berdan said:
Hi.
...
...which confuses me the most is the circuitry designed to keep steady temperature of the "hot wire" resistor.And in that research,they have the circuit that confuses the hell out of me.
Anyway,here it is :
http://img401.imageshack.us/img401/5583/ld52.jpg
I don't understand the choosing of circuitry.Why integral amplifier is used?Why the transistor afterwards?
Thanks in advance.
well i want to apologize for this being so long.
But i have a dear friend who is a mechanical and we used to spend hours on this type stuff - it's where i picked up my meager ME knowledge.
It just takes time to get a mechanism working in one's head.
So i hope you'll dwell on this and work the steps in your head one at a time. it'll take a few minutes. I think it'll help you with your talk..
Let's take the bottom amplifier, heater 1.
Do you understand that: the bridge is balanced at only one temperature of the wire?
And that: when the bridge is balanced, the voltage at left corner is same as voltage at right corner?
If so - good , we can address that 'integral amplifier".
If not say so, so we can get you to that point.
Accept for a moment that the amplifier attempts to hold the bridge balanced by controlling power applied to it. That's all it can do, it can't tweak R3 for there's no motor in that schematic. It can only control power applied.
Presumably R1, R2 and R3 have zero or very low temperature coefficient;
but heater wire Rh1 has a positive one.
So as the whole bridge gets heated by the power applied to it, only Rh1 changes resistance.
So there is some design temperature where Rh1 equals R3 and the bridge is balanced. R3 sets that temperature.
And power applied to bridge will be whatever is necessary to make power into Rh1 equal power lost to air flowing over it
at that design temperature!.
Aha ! More airflow = more power removed = more power must be applied.
So - how does the little amplifier go about that?
Fundamental rule of opamps is this - they will hold their two inputs equal. It is incumbent on designer to provide circuitry that allows the opamp to hold them equal.
The other important trait of opamps is their input is electrically quite 'stiff' - it won't accept any current.
Okay now back to the bridge
This is important - voltage at left corner is always fixed at some fraction of voltage applied by opamp. Let us assume half, for simplicity's sake. If R3 isn't = R1 it'll be some other fraction but let's stay simple with half.
The little opamp's + input is tied directly to left corner of bridge.
So the opamp will do whatever is within its power to hold its - input at same fraction of applied bridge voltage, half.
...Okay now it's thought experiment time...
Assume we are at equilibrium - steady airflow over wire and it's holding steady at design temperature so voltage on both corners is half applied bridge voltage.
Power is whatever is necessary to keep right corner at that voltage, and steady because we're at equilibrium...
There's no current through R4 so no voltage across it, so opamp's -input is also at half applied bridge voltage.. .
Now in your mind increase airflow ever so slightly.
Wire cools a teeny bit so Rh1 decreases, voltage at right corner of bridge drops ever so slightly.
R4 is in between that right corner and the opamp's - input. Opamp will do something to keep his - input at half applied bridge voltage - what can he do?
This is where the capacitor comes in.
There's only one thing opamp can do - move his output higher to draw some current through capacitor C1 and R4 , raising voltage at his - input back up to half applied bridge voltage.
Aha there's the mechanism: cause some current through R4.
How about putting a number on it?
Allow me license to assume a number for purpose of demonstrating a principle...
(Edit - I tried to clean up the arithmetic, edits in this color)
The voltage to be made up by current through R4 is, let us just guess- one millivolt, .001 volt.
Ohm's law says we need 0.001/R4 amps through R4. That's 1/R4 milliamps..
Well, that current flows also through C1.
Current through a capacitor is in proportion to rate of change of voltage across it, I = C X dv/dt.
( Usually we deal with sinewaves which are a mathematical curiosity. Let us stick with DC here.)
So, a current of .001/R4 amps through capacitance C1 requires what rate of change of voltage?
i = C1 X dv/dt
.001/ R4 amps = C1 X dv/dt
dv/dt = .001/ (R4C1) volts per seond
so opamp will begin raising his output by 1/R4C1
milli volts per second to hold his inputs equal...
and that's a constant rate of change in positive direction.., and more positive increases power to bridge...
AHA ! So an increase of airflow causes the little opamp to begin steadily raising his output voltage.
That will heat the wire more, making up for increased heat loss to airflow, thereby re-establishing bridge balance , thereby reducing need for current through R4,
so you should see a nice asymptotic approach to equilibrium at somewhat higher power.
That's what you'd expect for an increase in airflow, isn't it?
You should work that thought experiment in your head until it's intuitive both directions - both airflow increase and airflow decrease.
...end though experiment...
Now observe that if dv/dt = ki, v = k∫it.
That's why they call it an integrating amplifier.
In this circuit it's an integral controller.
Most electronic folks will understand if you just point to it and call it an integrator. Point out it's wired to hold the bridge balanced.
If your audience is mechanicals, tell them it's a 4 bar mechanism held in place by an integrator that tweaks the length of leg Rh1..
There are some genuine educators in the forum, I'm not one so if I'm off topic it wasn't by intent.
Sophie already pointed out the transistor let's them use an inexpensive low power (but precision if you're going to integrate) opamp.
old jim