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my problem is that in this circuit:

enter image description here

Where \$\pm V_{cc} =\pm 15V, R1=1k\Omega, R2= 10k\Omega\$, if I introduce a signal like: \$ v_i(t) = 2 \cos{(2\pi 1000 t)}\$, the output voltaje is this:

enter image description here

Where the green line is the input and the red one the output. The problem is that although the O.A. is working between \$\pm 15V\$, the output signal is saturated at 4V approximately and I don't know why that happens.

The netlist of Pspice is this:

enter image description here

Moreover, I went to the laboratory and the O.A. ua741 was also saturated with that input signal.

Thank you so much for your help.

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    \$\begingroup\$ If that's a simulation then the 10kO (ten-k-oh rather than ten-k-zero) might be a problem. Try removing the 'O'. If it is a simulation then your question should make it clear and state the software being used. Welcome to EE.SE. \$\endgroup\$ – Transistor May 16 '18 at 16:08
  • \$\begingroup\$ @Transistor good catch, that could be the issue. Looks like PSPICE to me from the output graph. \$\endgroup\$ – John D May 16 '18 at 16:11
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    \$\begingroup\$ post your spice netlist \$\endgroup\$ – Voltage Spike May 16 '18 at 16:49
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    \$\begingroup\$ I count 4 opamps in the netlist. Is it possible for you to create a separate netlist that actually reflects your lab arrangement so that we don't have to guess? \$\endgroup\$ – jonk May 16 '18 at 17:15
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    \$\begingroup\$ My PSpice 9.1 netlist look like this R_R1 0 $N_0001 1k R_R2 $N_0001 vo 10k V_V3 vi 0 DC 0 AC 0 +SIN 0 2 1k 0 0 0 V_V1 Vcc 0 DC 15 X_U1 vi $N_0001 Vcc -Vcc vo uA741 V_V2 0 -Vcc DC 15 R_RL vo 0 10k \$\endgroup\$ – G36 May 16 '18 at 17:18
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In addition to your possible syntax * errors, you have a noninverting gain of 1+10 times the input peak of 2 V, Which should result in +/-22 Vp but cannot.

If your R values 100R* are too small, then the internal current limiter will limit the output voltage. I-max is 40mA

You should allow the output to keep away from the supply rails by the amount given in the data sheet which may be 2 V in your case.

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    \$\begingroup\$ Input peak looks like 1 volt to me. \$\endgroup\$ – Andy aka May 16 '18 at 17:09
  • \$\begingroup\$ Looks like 2V in the actual equation of the question \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 '18 at 17:35
  • \$\begingroup\$ Firtsly, sorry about my syntax,but English is not my mother tongue. Apart form that, I think that the saturation is produced due to the value of the load resistance. For example, if I place a 100 ohms resistance, theoretically vo=22V, which implicates a current of 220mA. However, the maximum output current of the O.A is 20mA, which means an output voltage of 2V. \$\endgroup\$ – Josemi May 16 '18 at 17:44
  • \$\begingroup\$ That is correct if you are using typical values \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 '18 at 17:45
  • \$\begingroup\$ By now I hope are reading I-max and not I-typ \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 '18 at 18:42

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