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THE PROBLEMATIC CIRCUIT

i simulated the circuits in an mobile app. below is another circuit with negative feedback the negative feedback circuit

-I understand how the second circuit works(-ve feedback) but the problem is with the first circuit

-Here is how the -ve feedback works(i think): We apply 1 volt to inverting input the output keeps falling(since the input is given to inverting input) from 0V and as it approaches -1V the inverting input becomes 0V(approx)and becomes virtual ground.now the system is stable

-But in positive feedback circuit (the first one) the input is positive and so is the output when we feed it back the input only becomes more high and must take the output to +ve saturation but as you can see the simulator is not in agreement with it

-Is it the simulator fault or is my explanation?

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    \$\begingroup\$ You should move some of those capital letters out of the title and swap them with some lower case ones from the body of your question. \$\endgroup\$ – Majenko Jan 2 '15 at 15:35
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    \$\begingroup\$ If your simulator showed the first circuit as "workin" and not saturating at the positive rail, then scrap the simulator. (LT spice is free.) \$\endgroup\$ – George Herold Jan 2 '15 at 15:38
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    \$\begingroup\$ That's brutal. What simulator is that? (so I can warn people off it). \$\endgroup\$ – Spehro Pefhany Jan 2 '15 at 17:14
  • \$\begingroup\$ @Majenko what's so wrong with the title? \$\endgroup\$ – dushyanth Jan 3 '15 at 4:37
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    \$\begingroup\$ @SpehroPefhany the app is everycircuit \$\endgroup\$ – dushyanth Jan 3 '15 at 4:50
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You've got the explanation right, it's just that you don't understand the simulator. It has found a solution and presented it. What it has not done is a stability analysis, and in this case the circuit is not stable. The simulator has not varied the current, but rather found a set of values that "work" for a single, exact set of values.

Try this: replace your 1 mA current source with an AC current source. Let's say +/- 10 uA with a 1 mA DC offset. Perform a time-domain (transient) analysis on the circuit and watch what happens in each circuit as soon as the current varies from exactly 1 mA.

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  • \$\begingroup\$ I am afraid, an ac analysis (magnitude) will not reveal any instability. Instead, an analysis in the time domain is necessary. \$\endgroup\$ – LvW Jan 2 '15 at 17:55
  • \$\begingroup\$ Time domain was assumed. "Watch what happens in each circuit as soon as ..." \$\endgroup\$ – WhatRoughBeast Jan 2 '15 at 18:25
  • \$\begingroup\$ OK - my question was because of the "AC current source". Sorry - it seems I have overlooked the last sentence. Or have you modified the last sentence? \$\endgroup\$ – LvW Jan 2 '15 at 18:30
  • \$\begingroup\$ No, but I did insert the phrase before "watch what happens", since apparently I was unclear. \$\endgroup\$ – WhatRoughBeast Jan 2 '15 at 18:54
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The simulator has indeed found a "stable" operating point for the first circuit. A stable operating point exsists due to the fact that the simulator has a certain numerical tolerance when calculating voltages and currents.

In reality the operating point is not true. For instance if you look at the voltage accros the positive feedback resistor in circuit one, it's 0.99999V, which will give you 0.99999 mA of current flowing through it. The simulator shows 1mA - not true. If you would feed 1mA through the feedback resistor and start with -10uV at the non-inverting input, you would end up with -1.000010V at the output. This would then imply the OPAMP has a gain of 100001, which is not so.

The "stable" operating point is due to the relative tolerance of the simulator.

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It seems that the first circuit is stable - but, in fact, it is NOT.

But the simulator does not make any error. Each simulator will give the same results - for DC analysis! Why? Because this specific analysis does not contain any switch-on transient. (Also by hand calculation you can find a stable solution).

That means: The simulator assumes that the power as well as the input signal was connected at negative infinite times. The same will happen for an ac analysis (magnitude) of this circuit. However, analyzing the phase response you would see that the system has a continously rising phase which is an indication of instability.

Perform a transient analysis (time domain, starting at t=0) - and you will see that the output will go into saturation (as expected).

Analogy (mechanical): One ball riding upon another ball. This system can be stable (theoretically) in case it would exist since infinite times - and if there is no disturbing movement from the environment.

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