Confused about Inductors

[BlueText]

In this circuit the transistors are not conducting. If a diode is biased forward, the BE junction is biased in reverse. So the current through CE is 0A. If the diode is biased in reverse, then trivially current is 0A. So in any case the only valid current in this circuit is no current.

Apologies, I may have used the wrong transistor symbol. I am assuming a current going into the left side of the top PNP transistor would block current from circulating in the right hand side of the circuit.

[IanB]

For a transistor to turn on you have to apply a voltage to its base. If your circuit starts out with only a charge on the capacitor and both inductors "at rest", then where is the voltage to turn on either transistor? Both transistors will be off and no current can flow through them.

[MrAl]

Look up the Hartley oscillator.

Ah thanks for that. I didn't know this existed! 

Regarding the circuit, I admit I may have it wrong, but here is what it is supposed to do. The capacitor would have a starting charge with it, say, discharging its current in a downward direction and charging the left inductor. Then the inductor discharges into the capacitor in the same direction so it ends up being polarised in the opposite direction it started in. Same as a classic LC circuit. Then, when the capacitor discharges again, as the current is flowing in the opposite direction it charges the right inductor instead, which discharges as normal and the circuit returns to its original state. The diodes and transistors were designed to only allow current to circulate in one half of the circuit at a time, dependant on it's direction.

I am assuming, with the PNP transistors, that current circulating clockwise in the left hand part of the circuit would enter the top transistor and cut off the right hand part of the circuit from receiving any current?

As for the purpose of the circuit, well, I am interested in learning electronics and I thought the best way to do it would be to set myself a challenge and having an inductor rest while another is doing something would be interesting and force my brain to work to come up with a solution. If the circuit works, it could be interesting to have different kinds of inductors on each side so it would have two speeds of resonance or something, maybe producing an asymmetric AC current. Not sure if there are any real world applications for that though.

Regarding resistance, I would assume any actual resonant circuit built in the real world will have resistance and wouldn't run for ever, so would need an input current occasionally.

I think if you have initial energy in both inductors you can see some behavior like you are looking for which would be resonance for a little while then a drop off in energy.   I dont think it serves a good purpose for learning about circuits though it's just too esoteric to be useful in the respect.

What would be more useful would be a circuit that is more like a regular textbook circuit like with an inductor and cap and resistor, and then later a diode, then maybe later a transistor.

But back to the first line above.  What do you think you will get out of this if you prove that with initial energy in both inductors and a particular current direction for both you see some resonance followed by a complete damping out of all energy?
I just feel you wont get anything out of it because it's just too unusual.  Maybe in an advanced course in engineering you may, but even then you can probably do much better.

I've seen other circuits like this that are so far out in left field that nobody agrees on exactly how they work or how they should work.  It ends up being a never ending discussion on anything but electronics, but more on philosophy and how things with no parasitics work or how things with parasitics work or somewhere in between.

There are also numerous programming examples where people try to write the most obscure lines of code to perform a simple function and then see who can figure it out.  It ends up being a huge waste of time and is mostly just for the novelty of the thing.

My advice would be to first study the more common circuits with LRC and maybe some other things and learn the three types of resonance and how they are related, and the three possible types of responses: underdamped, critically damped, and over damped.  That will get you some really useful insight about LRC circuits and related that you will be able to use in many circuits that come at you in the future.  The circuit at hand will only get you as far as that circuit, if that, and that's it, as you will probably never encounter a circuit exactly like this again.

A this point i would have to ask if you studied the more common LCR circuits yet.  One thing you dont do is jump into more complicated circuit before you fully understand the more common ones and how to analyze them fully.

[BlueText]

For a transistor to turn on you have to apply a voltage to its base. If your circuit starts out with only a charge on the capacitor and both inductors "at rest", then where is the voltage to turn on either transistor? Both transistors will be off and no current can flow through them.

The transistors are the PNP type which allow current to flow through when the base signal is low and shut off the current when the base signal is high.

I think if you have initial energy in both inductors you can see some behavior like you are looking for which would be resonance for a little while then a drop off in energy.   I dont think it serves a good purpose for learning about circuits though it's just too esoteric to be useful in the respect.

What would be more useful would be a circuit that is more like a regular textbook circuit like with an inductor and cap and resistor, and then later a diode, then maybe later a transistor.

But back to the first line above.  What do you think you will get out of this if you prove that with initial energy in both inductors and a particular current direction for both you see some resonance followed by a complete damping out of all energy?
I just feel you wont get anything out of it because it's just too unusual.  Maybe in an advanced course in engineering you may, but even then you can probably do much better.

I've seen other circuits like this that are so far out in left field that nobody agrees on exactly how they work or how they should work.  It ends up being a never ending discussion on anything but electronics, but more on philosophy and how things with no parasitics work or how things with parasitics work or somewhere in between.

There are also numerous programming examples where people try to write the most obscure lines of code to perform a simple function and then see who can figure it out.  It ends up being a huge waste of time and is mostly just for the novelty of the thing.

My advice would be to first study the more common circuits with LRC and maybe some other things and learn the three types of resonance and how they are related, and the three possible types of responses: underdamped, critically damped, and over damped.  That will get you some really useful insight about LRC circuits and related that you will be able to use in many circuits that come at you in the future.  The circuit at hand will only get you as far as that circuit, if that, and that's it, as you will probably never encounter a circuit exactly like this again.

A this point i would have to ask if you studied the more common LCR circuits yet.  One thing you dont do is jump into more complicated circuit before you fully understand the more common ones and how to analyze them fully.

Yes, I would say I jumped in at the deep end here.  I guess I'm more into digital electronics like logic gates and stuff where you only have to think about whether things are on or off. Still I feel I've learned a lot here.

I still think the circuit would work as I advised. The top transistor cuts off current to the right hand side of the circuit when the left side has current and vice versa. A normal LC circuit with only 1 inductor will have AC resonating in it, where as this will have half the AC sine wave through the left inductor and half the AC sine wave on the right inductor and not at the same time. If that makes sense?

[wasedadoc]

I still think the circuit would work as I advised. The top transistor cuts off current to the right hand side of the circuit when the left side has current and vice versa. A normal LC circuit with only 1 inductor will have AC resonating in it, where as this will have half the AC sine wave through the left inductor and half the AC sine wave on the right inductor and not at the same time. If that makes sense?

You have already been told that it will not.  The circuit will not allow current to flow anywhere. Period.

[IanB]

The transistors are the PNP type which allow current to flow through when the base signal is low and shut off the current when the base signal is high.

And where will you obtain these magical transistors that behave this way?

I suggest you get yourself a handful of NPN and PNP transistors, a breadboard, some potentiometers, resistors and LEDs, and try some experiments to see how transistors really work.

The whole of science is about experiment. Things do not work the way you imagine they do, they work the way experiment shows they do.

[BlueText]

And where will you obtain these magical transistors that behave this way?

I suggest you get yourself a handful of NPN and PNP transistors, a breadboard, some potentiometers, resistors and LEDs, and try some experiments to see how transistors really work.

The whole of science is about experiment. Things do not work the way you imagine they do, they work the way experiment shows they do.

My apologies, clearly I thought I knew how a PNP transistor worked, when I didn't. Is there any kind of switch/transistor that closes when a current is sent to its base?

[MrAl]

And where will you obtain these magical transistors that behave this way?

I suggest you get yourself a handful of NPN and PNP transistors, a breadboard, some potentiometers, resistors and LEDs, and try some experiments to see how transistors really work.

The whole of science is about experiment. Things do not work the way you imagine they do, they work the way experiment shows they do.

My apologies, clearly I thought I knew how a PNP transistor worked, when I didn't. Is there any kind of switch/transistor that closes when a current is sent to its base?

I am not sure why some members are claiming that there will be no current flow.  If you set the initial current in the two inductors you get something like you wanted.  It doesnt make sense why you would need to see this happen though.

This is with each inductor being 1mH and the cap being 0.5uf, and the current in the left inductor being 1 amp flowing up, and the current in the right inductor flowing up and also at 1 amp initially.
So doesnt make any sense that anyone would say it does not do anything at all.
Also, an inductor with at least some initial current will force something to happen no matter what.  That's because they force the voltage to go to a level which will force conduction even if it is for only a short time.

In this snap shot the horizontal is set at 200us per division.
Clearly there is some ringing and then it damps out.

The circuit may make more sense to some people if you remove the cap then examine the circuit, then put the cap back in and think about what happens then.  Maybe replace the cap with a resistor and examine that circuit too first.

[wasedadoc]

Current in the direction of the arrow in the schematic can only flow through the inductor on the left if the bottom transistor is conducting.  To be conducting that transistor needs some base current.  That base current can only come from the right hand loop.  For current to flow in the right hand loop the top transistor must be conducting.  However, the current flowing in the left hand loop is because the anode of the top diode is more positive than its cathode. That means the base of the top transistor is more positive than its emitter.  As it is a PNP transistor that means its base-emitter is reverse biased.  No base current.  No collector current.  Which is the opposite of what has just been determined as its necessary state in the first sentence.

[IanB]

I am not sure why some members are claiming that there will be no current flow.  If you set the initial current in the two inductors you get something like you wanted.  It doesnt make sense why you would need to see this happen though.

Because the stated starting point is the capacitor charged and no current in the inductors. If you build the circuit on the bench and charge up the capacitor with an external power supply using flying leads, where will the current in the inductors come from?

[BlueText]

I am not sure why some members are claiming that there will be no current flow.  If you set the initial current in the two inductors you get something like you wanted.  It doesnt make sense why you would need to see this happen though.

This is with each inductor being 1mH and the cap being 0.5uf, and the current in the left inductor being 1 amp flowing up, and the current in the right inductor flowing up and also at 1 amp initially.
So doesnt make any sense that anyone would say it does not do anything at all.
Also, an inductor with at least some initial current will force something to happen no matter what.  That's because they force the voltage to go to a level which will force conduction even if it is for only a short time.

In this snap shot the horizontal is set at 200us per division.
Clearly there is some ringing and then it damps out.

The circuit may make more sense to some people if you remove the cap then examine the circuit, then put the cap back in and think about what happens then.  Maybe replace the cap with a resistor and examine that circuit too first.

OK, that's interesting, and weird. I'm not sure how the right inductor can have current flowing up when the bottom diode would prevent current flowing in that direction. Are you using some kind of circuit simulator? I wouldn't mind trying that as it might avoid me asking some embarrassing question in future. 

[golden_labels]

I am not sure why some members are claiming that there will be no current flow.  If you set the initial current in the two inductors you get something like you wanted.

Because there is no path for the charge to flow? Of course we may argue about some capacitance and leakage in real transistors and diodes, and that in a physical BJT the base-collector junction can actually conduct. Which should allow for some transient ringing. But for ideal components the situation is as in the picture I attached. If I am not wrong, of course — feel free to correct me.

Can you draw a loop in this picture? I can’t.

Note that I am already pretend I do not see the currents are opposing each other.

[BlueText]

Sooo, I may actually understand how PNP transistors work now.  Electrons build up and unless a small number of them are allowed to escape to ground through the base, no current can pass through the transistor. Given that, maybe the following will work?



Or maybe I have just submitted more circuit gore.  Not sure if the diodes are needed now though...

[wasedadoc]

Still will not work.  Assume current flowing in the direction of the arrow in the left inductor.  It goes through the top diode.  It cannot go through the base-collector of the bottom transistor. (The base-collector is a n-p junction.)  Two other possible routes are into the emitter of the top transistor or into the capacitor and thence into the emitter of the bottom transistor.  If it goes into the capacitor it charges the capacitor, its top plate becoming more positive than the bottom plate.  That causes reverse bias of the base-emitter junction of the bottom transistor so it cannot conduct.  But that capacitor voltage is forward bias of the base-emitter of the top transistor allowing it to conduct so let's imagine the current goes into its emitter, out of the collector and through the right side inductor.  Then through the bottom diode.  But there is no continuing path back to the left inductor because the bottom transistor is not conducting.  It isn't possible to have both transistors conducing simultaneously- if one has its base-emitter forward biased the other is reverse biased.

Taking the diodes out does not change the above,

So again no current flowing.

[golden_labels]

BlueText, please attach images to the posts. Do not drop them at random external servers. The forum has an attachment feature.

The circuit is symmetric with respect to the capacitor. Not only components, but also conditions present. Which means current through that capacitor must be zero. If it wasn’t, one of the sides would need to observe different conditions. That contradicts the initial assumptions. It would also raise a question of which side it would be, which by itself is not a problem, until one asks why in a perfectly symmetric situation one option would be preferred over the other.

If current through the capacitor is zero, it may be treated as open. The circuit becomes what is seen in the attachment. Let’s assume that current I₁ is present. That implies that there must be current at transistors’ bases (I_B). There are two paths available. The more obvious is loop through base-emitter junctions of both transistors. But there is nothing to drive that current. The other is two loops: one through L1, D1, base-emitter junction of Q1 and then emitter-collector of Q2, and a symmetric one for L2. However, that would induce impossible voltages. Assuming⁽¹⁾ voltage across an inductor is xv and each P-N junction drops v: D1 cathode is (x-1)·v, Q1 base is (x-2)·v, the rest is dissipated on Q2. But since the situation is perfectly symmetric, equivalent reasoning for current through L2 indicates that D2 cathode must be (x-1)·v. And that cathode is the same node as Q1 base, which leads to a contradiction. Of course voltages are relative and L2 is independent of L1, so we may neatly assume that voltage at D2 cathode is (x-2)·v, but then base of Q2 is at (x-3)·v, which is not the same as D1 cathode, leading to the same contradiction.

Another take is not involving currents at all. From KVL the sum of voltages on the inner loop, formed from base-emitter junctions of Q1 and Q2, must be 0. However, if there is current through those junctions, both see some positive voltage. And a sum of two positive numbers is always positive, which contradicts KVL.

⁽¹⁾ The argument itself doesn’t depend on P-N junction drops being equal. I used v for them merely for simplicity. You may conduct the same reasoning with any set of non-zero values.

[BlueText]

Thanks for the circuit analysis. It's very helpful.

I've played about with designs on the lushprojects.com circuit simulator and it looks like I might not actually need any transistors or gates and that just having the diodes are good enough.

Dual LC circuit

I was worried that 1 inductor would discharge straight into the other inductor, by-passing the capacitor altogether, but it seems current much prefers to flow into the capacitor first, as long as it doesn't get too saturated. There is still some degree of inductor to inductor transfer of current though, which I guess is damping down the oscillations a bit too much. Maybe some JFET transistors might help in this regard.

[BlueText]

Actually this is a better example of the circuit. The capacitor is precharged with 5v. It resonates, but dies down too quickly as sometimes one inductor discharges straight into the other.

Dual LC Circuit 2

[golden_labels]

To start with, be aware this is a simulator created by Paul Falstad and he maintains the up-to-date version of this circuit simulator webapp. It seems Lushproject hosts a copy endorsed by Paul, but Iain hisemlf notes he’s no longer able to maintain it.

Remember that any simulation program is a glorified calculator. It calculates a mathematical model, not represents reality. The components you draw are visual aids to construct mathematical equations. But it’s up to you to ensure those equations make sense. If you create an invalid model, you get invalid results.

In this case the model seems reasonable. Chiefly, because what you have created is a more complicated version of an LC tank. Let me redraw your circuit. See the attachment for the rearrangement. First two steps are not introducing any changes to the circuit: the diodes get shifted closer to inductors and then inductor-diode pairs are moved to one side. In the last step a change is introduced: D2 is swapped with L2. While that introduces topological change, the rest of the circuit will not see any difference. Can you now see how this is like a standard LC tank, with inductance on the left and capacitance on the right, but with each inductor handling current going into one direction only? The diodes, having non-zero forward voltage, will introduce some minor shifts in the behavior, but the general idea stays the same.

[MrAl]

Current in the direction of the arrow in the schematic can only flow through the inductor on the left if the bottom transistor is conducting.  To be conducting that transistor needs some base current.  That base current can only come from the right hand loop.  For current to flow in the right hand loop the top transistor must be conducting.  However, the current flowing in the left hand loop is because the anode of the top diode is more positive than its cathode. That means the base of the top transistor is more positive than its emitter.  As it is a PNP transistor that means its base-emitter is reverse biased.  No base current.  No collector current.  Which is the opposite of what has just been determined as its necessary state in the first sentence.

Well actually we have not yet established what the assumptions are about the components or operating conditions.
This is why i asked about the theoretical assumptions earlier in this conversation.
I am attaching a drawing that illustrates SOME of the possible assumptions that are either left out, or alternately ignored.  Depending on which you choose, the operation of the circuit will be entirely different.  That's why listing the assumptions is so important in a circuit of this nature.
Once we know the assumptions, we can come up with a definite conclusion.  Without that, there will continue to be very large differences of opinion.

[MrAl]

I am not sure why some members are claiming that there will be no current flow.  If you set the initial current in the two inductors you get something like you wanted.  It doesnt make sense why you would need to see this happen though.

Because the stated starting point is the capacitor charged and no current in the inductors. If you build the circuit on the bench and charge up the capacitor with an external power supply using flying leads, where will the current in the inductors come from?

I think the more basic question was can the circuit show any kind of resonance, and to that end we try to find different ways that it may do that.  In particular, the initial conditions.

There still has not been any attempt to establish any assumptions so there can be no definite conclusions yet.  In a circuit as unusual as this the assumptions are just as important as the components themselves because they are a large part of what makes up the components.

I post an attachment to show some of the things left out.

[MrAl]

I am not sure why some members are claiming that there will be no current flow.  If you set the initial current in the two inductors you get something like you wanted.  It doesnt make sense why you would need to see this happen though.

This is with each inductor being 1mH and the cap being 0.5uf, and the current in the left inductor being 1 amp flowing up, and the current in the right inductor flowing up and also at 1 amp initially.
So doesnt make any sense that anyone would say it does not do anything at all.
Also, an inductor with at least some initial current will force something to happen no matter what.  That's because they force the voltage to go to a level which will force conduction even if it is for only a short time.

In this snap shot the horizontal is set at 200us per division.
Clearly there is some ringing and then it damps out.

The circuit may make more sense to some people if you remove the cap then examine the circuit, then put the cap back in and think about what happens then.  Maybe replace the cap with a resistor and examine that circuit too first.

OK, that's interesting, and weird. I'm not sure how the right inductor can have current flowing up when the bottom diode would prevent current flowing in that direction. Are you using some kind of circuit simulator? I wouldn't mind trying that as it might avoid me asking some embarrassing question in future. 

Everything will be weird until you list the assumptions about the components.  You will continue to see wide differences of opinion among members until you specify every little tiny detail.  That's because these details change the behavior dramatically.
See attachment that shows some of the details left out, but not all.  There is still reverse leakage in the diodes and BE diode also.  There is also internal capacitances.

A lot of people use the LT Spice simulator used to be called "Switcher Cad".  It's a pretty good one.

[BlueText]

Well I would say I'm ready to finish this project. Thank you for all the help as I've learnt a lot from this! 

Although I couldn't get the circuit to perform with gates I did settle on this design as it pretty much does the job I was looking for:

Parallel LC circuit

Basically I dumped the right hand side inductor / diode combo and replaced it with a small dual direction inductor which seemed to perform better as there is far less current trying to bypass the capacitor from one inductor to the other and it allows the capacitor to reset itself to its original polarity. Then I added more smaller inductors in parallel so I could keep the frequency high but have enough enough power to cycle the larger inductor.

I guess I'll have to build it now and see if it can work at all in the real world. 

[wasedadoc]

Thank you for all the help as I've learnt a lot from this!

It appears that you have yet to learn that inductors in parallel behave as a single inductor.

https://www.omnicalculator.com/physics/parallel-inductors

[BlueText]

It appears that you have yet to learn that inductors in parallel behave as a single inductor.

https://www.omnicalculator.com/physics/parallel-inductors

If you replaced the parallel bi-direction inductors with a single bi-direction inductor of equal total strength, when the single bi-direction inductor is almost discharged you get one inductor feeding into the other, which is not good for keeping the osculations going. Making the bi-direction inductor a lot smaller reduces this. Then you just need to add more bi-direction inductors in parallel to increase the peak heights to something more desirable. At least that's what the simulation shows.

[golden_labels]

I assume that “bi-directional inductors” is used to contrast with the inductor-diode pair. If not, then you got something wrong: inductors are not directional devices.

A bunch of inductors connected between two nodes may always be replaced with a single inductor with no difference in behavior, no matter how they were originally arranged. It works the same as with capacitors or resitors.⁽¹⁾ If you swapped all those inductors with a single one and the behavior has changed in even a slightest manner, it means that you have calculated the value wrong.

As it happens, Falstad’s simulator has examples regarding that: Inductances in series, Inductances in parallel.

It doesn’t matter they push current into each other. All those currents sum to 0 and are not seen on the outside.

⁽¹⁾ In fact that doesn’t even depend on element type. They are all impedances and may be simplified to exactly one impedance. Because no basic passive element provides both real and imaginary part of the impedance, using more than one element may be required. But that’s it. Absolutely any subcircuit consisting of passives only and spanning between two nodes may be simplified that way without any change in behavior.