The said:
But isn't the current determined by the resistor itself? The closest
to this I've thought of was using the 1mOhm current sense resistor in
series with the load resistor and measuring the millivolt drop across
the sense resistor. But again, the accuracy of the method was a big
question mark in my own head.
The current will certainly be influenced by the resistance, unless the
current comes from a current regulated supply (which is a fine thing
to use). The point is that you do not assume some current, you
measure it by putting an amp meter in series with the resistor. At
the same time to put a separate volt meter across the resistor, and
you calculate its resistance (under those conditions, it you don't
want to make assumptions about temperature coefficient) by dividing
the voltage across the resistor by the current through it. You may
need to limit the current to a safe value for the source by also
having some other resistance in the circuit, like a large incandescent
lamp or a toaster.
Hmm, what would be the thermal characteristics of such a setup? As in,
how high can the temperature be allowed to go up to and such?
That depends on the melting temperature of the wire, and how much
thermal change in the resistance you can tolerate. But you can
measure the thermal effect with the same resistance measurement method
I spoke of above.
The big advantage of this method is that there is very little air drag
compared ot an aluminum heat sink, so you can move a lot of air with a
small, axial fan. The wire is just a spider web across the air flow.
They carry a thousand watts out of an ordinary hair dryer, this way.
And you know how small those are.
heehee, sorta, I did post a schematic a few days back. Only one person
replied and said it wouldn't do what I expect but when I asked why,
there was no response.
It didn't capture my attention, and I thought you had a conversation
going on with someone, so didn't butt in.
The circuit I did had a current sense resistor in line with the load
bank, e.g.
+ve ---+-Rcs-+----+--/ -R1---+
| | | |
| | +--/ -R2---+--- Gnd
| | | |
| | +--/ -R3---+
OA+ OA-
The idea was to feed OA+ into the input of the OpAmp and the OA- into
the -Vcc of the opAmp, thinking that that should make the OpAmp read
the drop over the current sense Rcs. Then I would apply a reference mV
to the +input of the OpAmp. Except of course, it didn't work.
Linear opamp circuit make the assumption that the two inputs stay at
essentially the same voltage for the whole range of output voltage.
The input signal and the feedback have to make that happen when you
are getting the desired output.
A subtractor is one configuration that amplifies a voltage difference
and outputs that amplified difference on an arbitrary reference
voltage (like zero volts). There are also current mirror circuits
that rereference a high side current (or small voltage difference) to
another voltage rail.
Section 3 of this circuit collection shows several ways to do this
from a full, 3 opamp instrumentation amplifier (top of page 14), a two
opamp version that can have inputs outside the amplifier input voltage
range (top of page 15), and a two opamp version that has to have the
input signals within the opamp input voltage range. But since both
your voltages come from a low impedance source you can just use the 1
opamp subtractor that is all the resistors and the right amplifier out
of the I.A. at the top of page 14.
http://www.national.com/an/AN/AN-31.pdf
Here is an example of moving the current shunt signal to a different
reference with a current mirror:
http://www.zetex.com/3.0/pdf/ZDS1009.pdf
There was also the issue of somehow tying the reference voltage to the
switches for R1 to R3. I thought a voltage adder might be the idea but
I'm stumped on exactly how to do it. EWB doesn't work the way I
expect, I can't use it as a test and try thing since I have no idea
why it's producing the voltages it is (generally full +Vcc or -Vcc)
Are you trying to select the load resistor based on a measurement of
the load current, or step the load resistors to hold the current to
some setpoint (as close as the switching will allow)?
Would greatly appreciate it if you could give me some hints about it
too! Thanks!!!
If the load resistors are all equal there is a different method than
if they have a binary multiple relationship. In the first case, any
resistor is equivalent to any other so you need a circuit that just
switches them on or off in a straight sequence. In the second case,
you need ot switch them on and off with the equivalent of a binary up
down counter. But in both cases, this selection circuit has to be
connected to a decision making function that does this selection in a
stable and orderly fashion (at a speed that does not exceed the the
decision making process's ability to evaluate its previous decision
before making another one). This can all be done with opamps and
comparators and discrete logic (fast but large), or with a program
running in a microprocessor (slower but smaller and more feature
rich).
