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i made an inverting amplifier circuit using CA3130, the supply just the positive voltage and GND for V- supply. the circuit is used for full wave rectifier because the signal is around 1 MHz and 500 mV and i don't find any good diode for that frequency. the problem is, i expected only the positive voltage signal is produced but then the negative signal was appeared too in oscilloscope, how is this even possible?

schematic

simulate this circuit – Schematic created using CircuitLab

Thank you

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  • \$\begingroup\$ AC coupled scope. \$\endgroup\$
    – Andy aka
    Aug 6, 2018 at 6:36
  • \$\begingroup\$ @Andyaka i am sorry? i have checked the oscilloscope, it is DC coupled \$\endgroup\$
    – Zahi Azmi
    Aug 6, 2018 at 6:37
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    \$\begingroup\$ Schematic diagram, please. There's a button on the editor toolbar to allow you to add an editable one and it's very easy to use. \$\endgroup\$
    – Transistor
    Aug 6, 2018 at 6:45
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    \$\begingroup\$ Maybe there is AC coupling in the path anyways... Add the schematic. \$\endgroup\$
    – Vladimir Cravero
    Aug 6, 2018 at 6:45
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    \$\begingroup\$ The schematic looks fine. Let's see where you connected the oscilloscope and a photo or screen-grab of the output waveform. If you can include the input waveform in the same trace it would be most useful. \$\endgroup\$
    – Transistor
    Aug 6, 2018 at 10:10

3 Answers 3

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Your 100 ohm resistor values are way too low and the input signal goes directly through them and to the output. 5V/100 ohms= 50mA. The output current of a CA3130 with a 10V supply is only about 4mA which reduces the output swing to 3V peak. Try 10k or 20k for the resistor values.

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I don't see any source of negative voltage other than the input signal. I suspect your op amp circuit isn't working at all, and you're seeing the input signal appear at the output. Connect a probe to the input and see if the output is getting inverted as expected.

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Instead of discovering wheel again, how about using Ideal Diode configuration?

Source: https://en.wikipedia.org/wiki/Operational_amplifier_applications#Precision_rectifier

The voltage drop VF across the forward biased diode in the circuit of a passive rectifier is undesired. In this active version, the problem is solved by connecting the diode in the negative feedback loop. The op-amp compares the output voltage across the load with the input voltage and increases its own output voltage with the value of VF. As a result, the voltage drop VF is compensated and the circuit behaves very nearly as an ideal (super) diode with VF = 0 V.

Source: https://en.wikipedia.org/wiki/Operational_amplifier_applications#Precision_rectifier

This way you will stay in safe side in means of absolute maximum rating for input which is Vss-0.5v (Ref: https://www.intersil.com/content/dam/intersil/documents/ca31/ca3130-a.pdf)

Edit: I didn't state my reasoning clearly. As far as i understand, the problem arises when 0.5V input voltage spec is taken into account. It's hard to find low Vf and high speed diode but this changes when low Vf spec is relaxed; since 1N4148 alike components become possible to use.

So consider this an alternative solution.

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    \$\begingroup\$ This doesn't answer the question, "... how is it [the negative output on a single-rail supplied op-amp] possible?" OP also explained the difficulty in getting a diode that works well at 1 MHz and you haven't addressed this. \$\endgroup\$
    – Transistor
    Aug 6, 2018 at 12:13
  • \$\begingroup\$ I didn't downvote. Tidy up the answer explaining that it's an alternative solution to the problem and I'll delete my comments and counteract the downvote. \$\endgroup\$
    – Transistor
    Aug 6, 2018 at 13:02

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