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Thermostat/Fan

Started by Hawk, April 15, 2015, 10:35:00 AM

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Hawk

Hope this is okay to post as this is an audio amplifier rather than a guitar amplifier, but same concepts. I've attached a schematic (see bottom of schematic)

Trying to understand how the thermostat works. It is supposed to turn on the fan when the fan gets too hot. The fan is rated at 12V but the voltage on one side of the fan at R108 is 18v (won't that fry the 12v fan?).
What does this thermostat look like? Is it a thermistor?  I find no voltages on Q101,102. Is it safe to assume that there is no voltage as the thermistor is an open circuit until  the temp gets too high and therefore no voltage on transistors? Hmmmm...
Thanks! :-\



g1

  Some of the 18V will be dropped across R108, some across Q102.
There should be some measureable voltage at the transistors.  Check that R108 is getting it's 18V.
  We do not have the amp so there is no way to say what the component used for RT101 physically looks like.
 

Roly

Quote from: HawkHope this is okay to post as this is an audio amplifier rather than a guitar amplifier, but same concepts.

Speaking for myself, I'm up for anything electrical/electronic.  Staging can include lots of "other" stuff like lighting desks, smoke generators, wireless links, big screen computer-driven displays, &c, covering a wide range of electronics and other technologies.


The component RT101 is a thermistor with a fairly steep temperature/resistance characteristic (but without the specific part number I can't find a datasheet).  Thermistors come in a very wide range of physical shapes so it could look like almost anything.


When the output of op-amp U101B is +ve diode D102 will be back biased, so there will be no current injection into Q101 or Q102 and the fan will be off.

For the output of op-amp U101B to be +ve, high, then the non-inverting input "+" must be higher than the inverting input "-".

As Q103 is a PNP transistor when the op-amp output is high it will also be off.

RT101 is marked as having a nominal resistance of 330 ohms at 25ºC, and this must rise as it gets hotter.  This means it must be a Positive Temperature Coefficient, PTC rather than the more common NTC, type.

When it gets to the same value as R103 the voltage on both inputs will be equal (this is a bridge arrangement), but the thermistor pushing the "-" input just a little bit higher than the "+" input will cause the output to go negative.  (The op-amp has no negative feedback and is operating in switching mode)

When this happens D102 will be forward biased and Q101, 102, and the fan will turn on.

Q103 will also turn on and cause the voltage at the "+" input to be pulled down a bit.  This has the effect of latching the op-amp in the new state, and requires that the thermistor cools down a fair bit before the opposite switch-over occur and the fan turns off again.  This difference between switch-on and switch-off temperatures is called hysteresis.

This comes under the general heading of a bang-bang controller, all or nothing, in contrast to a proportional controller that varies the fan speed over a range according to the temperature, like this one.
If you say theory and practice don't agree you haven't applied enough theory.

Hawk

Thanks Roly I see what you mean by a proportional controller. Your description makes sense. But what I'm trying to wrap my head around now is this: the fan is rated at 12V/ .15 A. The -ve is 18. Do we use ohm's law to calculate  the voltage drop across  R108 to see that we will get -12 ve to the other side of the resistor? And how does this tie into the fan working once D102 is forward biased and the transistors turn on and provide a voltage to the other side of the fan (should that be 12+ve?).

From what I get from your response is that, at resting, with no overheating issues, we will find zero voltage on Q101 and Q102 collectors and bases and zero voltage on the cathode of D102 (measured found no voltages so hopefully that is correct).... Thanks. (By the way this unit is operational)

Enzo

If you had zero voltage on the collectors of Q101,102, then the full 18v would be across the resistor/motor pair.

If the transistors are off, and thus the motor as well, then the transistors would not conduct.  Zero current.  Use Ohm's Law to calculate the voltage drop across the resistor at zero current.  Zero volts.  So when the motor is off, then the full 18v should appear at those collectors.

Stop thinking of the transistors as putting a voltage on the other side of the motor.  Think of them as a switch that completes the circuit through the motor.

If you get zero volts on the transistor collectors, then the fan should be on.  If it is not, then don't look for complex causes, start at the start.  You have 18v on one side of the fan, OK, so is there 18v on the other end?  If not, the fan is open or stuck.  Does the fan spin freely if you spin it with your finger?  Does the fan get HOT sitting there?

You can disconnect the fan wires and power it from a bench supply, does it work?  Hell, a 9v battery would probably spin a good fan.

Are there part numbers silk screened on the board?  Like R102, C216, etc?  Look at anything on the heat sink, do you see RT101?  The sensor needs to be on the heat sink somewhere.  In my head I see one of those little blobs with a mounting tab, a screw hole through the tab.  Look for a small thing with two thin wire leads screwed to the heat sink.  Other ones may look like a diode and could be in a hole in the heatsink or maybe just leaning on it in a blob of white goo.  Yet others look like a small transistor with two legs.

Roly

Quote from: EnzoStop thinking of the transistors as putting a voltage on the other side of the motor.  Think of them as a switch that completes the circuit through the motor.

e.g. mentally replace Q101/2 with a switch or relay; that's all they are doing, on-off.

When off the circuit is open at the bottom, so you will measure the -18V supply in both ends of the resistor, both sides of the fan, and on the collectors of Q101 & 102.

When on 150mA will be flowing, the collectors of Q101 & 102 will be ground (close enough), you will have 12V across the fan, and the other 6 volts across the fan dropping resistor.

{I think that the way this is drawn doesn't help understanding.  The British convention is to stack supply rails according to their voltage, so from the diode the switching stages would be drawn below the ground line to give a physical representation to the voltages, the -18V supply as the most negative coming in from the bottom.  It may not be obvious but the op-amp is running from a dual supply, so when its output goes low it will go below ground to -Vsup and turn on the transistor "switch".}




You may already get this, but a bit more detail.

Open loop (i.e. without NFB) an op-amp has a gain of at least 100,000 times, meaning that it only takes 10 micro-volts of input to produce 1 volt of output, or only about 300uV to swing the output rail-to-rail - mighty sensitive.  So when it's near its switching point it is highly sensitive to electrical noise.

One of the functions of applying hysteresis via Q103 is to spread the "on" and "off" temperatures, say on at 50ºC and off again at 30ºC (or whatever).

But there is also a fair bit of electromagnetic crud floating about, hum from mains, radio stations and whatall that can easily amount to 300uV, so at threshold the op-amp would be trying to respond to all this other crud and waggle about, producing what is called "chatter", or in a proportional controller "hunting".

By shifting the reference voltage via Q103 the circuit will flip over the first time any signal takes it over the threshold, and it will stay there until the voltage (+noise) falls enough to take it over the new lower threshold, where the action will reverse - nice clean switching both ways.

See also: Schmitt trigger.




Quote from: Hawkthe fan is rated at 12V/ .15 A. The -ve is 18. Do we use ohm's law to calculate  the voltage drop across  R108 to see that we will get -12 ve to the other side of the resistor?

Yes.  (ref: Enzo's excellent reply)

Around 12V we can consider the fan to be a resistor of R = E/I ohms, 12/0.15 = 80 ohms.


{Long rambling aside:

but we need to remember that if it's one of the almost universal "brushless" type then it's actually a tiny switch-mode controller in the hub that tries to run the fan at a fixed speed over a wide range of voltages, and may not behave like a resistor at all by the time you get to say 6 or 18 volts.

Despite this and needing a bit of a thermal kick to get it spinning, once started my proportional controller is surprisingly linear across the temperature range.  This is mainly because at low voltages (and reduced drive current) there is a lot of magnetic "slip" where the controller spins the field faster than the fan rotor can follow, and it settles to some balance between reduced drive and blade air drag.

It's a bit "second order" but electronics don't like to be thermally cycled, best when warmed up to a steady state and remain at that temperature (maybe for many years in the case of big transmitters and telephone exchanges).  With that in mind a proportional controller is better than a bang-bang controller in that it tends to keep the internal operating temperature more stable within a smaller range.

This generally comes under the heading of Control Theory.  I'd encourage you to at least skim this article since all the s.s. power amps we deal with are subject to these laws.

This is one of the earliest feedback controllers;

...which came to be needed early in the Industrial Revolution when steam engines were needed to run at constant speed for mine pumps, textile mills, and later electric generators.

Control Theory allows us to have systems that approach the required value in the shortest possible time with minimal overshoot, and to remain in a predictable band in normal operation.

As we have just seen Space-X are still having troubles getting the loop dynamics right for their Falcon-9 soft landing, and gives some insight to how well the Luna landers did.  They apparently have the vertical dynamics right, but not the lateral dynamics, so it was still moving sideways when it touched down, and fell over.

Vid - almost, but not quite;
https://www.youtube.com/watch?feature=player_embedded&v=BhMSzC1crr0
You can see the side-to-side sway is not stabilised before touchdown.

Good tech article;
http://www.wired.com/2015/04/analysis-falcon-9-crash-landing/

The function of the sideways motion;

y(t) = (-35.8)*t + (148)
{...ahem...}
y(t) = (-35.8 )*t + (148)

...may look scary but it's just y = kx+c from High School.

"Distance y as a function of time t is -38.5 metres per second, plus 148 metres" (presumably distance from the bullseye).  I'm not the first to observe that it looks more like -3.85m/s and 14.8m off, and that may be a clue to the problem, a simple decimal point error, and it wouldn't be the first.

I'm not laughing.  Getting this pencil to stand on its end in three axes is a pretty neat trick, particularly if its Centre of Gravity is below half the height.}
If you say theory and practice don't agree you haven't applied enough theory.

Hawk

QuoteAre there part numbers silk screened on the board?  Like R102, C216, etc?  Look at anything on the heat sink, do you see RT101?  The sensor needs to be on the heat sink somewhere.  In my head I see one of those little blobs with a mounting tab, a screw hole through the tab.  Look for a small thing with two thin wire leads screwed to the heat sink.  Other ones may look like a diode and could be in a hole in the heatsink or maybe just leaning on it in a blob of white goo.  Yet others look like a small transistor with two legs.
Enzo, I've attached an image. Between the two output transistors on right is the sensor. I removed the back panel. If you look at it from the circuit board end you see two little wires going to the heat  sink. The sensor is black (you can't see it from the image and I put this back together so I can't re-do the pic).
QuoteStop thinking of the transistors as putting a voltage on the other side of the motor.  Think of them as a switch that completes the circuit through the motor
.
Thank-you, I will.
QuoteIf the transistors are off, and thus the motor as well, then the transistors would not conduct.  Zero current.  Use Ohm's Law to calculate the voltage drop across the resistor at zero current.  Zero volts.  So when the motor is off, then the full 18v should appear at those collectors
. My bad, I didn't have the fan plugged in when measuring. So the full voltage is at the collectors. Thanks Enzo for taking the time to delve into this. As always, I've learned something new!

Hawk

Roly, thanks for your response. I would have gotten back to you sooner but your post pointed me in many directions so I've been reading all about Bang-bang controller, hysteresis, control theory, Schmitt Trigger, and Driver LM3914. Great info.

QuoteAround 12V we can consider the fan to be a resistor of R = E/I ohms, 12/0.15 = 80 ohms.
{Long rambling aside:
but we need to remember that if it's one of the almost universal "brushless" type then it's actually a tiny switch-mode controller in the hub that tries to run the fan at a fixed speed over a wide range of voltages, and may not behave like a resistor at all by the time you get to say 6 or 18 volts.
At first glance that made sense 18-6=12V but then I tried ohm's law to calculate the voltage drop across R108 and did not find 6 volts theoretically but instead found .15AX150 Ohms=22v. Isn't R108 connected in series with the fan, therefore .15A?  Is .15A a max. rating, or will the fan work at a much lower current/voltage. Or, as Enzo pointed out, "a 9V battery would probably spin a good fan." So I applied 9V and the fan spun, so it does work at a lower voltage/current. What am I missing in terms of ohm's law? Please see attachment. Thanks.

Roly

Quote from: Hawkyour post pointed me in many directions so I've been reading all about Bang-bang controller, hysteresis, control theory, Schmitt Trigger, and Driver LM3914.
:cheesy:


That's great.  I try to set search leaders in italics, "if it takes your interest, Google/Wiki this".

Control Theory is all around us every day.  There's a modern glass lift in a nearby shopping centre that not only has all the workings visible, it's a wonder of Control theory they way it arrests with almost no obvious G-change, to exactly level.


The only fan that is speed proportional to voltage is one with a DC motor (e.g. car heater/AC fan, Thermatic cooling fans, most car motors in fact).

These "computer" or Muffin fans are a very different animal.  The power goes into a small circuit board containing lot of silicon.  It operates a bit like a stepper motor in that it has three phases driven by a controller on the board.  This controller either runs because it is getting enough voltage, my guess is that dropout is about 3V, or it's not.

When it's running we can presume that its output pulse rate is constant, trying to run the rotor at full speed.  The reason it runs slow on lower voltages is because less current can flow in the motor coils and is overcome by the air drag on the blades.

I've never tried it but I presume that once the motor is up to speed, say at 12 volts, the speed would remain fairly constant for increased voltages, until the smoke came out.  I think.

Even at 12V it is likely that the 150mA is average, and that there will be impulsing at the commutation frequency, whatever that is, so it's anything but a pure 80 ohm resistor.

I certainly treated it as one in my ThermoFan design, but that mainly means putting up with an initial starting friction that requires the amp to get fairly warm.  Once it's away, after that it's almost purely proportional over say a 10%-90% range.


When in doubt, measure it, so out with your multimeter and measure the fan current on various voltages.   :dbtu:
If you say theory and practice don't agree you haven't applied enough theory.

Enzo

Hawk, I can appreciate wanting to know and understand the theory behind stuff.  However, in my case anyway, I am first and foremost a repair tech.  So if something doesn't look quite right on a schematic, I am curious, but I also have to ask myself if the circuit used to work.   Many people see something they don't understand and instantly decide it must be a "design flaw" as if the engineering team that designed the amp completely overlooked fundamental electronics.  Once in a while we see mistakes on schematics, but the circuits themselves are correct.  And once in a rare while we see an actual design error.  In the vast majority of cases though, the circuit you see is perfectly OK.  SO I have to focus on why it is not working, and worry later about the theory.  Often finding the problem puts the theory into a better focus.

My 9v battery spins the fan, and if you put 20v on the fan, I would bet my lunch money it would spin that way too.  In other words the fan doesn't really care all that much about the voltage, as long as sufficient current is there.  And more importantly, we know it works - the fan is good.

So far we have a good fan.  But the fan has to work in its context, our circuit.  You have the full supply voltage at those collectors..  Q102 is your switch transistor, it turns the fan on by conducting E-C.  In other words it grounds its own collector.  If you take a grounded clip wire and ground that collector, the fan should spin.  Does it?  If not something basic is wrong, but with the voltage there, it ought to spin.

Now to answer at least some of your questions, leave it clipped on and running.  Now measure voltage drop across the 150 ohm resistor.  That will tell you the current through the circuit, and it will also tell you what sort of voltage is actually dropped across the fan.

Remember, Ohm's Law works on resistors.  The fan is not a resistor.

Now, back to your problem.  If Q102 won't turn on the fan, then either it is defective, or it is not being turned on at its base.  Oh, and it is also possible its emitter to ground connection has failed so it cannot complet a circuit.  If Q102 is OK, then look at the base of Q101.  Is there some negative voltage across R107?  Follow the action back through the diode and the IC and see that the control voltage is happening or not.

And of course, heat applied to the sensor ought to start the fan.

J M Fahey

#10
QuoteAround 12V we can consider the fan to be a resistor of R = E/I ohms, 12/0.15 = 80 ohms.
...  At first glance that made sense 18-6=12V but then I tried ohm's law to calculate the voltage drop across R108 and did not find 6 volts theoretically but instead found .15AX150 Ohms=22v. Isn't R108 connected in series with the fan, therefore .15A?  Is .15A a max. rating, or will the fan work at a much lower current/voltage. What am I missing in terms of ohm's law?

1) with a 18V supply and 180 ohms series resistor that fan will never ever get 12V and pass 150mA at the same time.
2) ohms Law applies ALWAYS, but the resistor is linear, meaning it will always show 180 ohms , as in:
*with 18V it will pass: 18000mV/180 ohms = 100mA
* with 9V it will pass: 9000mV/180 ohms = 50mA
and so on, at any voltage applied, while the motor is guaranteed to pass 150mA with 12V applied but with a different voltage anything can happen.

Not "anything" either, I'm quite sure that the fan maker has nice graphs showing speed, current, air flow, temperature, RPM , noise, vibration, bearing life depending on speed and a host other parameters, but those are for the Gods, we mere mortals must be happy with just one pair of values,:current@12V , plus CFM (air flow) , and maybe expected life.

Point is, that amp designers are happy with the fan running at somewhat lower speed, and rather than long calculations (doubt they have all the factory data anyway)  , also considering they have raw 18V available, they simply applied said 18V through different resistors until the fan run at a speed they liked.

The rest of the circuit is just a thermally triggered ON/OFF circuit or maybe it acts like a proportional controller (doubt so) so the fan adjusts its speed to heatsink temperature.

Secret way to know actual current and voltages, TOP SECRET, If I tell you I'll have to Kill you:
connect the fan , black lead to ground, red lead to +18V through the 180V resistors, measure voltage across resistor (the only element you can trust there) and fan and post results.

We'll fire up our 18000 tube ENIAC Computer and calculate:
* current
* voltage across the fan
* power dissipated in the resistor.

..... or you can calculate that yourself :)

Please post results and don't overthink it.

IF you are interested, another day , preferrably a long windy boring Winter afternoon, connect the fan with lots of resistors, from 20 ohms to 2000 ohms, measure and note voltages and currents and when it either overheats or stalls, and YOU draw the curve.

Please post it, it will be useful and I'm not kidding at all.

FWIW in very hot tube amps, I connect a bridge rectifier and a filter cap (usually 1000uFx16V) to the 6.3V winding, get some 8 or 9V DC and use that to feed a 12V PC fan, .

It runs slower than usual, but moves air around and makes very very little noise.

Makes unbearable chassis quite bearable.

No, I never measured current, although it should be somewhat below nominal 150mA .

Fact is, it works like a charm, lasts forever, isn't annoying, what is not to like?

Roly

Quote from: J M FaheyIt runs slower than usual, but moves air around and makes very very little noise.

A friend worked at (then) Telecom Australia Research Labs and told me that the noise from fans in the new AXE's (computer exchanges) going in had become a problem, so they did a little project which came up with the astonishing discovery that "90% of the cooling effect comes in the first 10% of the rev range".  Since these were 240V fans the simple fix was to reconnect them is series pairs running about half speed.

Since then I've quietened down a number of older computer systems using capacitive droppers in series with the (mains) fans and can confirm half speed makes next to no difference to the internal temperatures.

A properly sized heatsink really only needs the assistance of a gentle breeze to push away the hot boundary layer of air clinging right against the fins, and almost full performance is restored.  All a full-speed all the time howling gale does is introduce more fluff faster.


If you say theory and practice don't agree you haven't applied enough theory.

Hawk

JM...wired up fan like you suggested. Voltage across fan--4.11V, voltage across 180 ohm res. 13.86, so 77ma through resistor, Total Voltage approx.18volts.  I tried measuring the fan resistance but I'm getting a really large number--177k--so I must be doing something wrong.


Quoteastonishing discovery that "90% of the cooling effect comes in the first 10% of the rev range
". Cool fact Roly, thanks, I wouldn't have know that.

Turns out the manufacturer has decided that the fan is not needed and the heatsink will be enough....guess that stuff happens all the time...

Roly

I was actually pretty dubious about the 90/10% thing at first, but it seems to have checked out over the years.  The 90/10 may be a bit extreme but there is still very wide scope for quieting down noisy equipment fans.


Quote from: HawkVoltage across fan--4.11V

:lmao:

Quote from: HawkI tried measuring the fan resistance but I'm getting a really large number--177k--so I must be doing something wrong.

With a DMM on Ohms range?

The short answer here is that the computer in the fan thinks it is being powered by a very flat battery (and I'm only surprised that the reading wasn't all over the place).

The key thing here is to understand that the fan contains a tiny computer, a very simple one, but that still needs a certain amount of voltage to bootup and operate correctly, then the motor coil drivers are activated.

It is because this fan is so UN-resistor like that we resort to measuring its current indirectly as the drop across the series resistor.  These fans are emphatically not a resistor and behave nothing like one, they do not obey Ohms Law because they are dynamically doing stuff inside, switching motor coils on and off.

A DMM measures resistance by applying a constant current and reading the resulting voltage directly as resistance.  This means the probe voltage is highly sensitive to loading, only microamps available.
If you say theory and practice don't agree you haven't applied enough theory.

g1

I'm a little confused about a couple things so will ask for clarification:

Quote from: Hawk on April 22, 2015, 08:48:06 AM
Turns out the manufacturer has decided that the fan is not needed and the heatsink will be enough....guess that stuff happens all the time...
I thought this was about a circuit that you had in front of you as you mentioned a heat sensor and showed a picture of an actual fan.  Are you now speaking of a manufacturer of some other unit?

Quote from: Roly on April 22, 2015, 12:32:05 PM
The key thing here is to understand that the fan contains a tiny computer, a very simple one, but that still needs a certain amount of voltage to bootup and operate correctly, then the motor coil drivers are activated.
Roly, I had assumed this to be a very simple fan, but you speak of something more sophisticated.  Did you deduce this based on the photo or the resistance measurement?