| Help needed from Electrical/Computer Engineers?

Help needed from Electrical/Computer Engineers?

rich- asked:

I’m taking an introductory course for electrical engineering. This question deals with inverting op amps:

http://www.freeimagehosting.net/image.php?6397d8cfa9.jpg

Design an amplifier with a gain of -7. Enter the circuit in PSPICE. Use the LF356/NS op amp for your design. The pin numbers are written near the square connection boxes. You will need to make connections to pin 2, 3, and 6 (the inverting input, non-inverting input, and output) for your circuit.

There is more to the instructions but this is where I’m having trouble. I designed the circuit on paper but I don’t understand what I have to connect to pins 2, 3, and 6.

Any tips would be appreciated.
stores

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Comments

4 Responses to “Help needed from Electrical/Computer Engineers?”

  1. Ghadahav Kaffee on April 8th, 2009 9:07 pm

    you want a negative output, so you’ll connect you input signal to the inverting terminal
    you want a gain of 7, so you want to connect a divider network between the output and inverting terminals. the ratio of that network should be 7
    good luck

  2. billrussell42 on April 10th, 2009 10:09 am

    put a 2k between the input and the – input, put a 14k between opamp output and the – input, put a 1.75k from + input to ground.

    the gain is the ratio of 14k/2k or 7

    .

  3. Deth f on April 11th, 2009 1:34 am

    if ignorance is bliss then im ecstatic

  4. Nick G on April 11th, 2009 12:59 pm

    Pin 2 is the inverting (-) input. Pin 3 is the non-inverting (+) input. Pin 6 is the output. Before you do this, you really should understand what an operational amplifier is, what is does, and how to use it. I will help as much as I can, but, without this knowledge, it will do little to help with your understanding.

    Between pin 2 and the input signal source connect a resistor of value R. Between pin 2 and pin 6 connect a resistor of value 7R. Connect pin 3 to ground.

    The operational amplifier (LF356) works by its output trying to force pin 2 to be at the same voltage as at pin 3. Since pin 3 in this case is grounded (at 0V), the LF356 output will go to whatever voltage is necessary to force pin 2 to be 0V. The values of R and 7R do not matter, except they should not be so large that the stray capacitances and these resistors do not cause excessive phase shift or reduce the frequency response below that needed and they should not be so small that the load is presents to the source is so low that distortion results.

    Let’s say that the input source is at 0V. Because pin 2 is also at 0V there is no current through the input resistor R. The LF356 has negligible input current, so consider it to be zero. Since no current enters the node at pin 2, no current leaves it. This implies that the LF356 output pin 6 is 0V. Now, if the voltage at the input goes up to +1V, the current through the input resistor is 1V/R. This same current flows through the feedback resistor between pins 2 and 6. The voltage across this resistor is the current through it times its resistance, or (1V/R) x 7R = 7V.

    Conventional current flow is from positive to negative, so the input current from the driving source is from that source toward pin 2. Since no current flows into pin 2, all of that current goes through the 7R feedback resistor. The voltage then drops lower from input source to pin 2 and drops more from pin 2 to pin 6. We know that pin 2 voltage is 0V and the drop across the feedback resistor is 7V, so the output voltage at pin 6 is 0V - 7V = -7V. The input went up by 1V and the output went down by 7V, so the gain is -7V/1V = -7 V/V, or we simply say -7.