Question for noninverting op-amplifier for AC

[goodphy]

Hello.

I need to confess that I'm noob for electronics and studying right now.

I'm in OP-AMP study and found some sentence which is hardly understood for me.

"If the signal source is ac-coupled, you must provide a return to ground for the (very small) input current, as in Figure 4.6." (Please see the attached image)

Why does AC signal require additional path to ground? What make difference this from DC which don't requires this path, according to textbook.

In addition, OP amp requires supply voltage of +15 V and -15 V, from textbook. How can I get negative voltage? I guess both numbers are with respect to ground voltage which is typically thought as zero voltage.

 

[Jeff Rosenbury]

The op-amp has a very high input impedance, but it isn't infinite. The capacitor is effectively a DC open. So if an input is left open, the (unknown) internal circuitry of the op amp will determine its DC level.

If the circuit is DC coupled, the output of the previous stage will determine the DC level.

 

[Svein]

Those two circuits do more or less the same thing in different ways. Very short analysis:

• Fig 4.6: The input capacitor removes DC. Thus, the gain at 0Hz is ≈0, the gain at 100kHz is ≈10
• Fig 4.7: No input capacitor, so the OpAmp has to deal with whatever signal arrives at the input. The feedback tries to cancel out any DC. The gain at 0Hz is ≈0, the gain at 100kHz is ≈10

The main difference between the two is that the OpAmp in Fig 4.6 does not have to deal with DC, the one in Fig 4.7 has to.

 

[Averagesupernova]

Those two circuits do more or less the same thing in different ways. Very short analysis:

• Fig 4.6: The input capacitor removes DC. Thus, the gain at 0Hz is ≈0, the gain at 100kHz is ≈10
• Fig 4.7: No input capacitor, so the OpAmp has to deal with whatever signal arrives at the input. The feedback tries to cancel out any DC. The gain at 0Hz is ≈0, the gain at 100kHz is ≈10

The main difference between the two is that the OpAmp in Fig 4.6 does not have to deal with DC, the one in Fig 4.7 has to.

The gain here is never less than one.

 

[jasonRF]

"If the signal source is ac-coupled, you must provide a return to ground for the (very small) input current, as in Figure 4.6." (Please see the attached image)

Why does AC signal require additional path to ground? What make difference this from DC which don't requires this path, according to textbook.

The issue is that both of the inputs of the op amp are connected to transistors that need a DC bias to operate properly. If the inputs do not have a DC path to DC current they will not work properly. Different op amps have different levels of bias current required - bipolar opamps typically have the highest. If the input bias current is large there can also be voltage drop from ground to the input (eg, across the 100k resistor in figure 4.6).

jason

 

[Averagesupernova]

Basic rule to remember: Don't let the inputs float.

 

[meBigGuy]

If the inputs are allowed to float the voltage will be set unpredictably by bias and leakage currents.

In figure 4.7 the + input is not floating because it is driven by a source with some finite impedance and fixed DC offset. Figure 4.7 has unity gain at DC. Figure 4.6 blocks input DC.

Many op-amp circuits require two power supplies, a positive and a negative voltage relative to ground. If you want to operate from a single voltage supply you need to create a "virtual" ground. This can be as simple as biasing fig 4.6's + input to half the power supply voltage and AC coupling the output to remove the DC bias.
There are other more complex virtual ground circuits you can search for.

 

[Svein]

The gain here is never less than one.

Yes, of course. Blame it on it being late (for me). The correct statement is:

• Fig 4.7: No input capacitor, so the OpAmp has to deal with whatever signal arrives at the input. The gain at 0Hz is 1, the gain at 100kHz is ≈10

 

[goodphy]

Those two circuits do more or less the same thing in different ways. Very short analysis:

• Fig 4.6: The input capacitor removes DC. Thus, the gain at 0Hz is ≈0, the gain at 100kHz is ≈10
• Fig 4.7: No input capacitor, so the OpAmp has to deal with whatever signal arrives at the input. The feedback tries to cancel out any DC. The gain at 0Hz is ≈0, the gain at 100kHz is ≈10

The main difference between the two is that the OpAmp in Fig 4.6 does not have to deal with DC, the one in Fig 4.7 has to.

Thanks for comments.

Could you tell me why feedback works such a way that input DC is canceled? I only know golden rule for feedback OP-AMP; 1st, voltage difference between inverting and non-inverting inputs becomes zero, 2nd, input of the op-amp doesn't draw current. Can I derive DC-canceling from this rule? And what is difference between AC and DC input applying to OP-AMP?

 

[Svein]

Thanks for comments.

Could you tell me why feedback works such a way that input DC is canceled? I only know golden rule for feedback OP-AMP; 1st, voltage difference between inverting and non-inverting inputs becomes zero, 2nd, input of the op-amp doesn't draw current. Can I derive DC-canceling from this rule? And what is difference between AC and DC input applying to OP-AMP?

Sorry, my error. See post #8.

 

[goodphy]

The issue is that both of the inputs of the op amp are connected to transistors that need a DC bias to operate properly. If the inputs do not have a DC path to DC current they will not work properly. Different op amps have different levels of bias current required - bipolar opamps typically have the highest. If the input bias current is large there can also be voltage drop from ground to the input (eg, across the 100k resistor in figure 4.6).

jason

Hello.

OP amp requires DC biasing on inputs? Maybe I have to ask more fundamental question. What happens to OP-AMP when it receives pure AC voltage? How does it behave differently from DC input? And Why OP-AMP is called DC-amplifier?

 

[goodphy]

Basic rule to remember: Don't let the inputs float.

Hm...Could you specify more? Do you mean non-inverting input of Fig 4.6's circuit is floated when there is only DC input on that?

 

[goodphy]

The op-amp has a very high input impedance, but it isn't infinite. The capacitor is effectively a DC open. So if an input is left open, the (unknown) internal circuitry of the op amp will determine its DC level.

If the circuit is DC coupled, the output of the previous stage will determine the DC level.

Hello.

Thus you mean,,,OP-AMP has its own default non-inverting input value and this value is applied when non-inverting input is left to open?

 

[goodphy]

Sorry, my error. See post #8.

Thanks for your reply!

Could you tell me more about Fig. 6? When DC is applying to V_in, capacitor block the DC so non-inverting input (+) of OP-AMP is at ground (no current flows to + input thus no voltage drop across resistor connect between + and ground.) Thus OP-AMP feedback network here is doing nothing for DC.

Is my analysis right? And why is resistor from ground to + input necessary? I guess this resistor is not required for AC input and even for DC, I don't see its importance.

 

[Jeff Rosenbury]

Hello.

Thus you mean,,,OP-AMP has its own default non-inverting input value and this value is applied when non-inverting input is left to open?

More or less. However that value is very weak. Almost anything, including noise, can change it.

In the case of the lack of a DC bias, I would expect it to drift to be near one of the rails. (flip a coin). Then we would experience clipping.

Op-amps are designed to be used in one way. That is to make sure both inputs have the same voltage. If they have different voltages, that's called a comparator. (Which is basically the same circuit, but typically it has a higher output impedance and less driven current.)

It is up to us engineers to set the values we want. Feedback to the negative input is typically how that's done, but there are lots of topologies to achieve the goal. Leaving a floating input (positive or negative) is not one of them.

 

[Svein]

Thanks for your reply!

Could you tell me more about Fig. 6? When DC is applying to V_in, capacitor block the DC so non-inverting input (+) of OP-AMP is at ground (no current flows to + input thus no voltage drop across resistor connect between + and ground.) Thus OP-AMP feedback network here is doing nothing for DC.

Is my analysis right? And why is resistor from ground to + input necessary? I guess this resistor is not required for AC input and even for DC, I don't see its importance.

It is very important. If you do not insert a resistor to ground, you get two effects:

• The typical input impedance is on the order of 10MΩ. A floating non-inverting input will therefore have an input impedance of the order of the impedance between two traces on the PCB.
  
• The RC network on the non-inverting input will have a very low cut-off frequency, at least two decades lower than expected. An input signal of 1Hz will pass right through the input capacitor.

 

[goodphy]

It is very important. If you do not insert a resistor to ground, you get two effects:

• The typical input impedance is on the order of 10MΩ. A floating non-inverting input will therefore have an input impedance of the order of the impedance between two traces on the PCB.
  
• The RC network on the non-inverting input will have a very low cut-off frequency, at least two decades lower than expected. An input signal of 1Hz will pass right through the input capacitor.

Thanks. I didn't remember RC network is actually high-pass filter which should have finite value of resistor. Thanks again!

 

[jim hardy]

Goodphy

you probably got the idea now
here's a memory aid for down the road...

All real amplifiers have at their input pins something called "pump out current"
real amplifiers are not ideal and small current flows into or out of their input pins.
We ignore it to simplify analysis
but we have to deal with it when transferring from thought experiment to hardware. experiment
you'll see the amount of that pump-out current given in the datasheet as "Input Bias Current" but i think pump-out is a better name.

That's why both inputs should see about the same DC resistance to circuit common, and that resistance must be smalll enough to not make significant voltage at bias current as bias current flows through it
Bias currents range from hundreds of nanoamps to femtoamps.
It's one of the things one looks for in analog computer circuits like integrators.

 

[jasonRF]

Hello.

OP amp requires DC biasing on inputs? Maybe I have to ask more fundamental question. What happens to OP-AMP when it receives pure AC voltage?

In Figure 4.6, if the 100k resistor were not there the opamp would still draw the bias current, which would charge the capacitor, which causes the input to either go up or down in voltage (depending on polarity of current). This forces a dc offset between + and - inputs, which is amplified by the opamp. Eventually the opamp with be pegged to either the positive or negative rail voltage and your circuit will not operate properly.

At least that is how I remember it going. One of the first hits I got on a google search is here:

http://www.analog.com/library/analogDialogue/archives/41-08/amplifier_circuits.html

It looks useful.

jason