Re: Can someone explain how this circuit operates


"Greg Hines" wrote in
news:OtjHe.492$ snipped-for-privacy@nnrp1.ozemail.com.au:
Hi all,
>
> Don't know if I have the correct Newsgroup for this type of question,
> and if not please let me know the best place to post it.
>
> I'm after an explanation of a circuit which in attached, or can be
> found at:-
>
>
formatting link

>
> It's a circuit that controls the speed of a air pump to maintain a
> constant airflow. It uses PWM for this purpose. The PWM test point
> at the far RHS of the circuit goes to the gate of a MOSFET that
> switches the pump.
>
> DELTA P is the output of the flow transducer that's in the pump air
> flow after it has been amplified. R28 sets the airflow for the pump
>
> My questions concern U5C and U5D and the associated components.
> What's the purpose of these 2 OpAmps? Is it adding some sort of
> hystersis and if so the 10K feedback on U5C provides an awful lot of
> hystersis? Why would you use 2 200K resistors in series as R30 and
> R31?
>
> Any insight to the above questions would be appreciated.
>
> Greg
> -
> snipped-for-privacy@bigfoot.com.NOSPAM
> Remove NOSPAM when replying
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> begin 666 circuit.jpg
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> Attachment decoded: circuit.jpg
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I'll have a quick go at explaining. U3B and U5A form an oscillator: note
the feedback from the output of U5A to U3B's non-inverting input ie +ve
feedback. This forms a sawtooth oscillator. U3A is simply a buffer of half
the supply voltage to set the DC operating point. Probably a single supply
or battery supply. U5B is a comparator without hysteresis. Now, the Delta P
signal is first amplified by U5C, a DC stage. And no, the 10k feedback
resistor does NOT add hysteresis but simply puts a single pole in the
response to roll off high frequency response, probably pulsations in the
flow or pressure sensor.
The next stage is U5D forming an integrator. Of Course this has open loop
gain at DC which means that the PWM is capable of going to 100% and 0%. I
am assuming here though that the sawtooth amplitude is less than the supply
rails, you can verify with calcs or simulation. When the system is in
balance, the set point voltage of R28 will equal the value of Delta P
average voltage and there will be a constant duty cycle value. Q2 is
obviously an enable function.
The two 200k resistors in series in the integrator would have been chosen
by the designer presumably because that was an available value, and/or to
minimise the number of different values for a SMT design to minimise the
number of different reels used.
Reply to
Geoff C
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"Geoff C"
** I think that Geoff has supplied very good explanation - especially as the nature of the feedback sensor signal at "Delta P" is unknown. The circuit is part of a feedback control loop so it is designed to interface with the specific electro / mechanical items involved.
The "Delta P" signal may well be a rectangular pulse train derived from an opto-interrupter reacting to tiny fan blades breaking a beam - or it may be something entirely different.
The OP should have supplied that info.
......... Phil
Reply to
Phil Allison
half
P
supply
Thanks Geoff for the great explanation it was very helpful and thanks for taking the time to answer.
Greg
Reply to
Greg Hines
half
P
supply
Geoff,
One more question if I can. Is not U5C a comparator whose output would be either High of Low depending on the setpoint voltage on its positive input and the DeltaP voltage. The 10K feedback providing lots of hysteresis.
Greg
Reply to
Greg Hines
"Greg Hines"
** U5C is a unity gain inverting stage formed by R27 and R29 - the actual gain reducing to zero at some high frequency due to the 0.1uF cap.
Dunno how you get any other idea.
.......... Phil
Reply to
Phil Allison
"Greg Hines" wrote in news:A9VHe.189$ snipped-for-privacy@nnrp1.ozemail.com.au:
See Phil's answer to this. I think you might be regarding U5C as a comparator, it is actually an inverting amplifier as Phil says. *If* the input terminals to the opamp were reversed it *would* be a comparator. Maybe this has confused you?
Reply to
Geoff C
"Greg Hines" wrote in news:niXHe.341$ snipped-for-privacy@nnrp1.ozemail.com.au:
You have drawn this differently to the first schematic. The feedback network must go to the inverting (-) input of the opamp
Reply to
Geoff C
"Geoff C"<
** Oh deary me !!
The OP has fallen for the old trap of **assuming** that op-amps are always drawn with the + input uppermost.
......... Phil
Reply to
Phil Allison
"Phil Allison" wrote in news: snipped-for-privacy@individual.net:
Yes, a trap for young players that one.
Reply to
Geoff C
"Geoff C"
** The OP must be off hiding in a corner.
He has pulled his last post from both ISPs I get plus GG.
........... Phil
Reply to
Phil Allison
You are absolutely correct, my drawing is incorrect. However, I do appreciate everyones input and time.
Thanks guys.
Greg
Reply to
Greg Hines

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