verify the negative impedance converter

[ws0822]

this is the circuit for the negative impedance converter:
**broken link removed**

how to verify that the circuit? can i just use different voltage to test the circuit and measure the current then plot the graph to so than the V-I graph having negative gradient (negative resistance) to verify it?? can anyone suggests any better method??

 

[BrownOut]

What do you mean by verify? Do you need to actually measure values, or do you only need to prove the negative resistance conversion, which can be done by analysis.

 

[MrAl]

this is the circuit for the negative impedance converter:
**broken link removed**

how to verify that the circuit? can i just use different voltage to test the circuit and measure the current then plot the graph to so than the V-I graph having negative gradient (negative resistance) to verify it?? can anyone suggests any better method??

Hello,

I guess you are going through a breadboard circuit verification? Then the answer is "yes", you can change Vs and plot Is vs Vs and note that for positive Vs you get negative Is.
For example, if R1=R2 and are reasonable values, and R3=1k, then:
for Vs=1v you should measure Is= -1ma, and for Vs=2v you should measure Is= -2ma, etc.

Note that it would also be a good idea to test for the limit points of operation. For example, with a power supply of 6v and a rail to rail output op amp and the same above values the max usable input voltage would be 3v because the gain of the circuit is 2 and that means the very output would go up to 6v with only 3v on the input.

 

[ws0822]

What do you mean by verify? Do you need to actually measure values, or do you only need to prove the negative resistance conversion, which can be done by analysis.

hi, I am going through a breadboard circuit verification. beside plotting graph any other method??

 

[ws0822]

Hello,

I guess you are going through a breadboard circuit verification? Then the answer is "yes", you can change Vs and plot Is vs Vs and note that for positive Vs you get negative Is.
For example, if R1=R2 and are reasonable values, and R3=1k, then:
for Vs=1v you should measure Is= -1ma, and for Vs=2v you should measure Is= -2ma, etc.

Note that it would also be a good idea to test for the limit points of operation. For example, with a power supply of 6v and a rail to rail output op amp and the same above values the max usable input voltage would be 3v because the gain of the circuit is 2 and that means the very output would go up to 6v with only 3v on the input.

hi, thx for the reply... i can get what u r trying to say... besides that any other limitation??

 

[BrownOut]

Nope. That's the way it is done. You might also plot the output voltage of the opamp, and use Is=Vs-Vout-amp to gain some insight to how the circuit works.

You can also try running the ckt on a simulator.

 

[ws0822]

any other limitation for this circuit? can i prove the limitation with practical results ??

 

[MrAl]

hi, thx for the reply... i can get what u r trying to say... besides that any other limitation??

Hi again,

Resistor values can not go below a certain value or the op amp will not be able to power them (not enough output current).

 

[ws0822]

Hi again,

Resistor values can not go below a certain value or the op amp will not be able to power them (not enough output current).

thx again.... haha... ok now, if i replace the Vin with a ac sice waveform and R1 with a capacitor or inductor the frequency become the limitation right?(cannot go too high and too low)... if I do that how can I verify the circuit capability as a negative impedance converter with experimental results?

 

[MrAl]

Hello again,

If you replace R1 with a capacitor then you should see the input Z look like the negative of the impedance of a capacitor, multiplied by R3/R2. With R3=R2 and since the impedance of a capacitor is:
Z=1/(j*w*C)
the negative of that is:
Z(-)=-1/(j*w*C)
so you would see a phase reversal in the current relative to a real capacitor between Vin and ground. In fact, if you connect a real capacitor of the same value between Vin and ground you can compare current phases.
With R2 and R3 different the impedance would simply be:
Z(-)=-R3/(j*w*C*R2)
and since R2 is in the denominator making R2 larger than R3 will make the capacitor look bigger.

You do have to watch your output to make sure it doesnt go too close to the either power supply rail. This means you may have to choose your cap value and test frequency carefully.

 

[ws0822]

Hello again,

If you replace R1 with a capacitor then you should see the input Z look like the negative of the impedance of a capacitor, multiplied by R3/R2. With R3=R2 and since the impedance of a capacitor is:
Z=1/(j*w*C)
the negative of that is:
Z(-)=-1/(j*w*C)
so you would see a phase reversal in the current relative to a real capacitor between Vin and ground. In fact, if you connect a real capacitor of the same value between Vin and ground you can compare current phases.
With R2 and R3 different the impedance would simply be:
Z(-)=-R3/(j*w*C*R2)
and since R2 is in the denominator making R2 larger than R3 will make the capacitor look bigger.

You do have to watch your output to make sure it doesnt go too close to the either power supply rail. This means you may have to choose your cap value and test frequency carefully.

hi again... i can understand the equations... and what u mean is for different capacitance different frequency will be used right?? so me must adjust the correct frequency so that the value of Vout is greater than Vin? how can i verify that the circuit can product negative capacitance if capacitor is used then compare Vin and Vout looking for 90degree phase shift??

 

[MrAl]

Hello again,

I just meant that you may have to adjust the frequency to fit the cap value so that you get normal output voltages that dont try to exceed the power supply rails.

With that negative impedance you will see a phase reversal in current relative to a real capacitor, and that means a 180 phase shift. This means the circuit input current Iin will be out of phase with the current of a real capacitor connected from Vin to ground. Thus if you call the real capacitor current 0 degrees then the input current Iin to the op amp circuit will be 180 degrees out of phase with that real cap 0 degrees current phase.

You cant compare Vin and Vout though.

Here's what you might do:

Connect a capacitor in place of R1. Adjust the signal generator voltage to 1v peak and adjust frequency until you get maybe 2.5v peak output (with plus and minus 5v supplies or higher). Turn off, then connect another cap directly from Vin to ground. Then after you turn back on, check the current of the real cap and compare that current to the input current of the op amp circuit (this will be the current through R3). Observe that the phase difference is 180 degrees (or thereabout) between the two current phases.

You might need a low impedance signal generator however. We may have to figure that impedance into the equation also if you dont have that equipment available. See that you get a nice clean sine wave voltage into the circuit.

You might start with two 10k resistors and two 0.001uf capacitors and see how it goes. Start the signal generator at say 10kHz.

If you dont want to use two capacitors then you can observe that the phase of the input current relative to Vin is 90 degrees lagging instead of 90 degrees leading.

 

[ws0822]

Hello again,

I just meant that you may have to adjust the frequency to fit the cap value so that you get normal output voltages that dont try to exceed the power supply rails.

With that negative impedance you will see a phase reversal in current relative to a real capacitor, and that means a 180 phase shift. This means the circuit input current Iin will be out of phase with the current of a real capacitor connected from Vin to ground. Thus if you call the real capacitor current 0 degrees then the input current Iin to the op amp circuit will be 180 degrees out of phase with that real cap 0 degrees current phase.

You cant compare Vin and Vout though.

Here's what you might do:

Connect a capacitor in place of R1. Adjust the signal generator voltage to 1v peak and adjust frequency until you get maybe 2.5v peak output (with plus and minus 5v supplies or higher). Turn off, then connect another cap directly from Vin to ground. Then after you turn back on, check the current of the real cap and compare that current to the input current of the op amp circuit (this will be the current through R3). Observe that the phase difference is 180 degrees (or thereabout) between the two current phases.

You might need a low impedance signal generator however. We may have to figure that impedance into the equation also if you dont have that equipment available. See that you get a nice clean sine wave voltage into the circuit.

You might start with two 10k resistors and two 0.001uf capacitors and see how it goes. Start the signal generator at say 10kHz.

If you dont want to use two capacitors then you can observe that the phase of the input current relative to Vin is 90 degrees lagging instead of 90 degrees leading.

hi again... emmm... what adjustment can make the oscilloscope measure the current waveform?? my lecturer just teachs us use it to measure voltage only...

 

[MrAl]

Hello,

Well in that case you would have to measure the voltage across resistor R3 and use Ohms Law to calculate the current. Note that the polarity is important here. You'd have to put the probe tip at the input (at Vin which is non inverting terminal) and the ground lead at the output of the op amp, and to measure Vin you'd put the other probe tip right on Vin and the ground on circuit ground, but this would require a scope with two isolated inputs, something we dont normally have on hand.

Conceptually however, that is what we are after, but to do the measurement we would have to put the ground at the top of Vin and one probe at the output of the op amp and the other probe tip at ground. This would give us the required waveforms, but they would both be inverted. You could imagine them being inverted when you look at the scope, but since they are both inverted the phase relationship should remain the same.

Sometimes putting a ground at the input of an op amp causes noise problems, so if that happens you will have to put one probe on Vin and one probe on the output of the op amp and ground lead on circuit ground, then subtract Vin-Vout to get vR3 and then use Ohms Law to calculate the current. If your scope has this subtraction function you can do it this way, but if it does not it will be hard to imagine the subtraction of Vin-Vout. If need be we can use another op amp to make the second measurement.

Another method for measuring current that is often used is we put a smallish value resistor in series with the path we want to measure the current in. This would mean putting a small value resistor in series with Vin and measuring the voltage across it, paying attention to polarity again. Doing it this way only causes a single inversion in the current waveform, which is easy to invert in your head or using the scope if the scope has its second channel invert function (many scopes have this function for one channel). The scope will have to have enough gain to do it this way, and hopefully it doesnt pick up too much noise which you would have to try to ignore. The resistor has to be large enough to allow proper viewing on the scope with its highest gain setting but low enough not to affect the circuit too much.

 

[ws0822]

Hello,

Well in that case you would have to measure the voltage across resistor R3 and use Ohms Law to calculate the current. Note that the polarity is important here. You'd have to put the probe tip at the input (at Vin which is non inverting terminal) and the ground lead at the output of the op amp, and to measure Vin you'd put the other probe tip right on Vin and the ground on circuit ground, but this would require a scope with two isolated inputs, something we dont normally have on hand.

Conceptually however, that is what we are after, but to do the measurement we would have to put the ground at the top of Vin and one probe at the output of the op amp and the other probe tip at ground. This would give us the required waveforms, but they would both be inverted. You could imagine them being inverted when you look at the scope, but since they are both inverted the phase relationship should remain the same.

Sometimes putting a ground at the input of an op amp causes noise problems, so if that happens you will have to put one probe on Vin and one probe on the output of the op amp and ground lead on circuit ground, then subtract Vin-Vout to get vR3 and then use Ohms Law to calculate the current. If your scope has this subtraction function you can do it this way, but if it does not it will be hard to imagine the subtraction of Vin-Vout. If need be we can use another op amp to make the second measurement.

Another method for measuring current that is often used is we put a smallish value resistor in series with the path we want to measure the current in. This would mean putting a small value resistor in series with Vin and measuring the voltage across it, paying attention to polarity again. Doing it this way only causes a single inversion in the current waveform, which is easy to invert in your head or using the scope if the scope has its second channel invert function (many scopes have this function for one channel). The scope will have to have enough gain to do it this way, and hopefully it doesnt pick up too much noise which you would have to try to ignore. The resistor has to be large enough to allow proper viewing on the scope with its highest gain setting but low enough not to affect the circuit too much.

ok.. thx.. i can get it... btw, help me to check this circuit capability...**broken link removed**

is it a voltage dependant current source??

 

[MrAl]

ok.. thx.. i can get it... btw, help me to check this circuit capability...**broken link removed**

is it a voltage dependant current source??

Hello again,

It does look like that, however there are some limitations but also something interesting.

The static equation for vRL (voltage across RL) is:
vRL=(Vin*R2*R4*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)
and then of course that means that iRL is:
IL=vRL/RL

The limitation, along with the limited Vout again, is:
k*dVout^2/d^2t=-((R2*R4*RL-R1*R3*RL+R1*R2*R4))
or written more simply:
K= -(R2*R4*RL-R1*R3*RL+R1*R2*R4)

and if K<0 then the circuit is stable but if K>=0 then the circuit latches up.

But the good news is that because of the equation for vRL (the voltage across RL):
vRL=(Vin*R2*R4*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)

we can note that we have two terms in the denominator that both contain R1, and one is negative and one is positive. This means with the proper selection of R3 and R4 (and RL) we can make the circuit independent from R1 so that R1 can be just about any reasonable value!
This is a good thing, because that means if R1 is some output from some other circuit it doesnt matter what output impedance it is, this circuit will not load it at all. That means the input current through R1 is negligible. Pretty cool huh?
If we look at the denominator alone:
D=R2*R4*RL-R1*R3*RL+R1*R2*R4
we have a couple choices with the main goal is to get:
D=R2*R4*RL
or equivalently make this equal zero:
R1*R2*R4 -R1*R3*RL=0
We can then factor out R1:
R1(R2*R4 -R3*RL)=0
and we are left with:
R2*R4 -R3*RL=0
so the goal now is simply to get R2 times R4 equal to R3 times RL:
R2*R4 =R3*RL
Since there are a couple ways to do this i'll leave it up to you to figure out what you would like to do.
Note that once this condition is met we get a simpler vRL:
vRL=Vin*RL/RL=Vin
and of course that means that IL is easy to calculate:
IL=Vin/RL

Also you can make a check on the output voltage with this:
Vout=(Vin*R2*(R4+R3)*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)
just to make sure Vout stays reasonable.

 

[ws0822]

Hello again,

It does look like that, however there are some limitations but also something interesting.

The static equation for vRL (voltage across RL) is:
vRL=(Vin*R2*R4*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)
and then of course that means that iRL is:
IL=vRL/RL

The limitation, along with the limited Vout again, is:
k*dVout^2/d^2t=-((R2*R4*RL-R1*R3*RL+R1*R2*R4))
or written more simply:
K= -(R2*R4*RL-R1*R3*RL+R1*R2*R4)

and if K<0 then the circuit is stable but if K>=0 then the circuit latches up.

But the good news is that because of the equation for vRL (the voltage across RL):
vRL=(Vin*R2*R4*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)

we can note that we have two terms in the denominator that both contain R1, and one is negative and one is positive. This means with the proper selection of R3 and R4 (and RL) we can make the circuit independent from R1 so that R1 can be just about any reasonable value!
This is a good thing, because that means if R1 is some output from some other circuit it doesnt matter what output impedance it is, this circuit will not load it at all. That means the input current through R1 is negligible. Pretty cool huh?
If we look at the denominator alone:
D=R2*R4*RL-R1*R3*RL+R1*R2*R4
we have a couple choices with the main goal is to get:
D=R2*R4*RL
or equivalently make this equal zero:
R1*R2*R4 -R1*R3*RL=0
We can then factor out R1:
R1(R2*R4 -R3*RL)=0
and we are left with:
R2*R4 -R3*RL=0
so the goal now is simply to get R2 times R4 equal to R3 times RL:
R2*R4 =R3*RL
Since there are a couple ways to do this i'll leave it up to you to figure out what you would like to do.
Note that once this condition is met we get a simpler vRL:
vRL=Vin*RL/RL=Vin
and of course that means that IL is easy to calculate:
IL=Vin/RL

Also you can make a check on the output voltage with this:
Vout=(Vin*R2*(R4+R3)*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)
just to make sure Vout stays reasonable.

wa... so amazing... thx for ur help... it really help me a lot... i appreciate it very much...thx

 

[The Electrician]

Hello again,

It does look like that, however there are some limitations but also something interesting.

The static equation for vRL (voltage across RL) is:
vRL=(Vin*R2*R4*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)
and then of course that means that iRL is:
IL=vRL/RL

The limitation, along with the limited Vout again, is:
k*dVout^2/d^2t=-((R2*R4*RL-R1*R3*RL+R1*R2*R4))
or written more simply:
K= -(R2*R4*RL-R1*R3*RL+R1*R2*R4)

and if K<0 then the circuit is stable but if K>=0 then the circuit latches up.

But the good news is that because of the equation for vRL (the voltage across RL):
vRL=(Vin*R2*R4*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)

we can note that we have two terms in the denominator that both contain R1, and one is negative and one is positive. This means with the proper selection of R3 and R4 (and RL) we can make the circuit independent from R1 so that R1 can be just about any reasonable value!
This is a good thing, because that means if R1 is some output from some other circuit it doesnt matter what output impedance it is, this circuit will not load it at all. That means the input current through R1 is negligible. Pretty cool huh?
If we look at the denominator alone:
D=R2*R4*RL-R1*R3*RL+R1*R2*R4
we have a couple choices with the main goal is to get:
D=R2*R4*RL
or equivalently make this equal zero:
R1*R2*R4 -R1*R3*RL=0
We can then factor out R1:
R1(R2*R4 -R3*RL)=0
and we are left with:
R2*R4 -R3*RL=0
so the goal now is simply to get R2 times R4 equal to R3 times RL:
R2*R4 =R3*RL
Since there are a couple ways to do this i'll leave it up to you to figure out what you would like to do.
Note that once this condition is met we get a simpler vRL:
vRL=Vin*RL/RL=Vin
and of course that means that IL is easy to calculate:
IL=Vin/RL

Also you can make a check on the output voltage with this:
Vout=(Vin*R2*(R4+R3)*RL)/(R2*R4*RL-R1*R3*RL+R1*R2*R4)
just to make sure Vout stays reasonable.

This analysis doesn't fully respond to the OP's question:"is it a voltage dependant current source??"

The full expression for the current in RL is:

[latex]I_L = Vin\frac{R_2 R_4}{R_2 R_4 R_L - R_1 R_3 R_L + R_1 R_2 R_4}[/latex]

Rewriting like this:

[latex]I_L = Vin\frac{R_2 R_4}{R_2 R_4 R_L + R_1 (R_2 R_4 - R_3 R_L)}[/latex]

we can see a rationale for making R2*R4 -R3*RL=0; then the relationship between Vin and IL is:

[latex]I_L = \frac{Vin}{R_L}[/latex]

IL is inversely proportional to the value of RL. While it is true that this "...means if R1 is some output from some other circuit it doesnt matter what output impedance it is, this circuit will not load it at all. That means the input current through R1 is negligible." and such behavior has its uses, for this choice of R2, R3, R4 and RL the circuit is not behaving as a voltage controlled current source.

If the circuit is to be a voltage controlled current source then the current IL must be independent of the value of RL. Let's rewrite the full expression for IL like this:

[latex]I_L = Vin\frac{R_2 R_4}{R_1 R_2 R_4 + R_L(R_2 R_4 - R_1 R_3)}[/latex]

If we let R2*R4-R1*R3=0, then the relationship between Vin and IL is:

[latex]I_L = \frac{Vin}{R_1}[/latex]

Now the current in RL is independent of the value of RL; this is the behavior of a current source driving RL.

Of course, for proper operation, the opamp's output must not saturate.

 

[MrAl]

This analysis doesn't fully respond to the OP's question:"is it a voltage dependant current source??"

The full expression for the current in RL is:

[latex]I_L = Vin\frac{R_2 R_4}{R_2 R_4 R_L - R_1 R_3 R_L + R_1 R_2 R_4}[/latex]

Rewriting like this:

[latex]I_L = Vin\frac{R_2 R_4}{R_2 R_4 R_L + R_1 (R_2 R_4 - R_3 R_L)}[/latex]

we can see a rationale for making R2*R4 -R3*RL=0; then the relationship between Vin and IL is:

[latex]I_L = \frac{Vin}{R_L}[/latex]

IL is inversely proportional to the value of RL. While it is true that this "...means if R1 is some output from some other circuit it doesnt matter what output impedance it is, this circuit will not load it at all. That means the input current through R1 is negligible." and such behavior has its uses, for this choice of R2, R3, R4 and RL the circuit is not behaving as a voltage controlled current source.

If the circuit is to be a voltage controlled current source then the current IL must be independent of the value of RL. Let's rewrite the full expression for IL like this:

[latex]I_L = Vin\frac{R_2 R_4}{R_1 R_2 R_4 + R_L(R_2 R_4 - R_1 R_3)}[/latex]

If we let R2*R4-R1*R3=0, then the relationship between Vin and IL is:

[latex]I_L = \frac{Vin}{R_1}[/latex]

Now the current in RL is independent of the value of RL; this is the behavior of a current source driving RL.

Of course, for proper operation, the opamp's output must not saturate.

Hello,

So you are saying that it doesnt fully respond to the OP's question because there were no parentheses in the denominator? I think that is very strange of you to say especially since R1 was factored out in later lines and the goal clearly pointed out. Perhaps you should re read my previous post.

 

[MrAl]

wa... so amazing... thx for ur help... it really help me a lot... i appreciate it very much...thx

Hello again,


Hey that's ok, no problem, you're welcome.