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phase splitter issue

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Kindly justify this. What is so special about the characteristic of balance?

Consider the characteristic of voltage amplification. If I short the plate of a triode voltage amplifier to measure its short circuit output current, it is no longer an amplifier. Rather, it is a voltage nullifier. So I can't do this, right, because I'm not measuring/analysing the right circuit?

Same comment. If you want to measure source Z, it's Voc/Isc, by definition.
 
You are not addressing the question, perhaps because you did not read the full thread. The comment I responded to was something to the effect of, you can't measure the Concertina impedance by a one port measurement because you destroy its balance, and it is no longer a Concertina. Please read my reply in that context.
 
Dave, nice sims. But please consider Alfred Centauri's post # 146. If you feel you must keep the circuit balanced, then you must segregate the ip-generated portion of Vp and divide that by ip, rather than dividing the entire Vp which is contributed to by both ip and ik. If you are a calculus person, this means that it is the partial derivative of Vp with respect to ip that determines the plate impedance Zp.

Hey.

In looking at post 146 the problem I have is the phase of the current supply. I did this all a few years back but if the phases are as he presents you indeed get different plate and cathode impedances, however you also get in phase sine waves at the plate and the cathode in the time domain which is decidedly not how the circuit is intended to work. This was a big red flag to my BFA degree so I looked beyond. :)

I found if you reverse the phase of one of the current supply (so the currents behave like they do in say a PP output transformer) you get the results that I feel are correct and as a bonus you also get inverted signals at the plate and cathode just like one would expect from a phase splitter.

dave
 
You are not addressing the question, perhaps because you did not read the full thread. The comment I responded to was something to the effect of, you can't measure the Concertina impedance by a one port measurement because you destroy its balance, and it is no longer a Concertina. Please read my reply in that context.

In that case, you may finally be getting it! :up:
 
The real question is not how we intend the circuit to operate, but what does the definition of impedance tell us about its ports?

I contend that there is one definition of impedance for all linear circuits. Some people might feel that it is necessary to add some additional constraints to how impedance is measured, as with the Cathodyne, that the impedance must be tested while the circuit is balanced. I do not wish to contest this at the moment, but I will say that these constraints must not violate the definition of impedance.

My definition of impedance is as follows. To measure the impedance in a circuit between nodes A and B, cause a current to flow into node A, out of node B, into some device outside of the circuit that enables the current, and back into A to close the loop. The impedance between node A and B is the change in voltage between A and B due only to that current, divided by that current. Do you agree? If not, what is your definition? Surely we shouldn't even try to talk about something until we can define it.

If you agree with my def, we must reject any voltages between A and B due to other currents that may be present. Hence Centauri's argument.

It's not about what we intend a circuit to do, but what it does when its impedance is tested that counts.
 
I hope not. I think it's catching!

Let's try it again. If I can't measure the impedance of the Cathodyne plate by loading just the plate because then it's no longer a Cathodyne, then by that logic I can't measure the impedance of a voltage amplifier output by measuring its output short circuit current, because then it's no longer a voltage amplifier.

Discuss.
 
The real question is not how we intend the circuit to operate, but what does the definition of impedance tell us about its ports?

I contend that there is one definition of impedance for all linear circuits. Some people might feel that it is necessary to add some additional constraints to how impedance is measured, as with the Cathodyne, that the impedance must be tested while the circuit is balanced. I do not wish to contest this at the moment, but I will say that these constraints must not violate the definition of impedance.

My definition of impedance is as follows. To measure the impedance in a circuit between nodes A and B, cause a current to flow into node A, out of node B, into some device outside of the circuit that enables the current, and back into A to close the loop. The impedance between node A and B is the change in voltage between A and B due only to that current, divided by that current. Do you agree? If not, what is your definition? Surely we shouldn't even try to talk about something until we can define it.

If you agree with my def, we must reject any voltages between A and B due to other currents that may be present. Hence Centauri's argument.

It's not about what we intend a circuit to do, but what it does when its impedance is tested that counts.

In this case, the current flowing into node A and out of node B is caused by a change in voltage at node C and also causes changes on node D and node E.

unless injecting a current across nodes A & B results in the same behavior in nodes C & D I'll suggest you are measuring the wrong thing.

with a triode that doesn't happen.

dave

dave
 
Suppose you are handed a circuit in a closed box you can't unseal. You know nothing about the circuit. Somebody pulls out two wires from it and asks you what the impedance is between them. How do you measure it?

There can be no direct answer on such question. There can be in multiple-choice exams such answers only, in the real life everything may be very different. For example, one wire can be connected to the coil of the relay, another to the contact. What can you get from such a measurement? Nothing, if you don't perform other experiments discovering properties of the device that are valid for this device only. Then you have to measure properties that have the meaning for this particular device, but not such meaningless like resistance between coil and contact. Or, the box can be transformer, and measuring resistance between wires of different coils you can't get parameters that are valid for this device, such as inductance, stray capacitance, and so on...
 
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Technically, that's not the output impedance of both that you've measured. As I mentioned before, the output impedance of a node cannot be measured in this way.

Since you have this circuit all set up in the simulator, would you be interested in trying something out? If so, then:

First, zero V3.

Second, Replace R6 and R7 with AC current sources of opposite polarities, i.e., positive current is towards C1 or C2 respectively. Label the source replacing R6 as "Itp" and the sourse replacing R7 as "Itk".

but currents of opposite polarities would result in zero output from the splitter since ground is at the series junction of the two and the signals at the plate and cathode would be in phase.

if measured in your method I know the results. I still maintain it is a flawed measurement technique since it isn't measuring the correct thing.
 
Suppose you are handed a circuit in a closed box you can't unseal. You know nothing about the circuit. Somebody pulls out two wires from it and asks you what the impedance is between them. How do you measure it?

we have three wires and the condition that one of them is ground, the other two have identical voltages 180 degrees out of phase with each other.

how do you measure the output Z?

Ignoring the third terminal and phase is like arguing that weight doesn't exist and we can only consider mass... boy I wish I could selectively ignore the obvious (gravity)

dave




dave
 
Dave, you didn't answer my question. How do you measure the impedance between any two nodes in any linear circuit? Surely every pair of nodes has an impedance between them. I don't care how many nodes or signals there are, or what the signals are. Thevenin tells you how to do this with any two nodes in any linear network, as long as all the impedances are resistors. But at a single frequency, Thevenin's single resistor can be extended to a single impedance with a magnitude and a phase.

Forgive me, but I see no proof whatsoever for your analogy. It seems an assertion without justification. Can you be more specific and relevent?
 
OK, then check out the habit of this linear circuit. Measure the voltage between the nodes. Next, insert insert a sine wave current at the frequency of your choice. Now measure the change in voltage and divide by the current.

I hope Dave answers the question too.

That is just what I did. Maybe we have a communication thing going on here.

I inserted a 1V 1Khz signal to the input and worked the change in voltage against the change in current for two different loads for both the plate and the cathode and everything agreed.

the black box concept of two terminals works great for two terminal devices but in order for this situation to work you need to consider the variable third terminal over time.

dave
 
Dave, you didn't answer my question. How do you measure the impedance between any two nodes in any linear circuit? Surely every pair of nodes has an impedance between them. I don't care how many nodes or signals there are, or what the signals are. Thevenin tells you how to do this with any two nodes in any linear network, as long as all the impedances are resistors. But at a single frequency, Thevenin's single resistor can be extended to a single impedance with a magnitude and a phase.

Forgive me, but I see no proof whatsoever for your analogy. It seems an assertion without justification. Can you be more specific and relevent?

the two terminal model assumes everything in the black box is fixed. This doesn't even come close to represent how things are behaving.

dave
 
OK, then check out the habit of this linear circuit. '

For checking the habit of some unknown device this instruction does not work:

Measure the voltage between the nodes. Next, insert insert a sine wave current at the frequency of your choice. Now measure the change in voltage and divide by the current.

Edit: At last, it is far from enough. I can tell such a way only one parameter that can be completely meaningless for this particular device, and says nothing about it's structure: it can be open contact of relay, it can be gate of mosfet, it can be just a technological leg for soldering to PCB. ;)
 
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but currents of opposite polarities would result in zero output from the splitter since ground is at the series junction of the two and the signals at the plate and cathode would be in phase.

if measured in your method I know the results. I still maintain it is a flawed measurement technique since it isn't measuring the correct thing.

Dave, the reference direction of the two current sources is irrelevant (you can assign both positive and negative AC values to the current sources) but does make it easy to keep track of things.

Also note that in the first two simulation scenarios, the sources are activated one at a time and in the third, the sources are acting as a single source.

However, you can activate both with opposite currents and you will get an output.

Regardless, I've already done the sim with an FET splitter and the results are just as I described in post #146.

It's odd that you call it a flawed measurement technique though especially since the answer you get to part (3) of the sim should more or less match the one you got earlier. Which is to say, you measured the differential output impedance of the phase splitter.
 
I hope not. I think it's catching!

Let's try it again. If I can't measure the impedance of the Cathodyne plate by loading just the plate because then it's no longer a Cathodyne, then by that logic I can't measure the impedance of a voltage amplifier output by measuring its output short circuit current, because then it's no longer a voltage amplifier.

Discuss.

Nonsequitur. You're comparing the means of measuring (calculating, really- for measurement, you normally just change the load, not actually short it) a two output device with a specified constraint with the measurement of a one output device.

Equal loads, equal source Z. If you can suggest ANY equal loads which show different source Zs (as determined by definition of Thevenin source impedance) from the two outputs, I will do the experiment and see if you're correct.
 
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