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I was going through a problem on inverting Amplifiers and got the output voltage of the op amp as -10V. Can someone explain how the load current flows upwards from a grounded terminal? I know the output is at -10V so the load current flows from higher to lower potential, but isn't the - sign there just to indicate the 180° phase shift between the input and output voltage? Practically how can a current flow from a ground terminal, upwards?enter image description here

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    \$\begingroup\$ " isn't the - sign there just to indicate the 180° phase shift between the input and output voltage?" No! -10Volts is -10 volts against ground. You should assume the op-amp has a positive and negative supply. \$\endgroup\$
    – Oldfart
    Commented Feb 17, 2019 at 14:48
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    \$\begingroup\$ Since the opamp output is at -10V that mean that the opamp output voltage is -10 volts below the ground voltage (0V) electronics.stackexchange.com/questions/392010/… \$\endgroup\$
    – G36
    Commented Feb 17, 2019 at 14:52
  • \$\begingroup\$ @Oldfart, if I were to make this circuit on a breadboard, would it mean the load current flows from a ground terminal to pin no 6 of an IC, say 741 for example? \$\endgroup\$
    – penguin99
    Commented Feb 17, 2019 at 15:00
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    \$\begingroup\$ Yes, that is what happens if the output pin voltage is below the ground voltage (negative): the current flows INTO the pin. \$\endgroup\$
    – Oldfart
    Commented Feb 17, 2019 at 15:02
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    \$\begingroup\$ And for this circuit to work, it needs a split-supply voltage electronics.stackexchange.com/questions/370488/… \$\endgroup\$
    – G36
    Commented Feb 17, 2019 at 15:33

3 Answers 3

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You obviously have DC input +1V at the left. To get the calculated output -10V you must have negative supply voltage connected to the opamp. There are many designs which are valid only when one has dual power supply. The negative side here will sink the the output current through the output transistors inside the opamp.

Phase inversion is same as multiplying by minus one. That's useful with sinusoidal AC signals. Your amp circuit as a whole will generate phase inversion and voltage gain=10. As well one can say "voltage gain=-10")

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I ... got the output voltage of the op amp as -10V.

Your circuit has a gain of -10 so that implies an input of +1 V.

Can someone explain how the load current flows upwards from a grounded terminal? I know the output is at -10V so the load current flows from higher to lower potential, ...

In this case GND is at a higher potential than the op-amp output, VO, so the current flow will be as shown by iL.

... but isn't the - sign there just to indicate the 180° phase shift between the input and output voltage?

"Phase shift" suggests a periodic signal being processed. For this steady-state analysis the term "inversion" would be better.

Your concern is valid but note that the same is true of i1. The inverting input is at virtual ground so i1 is also "flowing upwards" to the op-amp output. In a more general circuit analysis we might assume that the op-amp sources current so that iO and iL arrows were pointing the other way. If we completed the calculations we would get negative currents in each case indicating that the currents are in the direction opposite to that of the arrows we drew. As marked in the example, the currents will be positive as current is being sunk by the op-amp.

Practically how can a current flow from a ground terminal, upwards?

It will flow from higher potential to lower - no matter which way up we draw the schematic.


From the comments:

Could you expand on 'virtual ground' ...

Basic op-amp theory of negative feedback is that the op-amp will adjust its output until the voltage at the inverting input matches (very, very closely) that at the non-inverting input. Since your non-inverting input is grounded then and the non-inverting input is at the same potential it is called a "virtual ground". (It's got ground potential without being actually connected to ground.)

and "..current is being sunk by the op amp"? Does it mean that the op amp draws the load current into it? If so, how would I record any gain?

enter image description here

Figure 1. Internals of the ancient 741 opamp. Source: Wikipedia.

From the internal schematic of the 741 op-amp it should be clear that the output can source current from the \$ V_{S+} \$ rail via Q14 or sink current to the \$ V_{S-} \$ rail via Q20.

So, an op-amp can source or sink current.

You are getting a voltage gain of -10. Gain doesn't have to be positive and it doesn't have to be > 1. Gains of 1, -1, 0.5, 50, -0.5 and -50 are all valid.

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  • \$\begingroup\$ Could you expand on 'virtual ground' and "..current is being sunk by the op amp"? Does it mean that the op amp draws the load current into it? If so, how would I record any gain? \$\endgroup\$
    – penguin99
    Commented Feb 17, 2019 at 15:06
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Commented Feb 17, 2019 at 15:14
  • \$\begingroup\$ The op amp theory that you mentioned, about voltage adjustment, is it some requirement that has to be fulfilled for the op amp to work properly? \$\endgroup\$
    – penguin99
    Commented Feb 17, 2019 at 15:22
  • \$\begingroup\$ I think you've missed some basic understanding somewhere along the way. An op-amp is a very high-gain differential amplifier with various imperfections including non-linearity. This generally makes them unsuitable for open-loop use. Applying negative feedback "tames" the amplifier and brings the gain down to reasonable values and allows correction of non-linearities in their operation. There are millions of articles and probably thousands of YouTube videos which should be of help to you. \$\endgroup\$
    – Transistor
    Commented Feb 17, 2019 at 16:01
  • \$\begingroup\$ Thanks for this. I'm still getting myself familiarised with this topic. Just wish my teachers were as clear as this forum haha \$\endgroup\$
    – penguin99
    Commented Feb 17, 2019 at 16:02
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The directions of the arrows that represent currents drawn in a circuit schematic are arbitrary.

This means that you can define a certain current flowing through an element (like a resistor) in whatever direction you like.

Then, when you solve the circuit (i.e. work out all the currents), if such current is actually physically flowing in the same direction of the arrow you have chosen, you'll get a positive current (e.g. + 2 mA); otherwise, if such current is actually physically flowing in the opposite direction of the arrow you have chosen, you'll get a negative current (e.g. - 2 mA).

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  • \$\begingroup\$ "The verses of the currents ..." is not standard technical English and you have no location in your user profile. Is it a typo or an odd translation? \$\endgroup\$
    – Transistor
    Commented Feb 17, 2019 at 16:05

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