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Can we change op amp of a circuit and still work? I am testing AD822 op amp with a simple inverting

op amp

Rin I use 10K Rf I use 20K, Vin is square wave of 5v, Vout I am getting a flat signal.

Is correct Vout should be a square wave of 10v if my V+ is more than 10v from a power supply?

My Vin signal 0v is connected to power supply v--, which is also connected to v-- pin of op amp.

My +pin of op amp is connected to ground terminal of power supply

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  • \$\begingroup\$ You know that you're using the inverting input, right? \$\endgroup\$ Jun 4, 2015 at 7:23
  • \$\begingroup\$ yes my Vin is connect to pin2, which should be inverting according to datasheet \$\endgroup\$ Jun 4, 2015 at 7:26
  • \$\begingroup\$ Where is your signal 0V connected and what power supply rails are on the op-amp. \$\endgroup\$
    – Andy aka
    Jun 4, 2015 at 7:27

2 Answers 2

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Consider that an opamp (in a negative feedback configuration like this) will try to make the voltage between the + and - inputs zero. You apply 0 or 5 V to Vin (I consider only low and high state of square wave).

In case Vin = 0, then opamp will make Vout = 0, then opamp inputs voltage is 0, opamp is happy :-)

In case Vin = 5 V then Vout would need to be - 10 V (do the calculation yourself !) Note that that is minus 10 Volts ! Are you feeding the opamp with a negative supply voltage ? Or only + 10 V ?

If there is no negative supply voltage the opamp output will make the lowest voltage it can make which is 0V. Therefore all you see is 0V !

Solution: supply the opamp with positive and negative supply (two separate supplies !) or (easier) instead of connecting the + input of the opamp to ground, connect it to + 5V (use to resistors in series with 10 V supply)

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  • \$\begingroup\$ thank you, I thought v-- from power supply would give me negative volt. I created a voltage divider and now everything works \$\endgroup\$ Jun 4, 2015 at 8:11
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You are using the AD822 on a single supply with the negative power pin connected to 0 volts. This is not a problem but any signal has to be biased somewhere near halfway between positive supply rail and ground (0V) for correct operation.

You are feeding in a logic square wave changing between 0V and 5V AND the output of the op-amp will try and produce a level of -10V when the input is 5V.

This would be impossible if the negative rail of the op-amp were 0V and the signal were inputted as suggested in the paragraph above.

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    \$\begingroup\$ @kevin0228ca Yes I know \$\endgroup\$
    – Andy aka
    Jun 4, 2015 at 8:35

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