Analog to digital conversion: are natural number values possible with an amplifier?

I drew this circuit up for a project in Multisim.

The inputs for V1 are -0.25 to 0.25, and the outputs are 0v to 10v.

When -0.25v is inputted the output should equal 0v. But its giving me an output that's close to zero instead, like -19mv.

Question: Is it possible to get the output the the exact natural(positive whole number) values?

Any help will be greatly appreciated.

circuit design of two designs

simulated circuit

simulated circuit of just the summing amplifier.

• You want discrete outputs like [0,1,2,3,...,10]? Or you want 0V to be 0V exactly? It is impossible to get exact analogue values, you need to specify an acceptable range, e.g. 0V +- ? mV. Also, I may be missing something but where is the digital part? – geometrikal May 26 '15 at 8:48

No, it is not possible to achieve ideal op amp results when using a practical op amp. Remember the 741 is not an ideal op amp; Multisim is trying to simulate your circuit based on real, non-ideal device characteristics. Consider the specifications for the LM741 op amp -- Texas Instruments LM741 data sheet Electrical Characteristics table starting on page 2.

LM741C Input Offset Voltage (Vos) can be up to 7.5mV over temperature variation. This characteristic alone means that an input of -0.25V will not necessarily output 0V, because the inverting input sees an unpredictable voltage difference in addition to the actual voltage at the pin. As though the input was not -0.25V but somewhere in the range between -0.2575V and -0.2425V. Higher input offset voltage correlates to higher output voltage error. There do exist op-amps that have lower offset voltage; these are advertised as precision op-amps. Having low input offset voltage can require making design trade-offs, so this characteristic is optimized in some op amps but not all.

Large-Signal Voltage Gain (A) is typically 200V/mV (or 200000V/V), but can be as low as 10V/mV in some cases. If the gain was infinite, then the op amp's inverting input would be driven by feedback to meet the same voltage as the non-inverting input, just as the ideal op-amp would. However with a real op-amp, the open-loop gain is large but finite. So this finite gain correlates to a finite voltage difference between the two inputs. The less large-signal voltage gain, the greater the difference between the inverting and non-inverting inputs. This also contributes to output error voltage.

Another point to be aware of with 741 is the Output Voltage Swing can get within only about 2V of the power supply rails. In this case it's not a problem, because your circuit uses +/-15VDC supply rails. However it's worth being aware that the 741 will not correctly drive output voltages too close to the power supply rail voltages. Its output signal dynamic range is limited. There do exist op-amps that can drive output very close to (but not exceeding) the power supply rail voltages, these are advertised as Rail-to-rail op amps -- for low-voltage operation this becomes very important.

I realize you may have chosen the 741 op amp by default, intending to use an ideal op-amp. The 741 was the first integrated-circuit op-amp and was very popular in the early 1970's. Nowadays the 741 is mainly used as a teaching example, because its internal circuit diagram is small enough that an EE student can follow it and recognize the various internal structures, yet it is also a complete, proven design.

It is not possible for an ideal op amp to exist, because its infinite gain would cause zero voltage between the inverting and non-inverting inputs, and that in turn would always cause the output to be zero. Ideal op amp is a great heuristic concept for understanding how negative feedback works. But a real op amp circuit needs to have large but finite gain to be able to function, and that leads to finite errors. Much of the engineering work is determining what is acceptable error margin.

• Thank you for your detailed explanation of ideal and real op-amps. – Omuse May 27 '15 at 22:15