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  1. L

    building pH controller

    In your reference circuit, the first op amp provides gain for the probe signal for a range of +/- 700 mV or a total range of 1.4 volts. The second op amp provides signal inversion, range scaling and level shifting. Your final output signal will have a range of 5.0 volts and be level shifted so...
  2. L

    discrete photodiode amplifier

    Have you looked at the TI OPA847? It is billed as a wideband, ultra-low noise operational amplifier suitable for making a transimpedance amplifier for photodiodes. I've been trying to use it with a PIN photodiode as a high-gain sub-microsecond single event detector.
  3. L

    Bat help

    You might start here: http://www.batterysavers.com/Compare-Batteries.html
  4. L

    inductive capacitance???

    So just where did you hook up the LED? If you put a Faraday shield around the input section of the breadboard circuitry, does the effect diminish?
  5. L

    Another Driving LED question

    The LED driver outputs of the MAX6974 have a current sinking interface. They do not "output 5 to 7 volts" as stated. Each MAX6974 will drive at most 16 pixels with multiplexing enabled, so how will you drive 100 pixels "using a MAX6974"? Also, the MAX6974 is intended to drive each LED...
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    saturated operational amplifier

    In any normal op amp circuit with dual power supplies, the '+' input to the op amp is grounded. The input signal is fed through a resistor to the '-' input along with the output signal through a feedback resistor. The op amp works to keep the voltage difference between its inputs equal to...
  7. L

    Help with coil limiting Spark Duration

    Sounds like you are looking for a 'Capacitor Discharge Ignition' system. http://www.gill.co.uk/products/digital_ignition/Introduction/4_CDIvsIIS.asp Back in 1968 I built a CDI and wired it into the points and ignition coil on the car engine, but did it through a relay that would restore the...
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    op-amp difference amplifier

    If you will notice, my expression for Vo is in the form: Vo= -cV1+bV2+aV3 where (if my derivation is correct), c= (R2/R1) b= (R4R5(1+R2/R1))/(R3R4+R3R5+R4R5) a= (R3R5(1+R2/R1))/(R3R4+R3R5+R4R5) Your task, after verifying the derivation, is to select values for R1, R2, R3, R4...
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    op-amp difference amplifier

    Letting Vi be the voltage at the op amp inputs, there are 2 node equations: (Vi-V2)/R3 + (Vi-V3)/R4 + Vi/R5 = 0 (Vi-V1)/R1 + (Vi-Vo)/R2 = 0 Substituting to eliminate Vi gives: ((V1/R1 + Vo/R2)/(1/R1 + 1/R2))(1/R3 + 1/R4 +1/R5) = V2/R3 + V3/R4 Solving for Vo gives: Vo=...
  10. L

    input / output of op amp

    It's just that you started out by writing an equation for the op amp working in its linear region, which is only valid when the input signal is a few microvolts on either side of 4 volts. For all other values of input voltage the op amp is driven into saturation. To think like an engineer...
  11. L

    input / output of op amp

    If instead of using an op amp, you substituted a comparator into the same circuit, what would the output look like? How is an open-loop op amp different from a comparator?
  12. L

    phase shift

    For the equation sin(At) = -sin(At+B), find B. Does the original question make sense for non-sinusoidal waveforms such as a pulse train?
  13. L

    Keep relay actuated longer

    Something does not make sense there. The more light shining on the photo-transistor, the less current should be drawn by the circuit. The photo-transistor shunts current away from the base of the transistor, turning off the transistor in bright light. If more than 1 ma current is drawn by the...
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    Want to amplify voltage ? help please..

    @barathbushan: "....and amplifiers don't just add energy...." Since you use the example of the input being a (small AC) signal, it would be reasonable to expect that the amplifier input and output are both AC coupled. So you would see a small AC signal going in, and a big AC signal coming out...
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    Want to amplify voltage ? help please..

    @barathbushan: "....thinks that amplifiers are energy adding devices...." The fact is that amplifiers are energy adding devices. That is why amplifiers need a separate power supply to provide the additional energy. The problem with the original poster is that he described a power converter...
  16. L

    building a strong solenoid

    If you have room for a solenoid, then you probably have room for a small electric motor. I would pattern the drawer opener after the mechanism of a garage door opener - screw drive or chain drive. The linear gear drive on a CD tray is a possibility if the mechanical tolerances of the drawer...
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    Want to amplify voltage ? help please..

    @ Steve: To answer the question posed in your sig ..... It seems sometimes that government, specifically the politicians in government, exhibit wishful thinking that they can break the laws of physics with the same ease that they break (ignore) the laws of economics. Also, being basically...
  18. L

    Want to amplify voltage ? help please..

    I believe you are trying to break a fundamental law of physics regarding the conservation of energy. Although it is not a crime to break the laws of physics, trying to do so ought to be. The simple fact is that you cannot get more energy out of a process than you put into it. When you put 12...
  19. L

    Keep relay actuated longer

    The first thing I would look at is to add another transistor in a Darlington configuration to drive the relay. Add the capacitor from ground to the transistors' emitter-base connection. That way the resistance of the relay coil will be multiplied by the gain of the added transistor when seen...
  20. L

    Want to amplify voltage ? help please..

    What is the current rating supposed to be at 100 volts? Here is a circuit at: http://ecelab.com/circuit-high-dc-conv.htm that seems capable of providing the 100 volts, current rating unknown.
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