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I bought an LM358 amplifier module to boost the input signal of 0-1 Vpp 20 kHz sine signal to 0-3 Vpp 20 kHz sine signal (gain of 3.)

I tested the circuit with the input pin (non-inverting pin) connected to ground and Vcc to =+5V, the op-amp output to a multimeter and input, output ground shorted to the ground pin.

I measured 2.5 V DC on the multimeter.

I'm measuring half of the supply voltage as output on the multimeter as I increase the supply voltage.

why am I getting 2.5VDC at output for 0 input signal?

enter image description here

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    \$\begingroup\$ Did you expect something different? Why? \$\endgroup\$
    – Dave Tweed
    Commented Jul 15, 2022 at 11:13
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    \$\begingroup\$ There is not a question here, try editing this to ask a question. \$\endgroup\$
    – RoyC
    Commented Jul 15, 2022 at 11:29
  • \$\begingroup\$ 2.5V output is what U1B is biased for. \$\endgroup\$ Commented Jul 17, 2022 at 2:28

2 Answers 2

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This is perfectly normal behavior for this amplifier.

You only have a single rail supply so the quiescent output is half of the supply voltage. When you apply a signal to the input that voltage will move up and down to VCC and Ground. Having it at half the supply voltage gives it the maximum range to move up or down.

This would be fine to drive a small speaker, if the output is going into another amplifier it will have a capacitor on it's input which will remove the DC component.

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If what you mean is that you grounded the IN pin of the header, then half VCC is what you should get at the OUT pin of the header. There is no AC voltage present, so C4 disconnects U1B's input from the rest and U1B becomes a non-inverting amplifier with gain = 1 (basically like a buffer, only in buffers R2 is usually just a short). The input of this amplifier is pin 5, whose voltage is VCC/2.VCC/2 x 1 = VCC/2. This is all about DC vs. AC. Your amplifier gain of 3 is AC gain.

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