LM3914 Design help

Discussion in 'General Electronics Chat' started by DC_Kid, Feb 25, 2008.

  1. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    from the datasheets for lm3914 it looks like the lowest full scale i can get is 0-1.2v. can it be made to 0.5v or 0.2v full scale?

    also, i am confused slightly about how Ref(out) and Ref(adj) work when they are tied to each other via voltage divider (shown on the 1st page of LM3914 datasheet at National's website http://www.national.com/mpf/LM/LM3914.html). i see that the 2 resistors add up to 5k, and the scale is 0-5v. does the addition of the 2 resistors equal the upper limit in voltage? but i dont quite get why Ref(adj) is connected into the voltage divider. can anyone explain how these work.

    what i am wanting to do is to be able to change the scale using dip switch (hence changing some resistor values without changing my set LED currrent). as example, i want to scale it 0-2.5v full scale, then with a dip switch flip the scale becomes 0-5v, etc. can this be done? would i just change R2 from 3.83k to 8.79k (10k - 1.21k = 8.79k)? but according to the block diagram Ref(out) to ground load determines LED current, so if i'm changing R2 then i am also changing LED current at the same time. i want to keep LED current fixed no matter what scale i choose, etc.

    from the datasheets for lm3914 it looks like the lowest full scale i can get is 0-1.2v. can it be made to 0.5v or 0.2v full scale? as example, i am monitoring current of a inductor via a 0.1 ohm resistor in series with the inductor. my known currents (depending on inductor used) is 2amp and 10amp. with 2amps i'm only gonna see about 200mV across the resistor. do i need a diff op amp to boost this up some for lm3914? but as you can see from the current ranges i want to be able to adjust the LM3914's scale by a factor of 5 (using dip switch, etc)

    any ideas?
    thanks
     
  2. Audioguru

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    Dec 20, 2007
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    The LM3914 has a max input offset voltage of 10mV to the first comparator's voltage. Then if the max voltage is set to only 0.2V full scale with a voltage divider, the lowest LED will have a threshold error of up to plus or minus 50%. If an opamp is used to amplify the voltage 10 times and has an input offset null adjustment (or if an opamp with a very low input offset voltage is used) then the error is reduced to up to plus and minus 5%.

    The brightness of the LEDs is determined by the amount of current from the reference pin.

    The voltage of the reference is adjustable similar to the LM317 adjustable voltage regulator. There is 1.25V from the Ref Out pin to the Ref Adj pin. A resistor between them has a current 1.25V/R= current. This current flows in a resistor to ground from the Ref Adj pin to ground which creates a voltage. The Ref Out voltage is the sum of the two voltages.
     
  3. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    so in the classic 0-5v setup as shown on the LM3914 datasheet, could i just use a pot to ground on Ref(out) for LED brightness, and then a pot to between Ref(out)-Ref(adj)-ground with wiper on Ref(adj)? this gives me independent control of each current of Ref(out) and Ref(adj).

    now that i think i understand how it works, max V of the scale is = to Ref(out) since its tied to V(hi) of the chip...?
     
  4. Audioguru

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    You need to increase the values of the resistors that set the 5V because they are a load on Vref and cause the LEDs to be fairly bright.
    The voltage divider for the comparators (Rhi) is also a load on Vref that adds current to the LEDs.
    The datasheet for the LM3914 and LM3915 show an extra opamp added so that LED brightness can be adjusted without affecting Vref.

    Yes but Vref must be less than 1.5V less than the power supply voltage.
     
  5. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    sorry for all the questions, i'm just trying to do 100 things at once.

    so it looks like i need a op-amp setup as a differential amp with a gain of 10x to boost my small input signal large enough to. any suggestions on which op-amp to use? my raw monitored signal will be between 0-.14v and 0-.6v (representing 1.4amp and 6amp across a .1ohm resistor). in this case i could use a single 1.2v full scale setup of lm3914 and simply use a pot to trim the input signal (from the op-amp gain stage) down, etc. i'm thinking of this way because 1.2v over 10 comparators has more sensitivity than say a 5v full scale over 10 comparators.

    now onto a question about the op-amp itself. due to the nasty stuff that comes from inductors and large currents, i'm gonna isolate the inputs of the op-amp via a mosfet analog switching IC. let me explain further. the inductor will be driven via a transistor to handle the load, but transistor turns on/off by PWM signal, and, when in off state eveything up to the collector of the transistor is at Vs (12v battery or 14.4v alternator, etc). then when transistor goes off i need to handle the flyback voltage which can range from 30-100v if no clamping zener used, or around 33-36v if clamping zener is used. so i dont want the op-amp to try and amplifier a 36v signal (op-amp will just saturate its output to Vcc) and the lm3914 will get a full scale blip. so to isolate the op-amp i'll switch the op-amp's input leads "on/off" from the resistor via fet analog switch. when the op-amp leads are not connected they will be tied to ground via ~100k resistors (so the inputs dont float).

    also, you'll notice i mentioned "transistor" to control the inductor load. its typically a NPN bi-polar so there's gonna be ~.7v drop across the transistor. if i measure V from the resistor(in) to ground i'm gonna see ~.7v + voltage drop across the resistor. in my case, the resistor at my known max current of 2amp is .7v + .2 = .9v, which still is not enough to drive lm3914 on its lowest scale of 1.2v. so i'm thinking a diff op-amp with one side of the op-amp tied to zero and the other gets thos .9v via a pot tuned to knock out .7v of the signal, leaving a output in direct proportion to the voltage drop across the resistor. i thought of just placing op-amp inputs across the resistor but i dont see the operational math the op-amp would use to create output in direct proportion to the voltage drop of the resistor....


    my math logic doesnt quite add up to the functionality i am after, hence why i'm posting for ideas.... i think i need a special op-amp setup (peraps simple) that amplifies the differences of voltages on each side of the resistor but with respect to ground. as example, the differences betwen voltage on each side of the resistor = .7 - IR, however, if i use ground as the reference the difference is (IR+.7) - .7 = IR, so how do i design a diff op-amp that uses this latter logic?
     
  6. Audioguru

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    The FET analog switch will be fried by the flyback voltage from the inductor.
    A few opamps do not need to have a negative supply to measure signals close to ground and work fine with only a positive supply voltage. The MC34071 single opamp (a dual and a quad are available) works with its inputs at ground and is fast. An MC33171 is slower and uses less power and an old LM358 dual opamp is also slow and uses a low supply current. Then you can make the opamp a differential amplifier with as much gain as you want and the FET anaog switches can commect its inputs directly acoss your current sensing resistor.
     
  7. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    [A] page 12 figure 48 of the 34071 datasheet http://www.onsemi.com/pub_link/Collateral/MC34071-D.PDF, almost what i need. however, the ground side of Rs in my case is not to ground, it has a NPN inline with it, so the bottom side of Rs will be at 0.7v when NPN is conducting current. does this mean i bring the ground side of R2 to a 0.7 voltage reference?

    for another circuit i was doing i was using a Burr Brown (now TI) INA114 http://focus.ti.com/lit/ds/symlink/ina114.pdf
    any thoughts on this op-amp used in a current sensing schematic? Figure 1 page 8 of this datasheet shows a basic schematic. since Vo = G * (V+ minus V-) i could connect V+ to the "in" side and V- to the "out" side of my current sense resistor. this seems to indicate that V+ and V- is measured from that pin to ground, which is good for me. so when current flows V- will be pegged at whatever voltage drop there is across the transistor that controls inductor current, and the "hi" side of my resistor will be IR + ~.7v. since i would be making a circuit intrusinon by adding any resistor in series with the inductor load (its a fuel injector) i can then perhaps make it 0.05ohm 1% or .5%, making it less intrusive, etc.

    does part B here make sense.
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    The 0.7V from the transistor changes with the current in the transistor and changes with the temperature.
    You need to use a differential amplifier (it can have gain if you want) to measure the voltage only across the current-sense resistor.
     
  9. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    yeah, i stated some wrong info... my mind is too busy these days.. i was thnking of base-emitter bias voltage... typically .7v

    i'm gonna go ahead with INA114 testing to see what it yields on the bench. thanks for the pic...
     
  10. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    one last question.

    lm3914 will limit LED current. does this mean i can tied the LED supply voltage to 12V even if LED is a 2V LED??
     
  11. Audioguru

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    Dec 20, 2007
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    12V is too high because 10V times the LED current will create heat. If all the LEDs are turned on at 30ma then the toltal current in the LM3914 is 300ma and the heat dissipation is 3W!. Its absolute max allowed dissipation is 1.365W if the ambient air is room temperature and then the chip will be at its max allowed temperature.

    Use a resistor to feed the LEDs as is shown in the datasheet. Then the resistor shares the heat. The LEDs would need a decoupling capacitor to ground where they join the resistor.
     
  12. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    ahh, yes, makes perfect sense. with resistor and set current i need to knock down the voltage from Vled to Vf(typ) of LED.

    thanks
     
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