Nezz10

I need a hand designing a circuit using 741 op amps.

Background - We have an inverter that powers a motor which turns a pump. The pump pumps mud around a test rig we have.

The inverter speed is controlled by a 10K pot feeding it a 0-10v signal, the system is turned on and the pressure in the pipe of the test rig is set by adjusting the pot.

The pressure will rise or fall slightly over time and I need to design a circuit to automatically compensate for this and adjust the signal to the inverter accordingly.

What I have decided to do is use a 741 integrator circuit which will have one input set via a pot (not the 10K pot on the inverter control box) and one input from a pressure transducer. The output needs to be the difference between these signals wether it be + or -

The output from the amp will be zeroed at the beginning of the test using the pot in the op amp circuit

The diagram below will help explain:

click me

I have very limited knowledge of op amps and only know about using an integrator because it was suggested by a mate.

Any help welcomed!

Trashman

Very rusty but I think the set speed should be a reference also i.e. the pressure input on one input of the op amp, the speed on the other.

AndyC_772

iTrader: (2)

TBH you need to find a mate with an in-depth knowledge of electronics and buy him a few beers.

What you're trying to do is perfectly reasonable, and you're right to look at doing it with an op-amp - though the 741 is a dinosaur!

The difficulty is not in producing a circuit that does what you want, but in producing one that'll be stable and not be prone to oscillate nor over-react to small fluctuations in input or output. That's likely to take some trial and error, since you don't know how long it takes for a change in pump speed to be measured as a pressure difference.

An integrator is unlikely to actually be the configuration you want - I see the idea (use one 741 to produce an error signal, integrate with another, then use a third 741 to derive a compensating signal that drives the integrated error toward zero). However, that's overly complicated and not really what you want. All you want is a feedback loop that keeps the pressure constant.

It can certainly be done, but it's probably a day or two's work for someone who really knows their stuff, and (like most electronics problems that come up on SN), not within the scope of a post on a bbs. Sorry

If it were easy, they wouldn't pay me enough to run a scooby to do it!

hedgehog

You might be better to think about using some form of microcontroller for this. Some have onboard A-D conversion and so can easily read discrete voltages from both a pot, to set the control voltage level, and your transducer. They are also inexpensive and something like a picaxe might work well.

You can buy a picaxe and development board for about £30. All you need to do then is write some code to compare the output of the A-D and provide a control signal output from the micro. This method might give you more control and it should be much easier to make it stable as you can take different actions depending on the amount by which the two voltages differ. If they were wildly different, for example, then you might assume a serious fault condition and shut the pump down totally.

Just a thought and it might not suit you as I don't know your precise circumstances.

AndyC_772

iTrader: (2)

(Your 1000th post, and it's not "we're all doomed" - congrats!)

Not a bad idea. How's your assembly language?

Then you just need to produce a PWM output from the PIC (easy, honest!) and scale it appropriately to drive the inverter. Have you any idea how much current the inverter control input requires? Is there a lot of electrical noise around? What power supplies do you have available?

hedgehog

Many of the current crop of micros can be programmed in BASIC (of a sort) or C so it may not be necessary to use assembly language.

I think the Picaxe has some form of BASIC programming interface as does the BASIC Stamp, hence the name. The stamp doesn't have a built in A-D, though you can easily add an external chip to do the job, but it does have a nice version of BASIC now plus a nice environment.

Sprint Chief

If you wanted to achieve this, as Andy says, you really need someone who knows what they are doing. Electronics is not as easy as it first seems and a little knowledge can be a bad thing, and after blowing up the twentieth 741 in a row frustration can set in. I couldn't resist adding a quick commentary to your diagram though!

Most importantly, don't try and wire the output to the op-amp to the 10k potentiometer output as illustrated. If the op-amp is trying to drive a level of +0.5V into a level of (some other voltage), the voltages won't sum, the op-amp will provide as much current as it needs to force the line to 0.5V. As the pot is 10k, if you are near the middle of the pots range, the op-amp will probably succeed, if you are near the rails, there will be a brief "pop" noise and smoke will come out of the little crack which will have just formed in the 741 casing.

You don't need an integrator! Although the configuration you've drawn is actually not an integrator and not so far off from what you want - to subtract two voltages you need a differential amp, which is what you've drawn! Unfortunately this won't quite do what you need as the result of the circuit op-amp circuit you've drawn is

Vout = G(vplus - vminus)

The G is the open loop gain of the op-amp, which is usually of the order 10,000 or more. As you can see Vout is going to spend most of its time at the rails! But it is possible to add some external resistors to bring this gain to unity, a description can be found under this link, figure 8.2, the "differential amplifier" circuit. Set R1 = R2 and bob's yer uncle. You'll need to set R1 and R2 to values which won't draw sufficient current from the potentiometer or drive of the pressure transducer to make them misbehave (i.e. quite big, at least 10x the value)

Once you've done that you'll need to use a non-inverting mixer to mix in the signal from the 10k pot before driving the inverter. As mentioned above, your hard-wired solution is a dead end.

Once you've done that, you just need to check the current draws and bandwidths are all appropriate to your application and debug the circuit and, if you're lucky, it might just work! No guarantees though...

Nezz10

Thanks for the replies guys,

I know that I cannot just hookup the output of the amp to the 10K pot, i would send these thru a summer amp. I just drew it like that cause I am lazy.

I didnt bother putting full detilas like the feedback resistor, I was just trying to give an idea of what kinda thing I am trying to achieve.

I will look at the picaxe idea to see what that can do for me and will also speak to my electroincs mate again to see if he will help me.

I know what you mean about blowing the 741's, I fryed two before I realised I have the voltage supplies the wrong way round! DOH!!!! lol

cheers again everybody

Sprint Chief

Hi Nezz10,

Ahh okay. I took the diagram a little too literally.

I also realised that I was talking a degree of rubbish. Unless you can match the real world extremely accurately the interaction of your circuit with the real world will probably cause stability problems.

The output of the differential amp should be seen as a first derivative of the pressure (differentiator), to use this signal directly would be risky, the output of this should be passed through a gain and an integrator as your friend first suggested (to give a first order control loop). The magnitude of the gain will define the bandwidth of the control loop. You'd probably need to filter the output of the pressure transducer as well if it has a high bandwidth (to prevent instability).

Hmmm definitely more tricky than it sounds

Sprint Chief

<penny drops>
Sprint chief finally reads Andy's post properly and realises it was all in there in the first place! d'oh...
</penny drops>