistor creating 1mV at 1nA reaches 10V full scale
> at 10uA, so clearly you need a range-changing circuit or another
> approach... log amplifier... current-to-frequency converter...
Imagine a requirement for continuously monitoring a sensor over a
range of 1nA to 10mA. Logarithmic circuits have been eliminated
from consideration by personal preference or by management. :)
Our 1E7 dynamic range would require a 23 or 24-bit A/D converter.
However, we cannot get such a converter operating up to say 1 MHz
conversion rates. But what if we could simultaneous run two 16-bit
input A/D stages from the same signal? A sensitive one working
from 0.8nA to 50uA, and a fast one working from 0.15uA to 10mA.
Note the liberal 3.3x scale overlap.
How is it possible, little grasshopper, to do such a thing?
Consider a pair of matched transistors operating as a wide-range
. ,-- E C --- I/2 out
. | B
. | |
. I in ===+ +=== gnd
. | |
. | B
. '-- E C --- I/2 out
The change in offset voltage for matched transistors is typically
under 25uV over a wide range of currents (1uA - 1mA for LM394 or
MAT02). Let's say it's under 75uV over our range of interest.
This corresponds to a change in current ratio of e^dV/Vt = 0.3%
(see AoE page 91), which is certainly acceptable and useful as an
absolute accuracy over 8 orders of magnitude. Our first task is
simply to attach a transresistance amp to each of the two currents,
one with a 200k resistor for 5V out fs with 50uA in (remember the
1/2 factor), and the other with 1.00k for 10mA fs.
Our second task is to insure the sensitive-channel's transistor
doesn't saturate when its amplifier saturates off scale (if the
transistor were to saturate its Vbe-Ie voltage would change,
spoiling the 1/2 ratio). This can be accomplished either with a
bypassing "zener" style clamp on the amplifier or an active clamp
circuit on the transistor itself, e.g., a biased Schottky diode,
etc. OK, that's not too difficult a problem.
Can this circuit maintain a wide bandwidth at low currents? We
saw in my last posting that sub-0.2nA noise-level, 45pF Cin, 90kHz
bandwidths are easy to achieve with a 1M transresistance. What's
the bandwidth of our transistor divider at 1nA? We know r_e =
kT/qIc = 25M-ohms at 1nA, so our emitter resistance will be 50M
(remember the 1/2 factor). This means for 45pF Cin the bandwidth
of our input node will be only 70Hz at 1nA, the higher-frequency
currents being shunted by the capacitance. Bummer!
Aha! The rather high 45pF input capacitance used above accounted
for 30pF of multiplexer capacitance. We'll remove that, and we'll
assume a small enough input transistor so its Ceb capacitance is
under 10pF. That gets us up to 210Hz. We're making some progress.
Note that for sensor currents above 1nA * 1E5/210 = 476nA our
desired bandwidth of say 100kHz is achieved. Hmm, how to get that
extra 500x for sensor currents of 1nA?
Our quest is not lost, although it becomes a bit more difficult.
Let's create a standing 500nA bias current just for the purpose
of lowering r_e for our pair of NPN input transistors. Aha!
. ,-- E C --+------------- I/2 out
. <-- | B '--/\/\--,
. I in ===+ | 40M |
. | +=== gnd +-- +9 to 9.5 adjustable
. -10 --/\/\--+ | 40M |
. 20M | B ,--/\/\--'
. '-- E C --+------------- I/2 out
OK, now we have our bandwidth, but have we given up our carefully-
obtained low-noise capability by introducing a noisy bias current?
Surely the shot noise from a huge 500nA standing current will
overwhelm the 1nA shot noise from our quiet 1nA sensor?
What to do little grasshopper?
Nothing! Resistors biased by quiet voltage sources don't make
shot noise! (See AoE page 432.) We do need to avoid resistors
with "excess noise" (same page), but we don't really even need
to filter the +/-10V power supplies! Showing that is left as an
exercise for the reader. However we do need to carefully adjust
the positive bias voltage to cancel the negative current, less the
base current. This ends today's lesson.
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