verify the negative impedance converter

MrAl said:

Hello,

So you are saying that it doesnt fully respond to the OP's question because there were no parentheses in the denominator?

Click to expand...

It wasn't because of "no parentheses" in the denominator. You said:

"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"

You then made D=R2*R4*RL, showing your choice with parentheses. You said:

"We can then factor out R1:
R1(R2*R4 -R3*RL)=0
and we are left with:
R2*R4 -R3*RL=0"

That choice leads to a relationship between Vin and IL, namely "IL=Vin/RL". This condition (R2*R4 -R3*RL=0) provides that for a fixed Vin, the current IL depends on the value of RL. This is not how a current source driving RL should behave.

The OP asked if the circuit was a "voltage dependant current source" and you didn't show the condition (R2*R4-R1*R3=0) for the current in RL to be independent of the value of RL. For this condition the relationship IL = Vin/R1 obtains; IL does not depend on the value of RL.

Also, for that independence to hold, given an ideal opamp and R2*R4-R1*R3=0, if we remove RL and calculate the driving point impedance at the node which drives RL (with the Vin node grounded), that impedance should be infinite--the output impedance of a current source.

The condition you analyzed (R2*R4 -R3*RL=0), while interesting in its own right, does not provide an (ideally) infinite impedance at the node which drives RL when RL is removed. For a fixed Vin the current in RL is not independent of the value of RL; it varies inversely as the value of RL.

The OP asked if the circuit was a "voltage dependant current source"; it seemed relevant to point out the condition which would make it so, which you did not do. That's what I was referring to when I said that you did not "fully respond" to the OP's question.

MrAl said:

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.

Click to expand...

It still reads the same as when I first read it.

Hello again,


I see now what you are trying to get at, but you do have to realize that this circuit is never going to be a true voltage controlled current source, only a pseudo one because there will always be one trade off for another...you can't have both.
I assumed that if someone saw RL in the equation they would know right away that RL had to be constant, otherwise that condition could not be met. That should be obvious from the equation and also the circuit was drawn with a constant RL.
If you do want to impose a variable RL then we do have to impose different requirements, that's should be a given. No problem with that as far as i can see, except as the explained above trade off which may or may not be significant either way.

I have provided plenty of groundwork in my previous posts if you would like to take it farther that's fine with me.

The Electrician said:

It wasn't because of "no parentheses" in the denominator. You said:

"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"

You then made D=R2*R4*RL, showing your choice with parentheses. You said:

"We can then factor out R1:
R1(R2*R4 -R3*RL)=0
and we are left with:
R2*R4 -R3*RL=0"

That choice leads to a relationship between Vin and IL, namely "IL=Vin/RL". This condition (R2*R4 -R3*RL=0) provides that for a fixed Vin, the current IL depends on the value of RL. This is not how a current source driving RL should behave.

The OP asked if the circuit was a "voltage dependant current source" and you didn't show the condition (R2*R4-R1*R3=0) for the current in RL to be independent of the value of RL. For this condition the relationship IL = Vin/R1 obtains; IL does not depend on the value of RL.

Also, for that independence to hold, given an ideal opamp and R2*R4-R1*R3=0, if we remove RL and calculate the driving point impedance at the node which drives RL (with the Vin node grounded), that impedance should be infinite--the output impedance of a current source.

The condition you analyzed (R2*R4 -R3*RL=0), while interesting in its own right, does not provide an (ideally) infinite impedance at the node which drives RL when RL is removed. For a fixed Vin the current in RL is not independent of the value of RL; it varies inversely as the value of RL.

The OP asked if the circuit was a "voltage dependant current source"; it seemed relevant to point out the condition which would make it so, which you did not do. That's what I was referring to when I said that you did not "fully respond" to the OP's question.




It still reads the same as when I first read it.

Click to expand...

MrAl said:

Hello again,


I see now what you are trying to get at, but you do have to realize that this circuit is never going to be a true voltage controlled current source, only a pseudo one because there will always be one trade off for another...you can't have both.
I assumed that if someone saw RL in the equation they would know right away that RL had to be constant, otherwise that condition could not be met. That should be obvious from the equation and also the circuit was drawn with a constant RL.
If you do want to impose a variable RL then we do have to impose different requirements, that's should be a given. No problem with that as far as i can see, except as the explained above trade off which may or may not be significant either way.

I have provided plenty of groundwork in my previous posts if you would like to take it farther that's fine with me.

Click to expand...

hi guys,thx for u guys helps... the previous assignment i done edi. With MrAl hints, i finally get a formula of IL= Vin/R1 when R2/R3=R1/R4 in order to become a VDCS.
And now, maybe i need helps again... this time i face with this gyrator circuit ...**broken link removed**

i have verified that the Zin = r^2/Z where r is the gyration resistance.... according to the formula, the circuit can produce a inverse impedance... I don have problem with the circuit analysis part but when come to practical part which is to verify the circuit capability i don have any idea to prove that this circuit can product a inverse impedance with experimental result... can anyone gv me some idea about this? thx...

Hello,

What does the phrase, "inverse impedance" mean to you? That means if you have Z and you have 1/Z when you multiply them together you get 1.
Since your circuit obviously does R^2/Z rather than 1/Z then you have to figure that R^2 in with it.

For example, with R=1k then R^2 is 1000k (1 meg), and if Z=100 then we have:
Zin=R^2/Z=1000000/100=10000, so with a 1v input source you should see 1/10000 amps flowing from the Vin source.

For Z either capacitance or inductance, you know that the cap is 1/(s*C) and the inductor is s*L. Notice that if you replace C with L or L with C you get the inverse of the other. Can you figure out what to do with Z being a cap or inductor now?
For more complex impedances (more elements involved in the impedance) you would have to do a little more work to figure out what to look for.

Remember when dealing with real life circuits like this to always check your limits. Output voltage, output current, etc. If the op amp can not handle some component values or the error introduced by a real life op amp is too high compared to the normal current or voltage values then the circuit will not behave nearly close to the theoretical and you might see wild results that appear to have nothing to do with the intended circuit. This is where lab experience or even careful spice analysis can be very valuable.

MrAl said:

Hello,

What does the phrase, "inverse impedance" mean to you? That means if you have Z and you have 1/Z when you multiply them together you get 1.
Since your circuit obviously does R^2/Z rather than 1/Z then you have to figure that R^2 in with it.

For example, with R=1k then R^2 is 1000k (1 meg), and if Z=100 then we have:
Zin=R^2/Z=1000000/100=10000, so with a 1v input source you should see 1/10000 amps flowing from the Vin source.

For Z either capacitance or inductance, you know that the cap is 1/(s*C) and the inductor is s*L. Notice that if you replace C with L or L with C you get the inverse of the other. Can you figure out what to do with Z being a cap or inductor now?
For more complex impedances (more elements involved in the impedance) you would have to do a little more work to figure out what to look for.

Remember when dealing with real life circuits like this to always check your limits. Output voltage, output current, etc. If the op amp can not handle some component values or the error introduced by a real life op amp is too high compared to the normal current or voltage values then the circuit will not behave nearly close to the theoretical and you might see wild results that appear to have nothing to do with the intended circuit. This is where lab experience or even careful spice analysis can be very valuable.

Click to expand...

if Z being a cap or inductor , what i have to do is what u mention before just compare the phase shift of Vin and Iin right??
what u mean by the complex impedances is z include the combination of RLC??

maybe this will help...
i've seen NICs in use, and for a specific purpose. i've seen it in use in an audio preamp. if the output impedance of a preamp is1k, for instance, and a number of inputs are fed with this output, the output level drops with each added load impedance. if the output is loaded with a 1k load, the output signal level will be 1/2 of what it was unloaded. to minimize this effect, an NIC is added to boost the output level as the load gets lower in impedance. this in effect lowers the output impedance of the preamp without altering the preamp itself. so one way to "verify" the operation of an NIC is to a) treat it as a two terminal "black box" that will eventually be connected across Zl (load impedance), b) treat the signal source as a pure signal generator in series with the source impedance. first you would measure the output voltage unloaded, then measure the output voltage with Zl connected. the voltage will be lower, because the source impedance and load impedance act as a voltage divider. next connect the NIC across Zl, and the effect of Zl is counteracted by the action of the NIC, and the output voltage will be closer to the open circuit value. if, as in the example somebody gave above, Zl is capacitive, the effect of the NIC will be inductive (which might not be such a good idea, because at some frequency Xc (capacitive reactance, or the "impedance" of the capacitor) and -Z(NIC) will be of equal magnitude, but opposite phase, and will likely become an oscillator... not exactly the kind of condition you want at the output of a preamp.

ws0822 said:

if Z being a cap or inductor , what i have to do is what u mention before just compare the phase shift of Vin and Iin right??
what u mean by the complex impedances is z include the combination of RLC??

Click to expand...

If Z is a capacitor you might try to resonate the inductive reactance at the input with another real capacitor. Measure the resonance frequency and see if the apparent value of the simulated inductance is what you expect.

The gyrator circuit you have here is more complex than it needs to be. See:

The Gyrator

ws0822 said:

if Z being a cap or inductor , what i have to do is what u mention before just compare the phase shift of Vin and Iin right??
what u mean by the complex impedances is z include the combination of RLC??

Click to expand...

Hi again,

Yes that is more or less correct, and you'll see the input looking like a capacitor to ground with an inductor for Z and the input looking like an inductor to ground for Z being a capacitor. That is probably very easy to verify and you shouldnt get into problems. I will caution you however that i have not attempted to check out the exact operation with a capacitor or inductor, it's all purely theoretical at this point.
There are ways to make the impedance float too (without using floating supplies) but i'd have to think back a long time ago.

By complex impedances, i meant just that yes, a combination of R and C, R and L, or R,L, and C too as you mentioned. You can choose R such that you get a damped response to make it easier.

As "The Electrician" pointed out, there are other circuits that are simpler for making gyros. There are single stage op amp circuits too that only use one section of the op amp instead of two. They are based on:
V=L*di/dt
I=C*dv/dt
and the idea is to interchange the function of voltage and current rather than do the actual inverse impedance as 1/(s*L) and s*C.
So there are at least two ways to make gyrators that i know of at least that work with inductors and capacitors

These things were more popular a long time ago.

i have a question, is it the ground of the oscilloscope probe is shorted 2getther??
Let take an example,
**broken link removed**

For the circuit above, if i have 2 probe for the probe1 i put the ground at point B and probe tip at point A and probe2 ground at circuit ground and tip at point B then will short circuit R2??
so, the current will flow through R1 and then to the ground of the probe1 then directly to the negative terminal of the source and cause the reading or probe1 become 12V??

oscope inputs are usually 1Meg input impedance, so no, the probes will not act as a short but would begin to have an effect on the circuit if your resistors were in the 100k-10Meg range . x10 probes are 10Meg impedance, and have even less effect on the circuit. the scope channels use separate diff amp inputs and do not interact.

actually, gyrators are still used a lot. the most common use is in audio graphic EQ devices, where a filter is needed for each frequency band. since inductors are large and expensive, it's much easier and compact to use a gyrator in place of inductors for each filter. of course a some audio processing equipment is now being implemented with DSP chips, and the filters are implemented in software, but most graphic EQs on the market are still analog and use gyrators. gyrators are also used in a lot of RF equipment too, since an amp and a couple resistors and capacitors are cheaper than the copper coil they replace, at least in low-level signal work up to about 50Mhz. with the newer op amp technology which can produce op amps with 300-500Mhz bandwidth, expect the use of gyrators to increase.

ws0822 said:

i have a question, is it the ground of the oscilloscope probe is shorted 2getther??
Let take an example,
**broken link removed**

For the circuit above, if i have 2 probe for the probe1 i put the ground at point B and probe tip at point A and probe2 ground at circuit ground and tip at point B then will short circuit R2??
so, the current will flow through R1 and then to the ground of the probe1 then directly to the negative terminal of the source and cause the reading or probe1 become 12V??

Click to expand...

Hello again,


Yes! Well, at least in most scopes unless they specifically specify that the grounds are isolated. You can easily check your scope with an ohm meter.

unclejed613 said:

actually, gyrators are still used a lot. the most common use is in audio graphic EQ devices, where a filter is needed for each frequency band. since inductors are large and expensive, it's much easier and compact to use a gyrator in place of inductors for each filter. of course a some audio processing equipment is now being implemented with DSP chips, and the filters are implemented in software, but most graphic EQs on the market are still analog and use gyrators. gyrators are also used in a lot of RF equipment too, since an amp and a couple resistors and capacitors are cheaper than the copper coil they replace, at least in low-level signal work up to about 50Mhz. with the newer op amp technology which can produce op amps with 300-500Mhz bandwidth, expect the use of gyrators to increase.

Click to expand...

Hello,

Oh still used a lot in audio equalizers? That's quite interesting really. Perhaps you can show a schematic for one with a gyrator, i wouldnt mind taking a look at one.

Inductors large and expensive for equalizers? Are we talking about the typical equalizer built into some amplifiers or are we talking about an equalizer that has to connect to the output of a somewhat higher powered amplifier and drive the speaker directly?

for instance a typical 10-band eq is shown below...
this is a common design and still in use

Hello again,

Thanks for the pic. I'll have to take a look at it soon. Looks nice the way they did it. Would you happen to know the year in which it was designed?

Here you have some info abut this active inductor circuit, use in equalizer.
https://www.electro-tech-online.com/custompdfs/2011/02/avt08063.pdf
https://www.linkwitzlab.com/images/graphics/inductr1.gif
https://www.linkwitzlab.com/images/graphics/inductr2.gif

MrAl said:

Hello again,

Thanks for the pic. I'll have to take a look at it soon. Looks nice the way they did it. Would you happen to know the year in which it was designed?

Click to expand...

looks like 2001 according to the date in the lower right...

a gyrator is a "special case" of the NIC. the "generic" NIC is a negative impedance at all frequencies, while a gyrator is actually a "negative capacitance".

i think what the OP was looking for was how the NIC behaves, and all the gyrator info was an unintended rabbit trail.

the usual way of measuring the output impedance of a signal source is to measure the open circuit voltage (Vo), then measure the voltage with a load (Vl). the difference in voltages and the known load impedance can be calculated to find the source impedance Zs=Zl((Vo/Vl)-1). a NIC across the source adds a negative Z to the source impedance and lowers the total source impedance. it acts as a kind of "helper" amplifier in parallel with the source.

unclejed613 said:

looks like 2001 according to the date in the lower right...

a gyrator is a "special case" of the NIC. the "generic" NIC is a negative impedance at all frequencies, while a gyrator is actually a "negative capacitance".

i think what the OP was looking for was how the NIC behaves, and all the gyrator info was an unintended rabbit trail.

the usual way of measuring the output impedance of a signal source is to measure the open circuit voltage (Vo), then measure the voltage with a load (Vl). the difference in voltages and the known load impedance can be calculated to find the source impedance Zs=Zl((Vo/Vl)-1). a NIC across the source adds a negative Z to the source impedance and lowers the total source impedance. it acts as a kind of "helper" amplifier in parallel with the source.

Click to expand...

Hello again,


Hey it's nice to see someone else interested in this stuff

I didnt realize there were so many graphic equalizer schematics on the web either, and the 'gyrator' we are talking about from your schematic looks like what i remember from way back when (years and years ago ) so it was even more interesting for me to see this circuit surface again.

From the standpoint of looking for a simplified analysis, it looks like you have it well in hand. However, i prefer to take the stance that the op amp with the two resistors and one lower capacitor constitutes a simulated inductor with series resistance, and seeing it in series with the second (upper) capacitor, that constitutes a tuned RLC circuit who's parameters can be easily calculated and thus make the analysis short and sweet
In fact, that op amp, two resistors, and one lower cap, come very close to a true RL circuit with the error being greatest at the higher end of the audio spectrum (where we get to 20kHz) and then the denominator starts to reduce the gain, but only by about 0.77 and so the impedance of that part of the circuit looks very close to R+s*L, the exact same impedance of an RL series circuit. For the lower frequencies it is so close to an equivalent RL series circuit that the error is way down in the thousandths of 1 percent or so.
In fact, if we call the feedback resistor R2 and the lower resistor R1 and the lower capacitor C1, we get an impedance very close to this:
Z=R2+s*C1*R1*R2
where we can immediately spot that R2 is the series resistance and C1*R1*R2 is the equivalent inductance.
That makes calculating the center frequency very simple as usual:
Fc=1/(2*pi*sqrt(L_eqiv*C_upper))

I like to look at it that way because it makes the circuit look just like an equalizer with a true inductor with series resistance in it.

Very interesting if you ask me.

unclejed613 said:

a gyrator is a "special case" of the NIC. the "generic" NIC is a negative impedance at all frequencies, while a gyrator is actually a "negative capacitance".

Click to expand...

How do you figure that a gyrator is a "negative capacitance"?

unclejed613 said:

i think what the OP was looking for was how the NIC behaves, and all the gyrator info was an unintended rabbit trail.

Click to expand...

The gyrator info wasn't an "unintended rabbit trail". The OP himself asked about a gyrator in post #23.