Trouble with op-amp fundamentals

Discussion in 'The Projects Forum' started by Not_Who_You_Think, Jan 1, 2013.

  1. Not_Who_You_Think

    Thread Starter New Member

    Jan 1, 2013
    5
    1
    Greetings -

    I'm working on a project to convert my 300-watt halogen torchiere lamp to LED's: 40% of the power consumption, twice the lumens.

    I have a Ph.D. in mechanical engineering, but I'm far less knowledgeable about electronics. I'm struggling to make a common op-amp work its magic for me, and I'm hoping someone here can toss me a clue.

    The lamp will be dimmable via FlexBlock modules, and it will also have a fan to cool the LED's. The dimmer feature on the FlexBlocks is such that when 0-2 volts is applied, the LED's are off, and from 2-10 volts, the LED's increase from about 5% of full power to 100% of full power. I'm trying to come up with a fan control module (FCM) that will turn the fan on when the dimmer voltage exceeds 1.5 volts, i.e. just before the LED's turn on.

    The FCM uses a variable DC converter so I can dial up a suitable fan RPM (i.e. as quiet as possible). That DC converter accepts a run/inhibit input signal, and I'm using an LM741 op amp to provide that signal. The op amp is supposed to compare the dimmer voltage with a 1.5-volt reference voltage and switch the output between zero and five volts, depending on whether the dimmer voltage is above or below 1.5 volts.

    The op amp chip I'm using is LM741CNNS:

    http://www.ti.com/lit/ds/symlink/lm741.pdf

    I have the following connections to it:

    Pin 1, offset null: no connection

    Pin 2, inverting input: I have set this to 1.5 volts (reference voltage) via a pot.

    Pin 3, non-inverting input: this will be the input from my dimmer, anywhere from 0-10 volts. Idea is that op-amp output should go high (and run fan) when dimmer voltage exceeds 1.5 volts.

    Pin 4, V-: ground

    Pin 5, offset null: no connection

    Pin 6, output: This is connected to ground through a 1M resistor, and also connected to the “inhibit” pin on the variable output power supply.

    Pin 7, V+: 5 volts DC supply.

    Pin 8, NC: no connection

    My problem is that pin 6 is consistently at 4.35 volts, regardless of whether Pin 3 (non-inverting input) is grounded or connected to 9 volts (a 9-volt battery). As I understand it, bringing Pin 3 to 9 volts should cause Pin 6 to go to ~5 volts (since V3>V2), and bringing Pin 3 to ground should cause Pin 6 to go to ~0 volts (since V3<V2). All of this is true regardless of whether Pin 6 is connected to the run/inhibit input of the variable DC converter for the fan.

    Any idea what I’m doing wrong? Is 1M too big of a pull-down resister for the op amp output? If so, how big should I be using?
    Thanks for any advice...
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,436
    3,360
    How many years did it take you to get a Ph.D. in mechanical engineering?
    Well you cannot learn electronics overnight.
    Even something as basic as an opamp. You don't have to start from scratch but you do have to start somewhere.
    I suggest you start with Ohm's Law and the tutorials found on the top of the All About Circuits page.
    The LM741 is the wrong opamp to use. It cannot work on a single 5V supply.
    The 1MΩ resistor is not too big. In fact it has nothing to do with the operation of the opamp. You cannot connect the inverting and non-inverting inputs independently unless you are building an analog comparator. If so, then select a comparator circuit.

    Not to discourage you. State what you are trying to do in non-electronics terms and let an expert here provide you with a solution.
    Edit: You have stated this in the first half of your post. Let someone else design the circuit for you.
     
  3. justtrying

    Active Member

    Mar 9, 2011
    329
    349
    If I am not mistaken, the attempt here is to build a comparator with the dreaded 741. We did it in school. It seems to me that the problem is exactly what Mr. Chips has pointed out - single rail power supply at 5V.

    I would try running the op-amp as wired but power at +/-15V and see what happens. However, since it seems that the output needs to swing between 0 and 5, definitely need to change the op-amp. There is another thread on similar topic with a list of op-amps that swing between 0 and 5V.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    1. The 741 input range starts at about 2 or 3 volts above the negative supply rail (ground in your case), so it won't work with 1.5V.
    2. The inputs must not exceed the supply rails. You may have damaged you device by applying 9V.
    3. The output will only swing within a couple of volts of the supply rails. With a 5V supply, you will only get the output to move by about a volt or two.

    You need to use a comparator. The symbol looks just like an op amp, but it is optimized for comparing voltages. Op amps are designed to be used in closed loop negative feedback circuits, and don't switch as rapidly as comparators. They also are not designed to output logic level voltages.
    LM393 is a dual comparator that should work for you. The input range includes ground, and can go as high as 36V without damage, even if the positive supply rail is less than that. The output will swing very close to ground. It is open collector, which means it needs a pullup resistor to a positive supply in order to pull up the output when the output transistor is off.

    Give us more info about the converter you are wanting to control, like a link to a datasheet.
     
  5. Not_Who_You_Think

    Thread Starter New Member

    Jan 1, 2013
    5
    1
    Thanks for explaining the details. I was under the impression that a comparator was just an op amp configured without any negative feedback(maybe because they use the same symbol :D); looks like there's more to it than that.

    Thanks for the recommendation; I will look into that.

    The fan PS is a PTN78000H. The run/inhibit feature seems to work fine: if I ground pin 3 (see page 5), the fan shuts down.
     
  6. t06afre

    AAC Fanatic!

    May 11, 2009
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  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    If you use the LM393, the datasheet recommends a small MOSFET between the comparator output and the inhibit input (Fig. 29). If you do this, remember that the MOSFET inverts, and it will need to be specified as a logic level device, with Rds(on) specified when Vgs is less than or equal to 5V (assuming you are using a 5V supply). The datasheet suggests BSS138. You could also use 2N7000 or 2N7002. If you use the MOSFET, you will need a 10k pullup from the comparator output to +5V.
    Having said all that, you should be able to connect the comparator output directly to the inhibit input. You won't need a pullup resistor, because the regulator module has one internally.
    The issue here is that the inhibit input has to be between 0V and 0.3V. The input current when the voltage is low is only about 0.25mA (coming out of the pin). A small MOSFET has a resistance of a few ohms, so it satisfies the low voltage requirement. If you look at Fig. 3 in the LM393 datasheet, you will see that the low output voltage is only about 25mV when the current is 0.25mA. I generally don't design to typical specs, but since typical in this case is an order of magnitude lower than the required 0.3V, I'm sure it will work OK.
     
  8. Not_Who_You_Think

    Thread Starter New Member

    Jan 1, 2013
    5
    1
    Thanks for your help. I tried an LM393 this morning, plugging the output directly to the run-inhibit pin of my variable DC-DC converter, and it works perfectly. As you noted, no pullup resistor is needed.

    Now that I've got my fan control system working, I can go ahead and order the expensive parts for the next phase of the project. :D

    Thanks again.
     
    Ron H likes this.
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