And the current flow in the 2K ?:-)
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |
Is the difference between the two diode voltages over 2k, very small comapred to the current comming down the 10k from +5. Probably not enough the add significant non-linearity overall.
Luhan
True, if only a few degrees, FALSE if a significant difference:
Bias current is ~430uA
For a 1°C difference the current is ~1uA
For a 50°C difference the current is ~50uA
Was it Jim Williams (or Bob Pease) who did the 10*I/1*I switching trick? THAT'S the way to do temperature measurements... accurately.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
Old Latin teachers never die...they just decline
It will go to the rail long before that...
At -2mv / deg-C and a gain of 235, 5 volts/235=3D 21mv max at the diode.
21mv/2mv =3D 11.5 deg-C. So the maximum error current is really about 11 uA.out of 4.5v/10k =3D 450 uA. makes the error about 2.5 % ??
May be still useful in some designs, not in others.
Luhan
Are you sure. If you mean a 2k to ground from the (+) input, I believe that would unballance the this circuit. That only works when there is no common mode offset at the inputs.
Luhan
You're assuming the circuit is fully balanced with equal currents going through the diodes when they're at the same temperature, and the opamp output is at 0 volts? Look at the circuit again, you'll see the missing resistor.
--
Thanks,
- Win
The opamp output is at the diode voltage. There is no current thru the
2k resistor because the minus input is at the same voltage.Where would put that extra resistor, and why?
Luhan
You said to me...
"The reason this makes me a real engineer is simple - they pay me to be one! You would not be the first 'old guard' engineer that was somewhat distressed at this situation."
Does that also apply to Win ?:-)
BTW: I've knocked off more 'old guard' engineers than you even know exist... starting with Huelsman before an auditorium of listeners ;-)
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
Old Latin teachers never die...they just decline
The payed part or the distressed part?
I think I get the picture here. Some engineers are bit over inflated ego wise; then there are a few of us running around with hat pins. You may scare them plenty, they seldom see me comming.
Luhan
I was just curious why this sudden display of "attitude"?
I'm certainly not concerned about you in the slightest.
As they say in the trade, you're a "light weight" ;-)
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
Old Latin teachers never die...they just decline
That's awkward. Didn't you say, "It uses a second diode to ballance out the one sensing temperature." That would mean your output should conveniently show a voltage proportional to temperature difference, show zero volts when both are at the same temperature, and go negative or positive when following a negative or positive difference. Right?
Your circuit doesn't balance, so you're off to a bad start.
See above. Study your circuit. Report back.
--
Thanks,
- Win
This one should work ok I think:
Chris
I guess my particular experience is quite the opposite. I get called in when all of the 'regular' engineers are stuck, or when the field is so new that there is no one with any experience in it at all. Its not so much technical expertise thats called for sometimes, its inventive solutions. I very much respect whatever knowledge my cohort engineers have, but solutions often have to take into account schedules, personalities, cash flow, available resources, whatever. They require someone to step back and rethink the problem from a larger perspective.
All of which has taken my career in some very interesiting directions. Like the guy who was designing a laser interferrometer (visar) in his garage. He needed somebody to make a slick graphical interface on a PC. This is before Windows came allong. Or the guy who wanted to actually build an 'air guitar' - one that could be played without knowing anything about music.
Things like this do give me an attitude. Hey, thats life ain't it?
Luhan
You are incorrect. When ballanced, the output is not zero, it has to be the same as the voltage at the + input, the diode drop voltage. You don't believe me? My oscilloscope says you're wrong.
I have this thing running on my bench.
Luhan
Very interesting, but very complicated. I was just wondering what could be done with a few standard, cheap parts. May have some use in the toy industry where cost cutting is insanely critical. I have a client who works in this arena.
Luhan
A precision thermistor usually yields the lowest system cost in such applications. There's no need for a voltage reference or even a regulated supply voltage, only a single precision resistor, everything else can be crap Shenzhen street-sweepings.
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
You are right, of course. This is basically a technical exercise that may not have much practical application. However, persuing ideas like this sometimes does lead to something new and useful.
Regards, Luhan
I'm sure your do. Luhan, stop being daft. To repeat what I said that you snipped, you wrote, "It uses a second diode to ballance out the one sensing temperature." That would mean your output should conveniently show a voltage proportional to temperature difference, show zero volts when both are at the same temperature, and go negative or positive when following a negative or positive difference. Ideally that's what your described design should do.
OK, clearly I was interpreting what you said. Balance has a meaning to most of us, and it _doesn't_ mean doing nothing.
It would make sense for your circuit to be balanced, because then you could see a difference. But because of your missing part, it simply follows one diode, then adds in the effect of another, so there's no way when you see the output voltage to determine whether it's the first diode or the amplified difference between the two that's responsible for the signal. See why that's a problem? That's close to being useless.
. ,---/\\/---, 470k . +5 --+--/\\/--, | ,--/\\/--, roughly, . | | 2k | __ | Vo = Vd1 + G(Vd1-Vd2) . ,--|
Please sir! [hops up & down excitedly] :-)
My arm-waving notion is that putting another 2k from the lower diode (on the non-inverting input) to ground will make the circuit balance i.e. the output will be 0V for 0 temperature difference (and perfectly matched parts).
Seems reasonable enough to me if the modification I suggest does what I think it will (I don't have components to hand to try it) but I'm interested to hear "what's still wrong" and "further improvements".
Oh no, no!! You stumbled (sorry, I couldn't help myself). Close but no cigar, it's back to the drawing board for you!
--
Thanks,
- Win
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