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Strange opamp's output impedance in spice

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4

1

$begingroup$

I'm simulating this circuit in Micro-Cap, which is the clipping stage of a guitar effect. The opamp model is the "NE-5532"

I want to measure the input and the output impedance. I expected an output impedance closer to zero Ohm, and an input impedance of about 10kOhm, with an "infinite" impedance at 0Hz due to the decoupling capacitor at the input.

Here it is the analysis in Micro-Cap.

As you can see the input impedance (the blue graph) is close to what i expected, but the red graph, which is the output impedance, it's really strange. It's almost 10kOhm, with a peak of almost 1MegOhm, and i can't really explain why.
If i switch the model to a "LF-155" i get a more "reasonable" results, with an output impedance of 1.680E-68 Ohm, which is also strange.

Can you help me? This thing is driving me crazy.

operational-amplifier impedance spice input-impedance single-supply-op-amp

share|improve this question

edited Mar 9 at 15:40

RawCode

asked Mar 9 at 15:31

RawCodeRawCode

194

New contributor

RawCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You got the first two graphs from a single run of the simulator?
  $endgroup$
  – The Photon
  Mar 9 at 15:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  something's fundamentally broken with this simulator or its NE5532 model. You physically can't have an output voltage of 1 MV
  $endgroup$
  – Marcus Müller
  Mar 9 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThePhoton Yes, this is a single run of the ac analysis
  $endgroup$
  – RawCode
  Mar 9 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is that a 10 ohm resistor from output pin to ground (R11?) The op-amp will try to maintain 4.5V across that resistor: too much DC current will flow for the op-amp (smoke would ensue). Try returning that resistor to the 4.5V supply instead of ground.
  $endgroup$
  – glen_geek
  Mar 9 at 16:21
  
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

I'm simulating this circuit in Micro-Cap, which is the clipping stage of a guitar effect. The opamp model is the "NE-5532"

I want to measure the input and the output impedance. I expected an output impedance closer to zero Ohm, and an input impedance of about 10kOhm, with an "infinite" impedance at 0Hz due to the decoupling capacitor at the input.

Here it is the analysis in Micro-Cap.

As you can see the input impedance (the blue graph) is close to what i expected, but the red graph, which is the output impedance, it's really strange. It's almost 10kOhm, with a peak of almost 1MegOhm, and i can't really explain why.
If i switch the model to a "LF-155" i get a more "reasonable" results, with an output impedance of 1.680E-68 Ohm, which is also strange.

Can you help me? This thing is driving me crazy.

operational-amplifier impedance spice input-impedance single-supply-op-amp

share|improve this question

edited Mar 9 at 15:40

RawCode

asked Mar 9 at 15:31

RawCodeRawCode

194

New contributor

RawCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You got the first two graphs from a single run of the simulator?
  $endgroup$
  – The Photon
  Mar 9 at 15:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  something's fundamentally broken with this simulator or its NE5532 model. You physically can't have an output voltage of 1 MV
  $endgroup$
  – Marcus Müller
  Mar 9 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ThePhoton Yes, this is a single run of the ac analysis
  $endgroup$
  – RawCode
  Mar 9 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is that a 10 ohm resistor from output pin to ground (R11?) The op-amp will try to maintain 4.5V across that resistor: too much DC current will flow for the op-amp (smoke would ensue). Try returning that resistor to the 4.5V supply instead of ground.
  $endgroup$
  – glen_geek
  Mar 9 at 16:21
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm simulating this circuit in Micro-Cap, which is the clipping stage of a guitar effect. The opamp model is the "NE-5532"

I want to measure the input and the output impedance. I expected an output impedance closer to zero Ohm, and an input impedance of about 10kOhm, with an "infinite" impedance at 0Hz due to the decoupling capacitor at the input.

Here it is the analysis in Micro-Cap.

As you can see the input impedance (the blue graph) is close to what i expected, but the red graph, which is the output impedance, it's really strange. It's almost 10kOhm, with a peak of almost 1MegOhm, and i can't really explain why.
If i switch the model to a "LF-155" i get a more "reasonable" results, with an output impedance of 1.680E-68 Ohm, which is also strange.

Can you help me? This thing is driving me crazy.

operational-amplifier impedance spice input-impedance single-supply-op-amp

share|improve this question

edited Mar 9 at 15:40

RawCode

asked Mar 9 at 15:31

RawCodeRawCode

194

New contributor

RawCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited Mar 9 at 15:40

RawCode

asked Mar 9 at 15:31

RawCodeRawCode

194

New contributor

RawCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited Mar 9 at 15:40

RawCode

asked Mar 9 at 15:31

RawCodeRawCode

194

New contributor

RawCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 15:31

RawCodeRawCode

194

New contributor

RawCode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 15:31

RawCodeRawCode

194

3 Answers
3

active

oldest

votes

7

$begingroup$

In comments you added this information,

this is a single run of the ac analysis

This method won't allow you to measure the input or output (especially the output) impedance accurately.

You need to test the output impedance by applying a source to the output with the input zero'd and vice versa. You will need two separate runs of the simulator to do this.

share|improve this answer

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You saved my day!
  $endgroup$
  – RawCode
  Mar 9 at 16:27
  

  
  
  
  

add a comment | 

0

$begingroup$

Another thing you should keep in mind, is that both input and output impedances are defined for small signal operation.
In this case, the circuit makes use of the rectifying properties of the diodes to clip the signal.
When you run a AC analysis, spice calculates the small signal model of every non linear component and proceeds as if they were linear using said model.
But in reality you'd be expecting a non linear behavior.
I encourage you to run a transient analysis with a sine wave (for a specific frequency) and compare both results

share|improve this answer

answered Mar 10 at 1:17

FrancoFranco

1

New contributor

Franco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

The opamp will have an open-loop rising Zout, looking inductive. Again, this is the OPEN LOOP Zout.

What happens with an inductor in a feedback loop? depends on the presence of capacitors and dampening resistors.

====== Here is what happens to a 45MHz OpAmp, closed-loop gain of 26dB (20X), and loaded by 1uF (1,000 nanoFarad) capacitor

Why this ringing (oscillation!) frequency, of 120,000Hz? Can we predict this?

Consider the OpAmp's R_out is 25 ohms. And Unity Gain Bandwidth is 45MHz. What inductance will have j25 ohms at 45Mhz?

1nanoHenry at 1GigaHertz is j6.3 ohms. Thus 1uH at 1GHz is j6,300 ohms. At 50MHz, the 1uH is 6,200 / 20 = j310 ohms. Our opamp, with its rising-with-frequency Zout, looks like 25ohms at 45MHz, or about 0.1uH.

Now, what is the Fring of 0.1uF and 1uF? 1uH and 1uF ring at 0.16MHz. Open Loop.

This circuit has ClosedLoop gain of 20X (26dB). With F3dB at 4.5MHz. What does this tell us?

{more later}

share|improve this answer

edited yesterday

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719

$endgroup$

add a comment | 

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7

$begingroup$

In comments you added this information,

this is a single run of the ac analysis

This method won't allow you to measure the input or output (especially the output) impedance accurately.

You need to test the output impedance by applying a source to the output with the input zero'd and vice versa. You will need two separate runs of the simulator to do this.

share|improve this answer

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You saved my day!
  $endgroup$
  – RawCode
  Mar 9 at 16:27
  

  
  
  
  

add a comment | 

7

$begingroup$

In comments you added this information,

this is a single run of the ac analysis

This method won't allow you to measure the input or output (especially the output) impedance accurately.

You need to test the output impedance by applying a source to the output with the input zero'd and vice versa. You will need two separate runs of the simulator to do this.

share|improve this answer

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You saved my day!
  $endgroup$
  – RawCode
  Mar 9 at 16:27
  

  
  
  
  

add a comment | 

$begingroup$

In comments you added this information,

this is a single run of the ac analysis

This method won't allow you to measure the input or output (especially the output) impedance accurately.

You need to test the output impedance by applying a source to the output with the input zero'd and vice versa. You will need two separate runs of the simulator to do this.

share|improve this answer

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

$endgroup$

share|improve this answer

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

answered Mar 9 at 16:24

The PhotonThe Photon

86.1k398198

0

$begingroup$

Another thing you should keep in mind, is that both input and output impedances are defined for small signal operation.
In this case, the circuit makes use of the rectifying properties of the diodes to clip the signal.
When you run a AC analysis, spice calculates the small signal model of every non linear component and proceeds as if they were linear using said model.
But in reality you'd be expecting a non linear behavior.
I encourage you to run a transient analysis with a sine wave (for a specific frequency) and compare both results

share|improve this answer

answered Mar 10 at 1:17

FrancoFranco

1

New contributor

Franco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

Another thing you should keep in mind, is that both input and output impedances are defined for small signal operation.
In this case, the circuit makes use of the rectifying properties of the diodes to clip the signal.
When you run a AC analysis, spice calculates the small signal model of every non linear component and proceeds as if they were linear using said model.
But in reality you'd be expecting a non linear behavior.
I encourage you to run a transient analysis with a sine wave (for a specific frequency) and compare both results

share|improve this answer

answered Mar 10 at 1:17

FrancoFranco

1

New contributor

Franco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Another thing you should keep in mind, is that both input and output impedances are defined for small signal operation.
In this case, the circuit makes use of the rectifying properties of the diodes to clip the signal.
When you run a AC analysis, spice calculates the small signal model of every non linear component and proceeds as if they were linear using said model.
But in reality you'd be expecting a non linear behavior.
I encourage you to run a transient analysis with a sine wave (for a specific frequency) and compare both results

share|improve this answer

answered Mar 10 at 1:17

FrancoFranco

1

New contributor

Franco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this answer

answered Mar 10 at 1:17

FrancoFranco

1

New contributor

Franco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 10 at 1:17

FrancoFranco

1

New contributor

Franco is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 10 at 1:17

FrancoFranco

1

0

$begingroup$

The opamp will have an open-loop rising Zout, looking inductive. Again, this is the OPEN LOOP Zout.

What happens with an inductor in a feedback loop? depends on the presence of capacitors and dampening resistors.

====== Here is what happens to a 45MHz OpAmp, closed-loop gain of 26dB (20X), and loaded by 1uF (1,000 nanoFarad) capacitor

Why this ringing (oscillation!) frequency, of 120,000Hz? Can we predict this?

Consider the OpAmp's R_out is 25 ohms. And Unity Gain Bandwidth is 45MHz. What inductance will have j25 ohms at 45Mhz?

1nanoHenry at 1GigaHertz is j6.3 ohms. Thus 1uH at 1GHz is j6,300 ohms. At 50MHz, the 1uH is 6,200 / 20 = j310 ohms. Our opamp, with its rising-with-frequency Zout, looks like 25ohms at 45MHz, or about 0.1uH.

Now, what is the Fring of 0.1uF and 1uF? 1uH and 1uF ring at 0.16MHz. Open Loop.

This circuit has ClosedLoop gain of 20X (26dB). With F3dB at 4.5MHz. What does this tell us?

{more later}

share|improve this answer

edited yesterday

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719

$endgroup$

add a comment | 

0

$begingroup$

The opamp will have an open-loop rising Zout, looking inductive. Again, this is the OPEN LOOP Zout.

What happens with an inductor in a feedback loop? depends on the presence of capacitors and dampening resistors.

====== Here is what happens to a 45MHz OpAmp, closed-loop gain of 26dB (20X), and loaded by 1uF (1,000 nanoFarad) capacitor

Why this ringing (oscillation!) frequency, of 120,000Hz? Can we predict this?

Consider the OpAmp's R_out is 25 ohms. And Unity Gain Bandwidth is 45MHz. What inductance will have j25 ohms at 45Mhz?

1nanoHenry at 1GigaHertz is j6.3 ohms. Thus 1uH at 1GHz is j6,300 ohms. At 50MHz, the 1uH is 6,200 / 20 = j310 ohms. Our opamp, with its rising-with-frequency Zout, looks like 25ohms at 45MHz, or about 0.1uH.

Now, what is the Fring of 0.1uF and 1uF? 1uH and 1uF ring at 0.16MHz. Open Loop.

This circuit has ClosedLoop gain of 20X (26dB). With F3dB at 4.5MHz. What does this tell us?

{more later}

share|improve this answer

edited yesterday

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719

$endgroup$

add a comment | 

$begingroup$

The opamp will have an open-loop rising Zout, looking inductive. Again, this is the OPEN LOOP Zout.

What happens with an inductor in a feedback loop? depends on the presence of capacitors and dampening resistors.

====== Here is what happens to a 45MHz OpAmp, closed-loop gain of 26dB (20X), and loaded by 1uF (1,000 nanoFarad) capacitor

Why this ringing (oscillation!) frequency, of 120,000Hz? Can we predict this?

Consider the OpAmp's R_out is 25 ohms. And Unity Gain Bandwidth is 45MHz. What inductance will have j25 ohms at 45Mhz?

1nanoHenry at 1GigaHertz is j6.3 ohms. Thus 1uH at 1GHz is j6,300 ohms. At 50MHz, the 1uH is 6,200 / 20 = j310 ohms. Our opamp, with its rising-with-frequency Zout, looks like 25ohms at 45MHz, or about 0.1uH.

Now, what is the Fring of 0.1uF and 1uF? 1uH and 1uF ring at 0.16MHz. Open Loop.

This circuit has ClosedLoop gain of 20X (26dB). With F3dB at 4.5MHz. What does this tell us?

{more later}

share|improve this answer

edited yesterday

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719

$endgroup$

share|improve this answer

edited yesterday

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719

answered Mar 10 at 2:35

analogsystemsrfanalogsystemsrf

15.1k2719