SpotTheMistake Critiques Roman Black's Two Transistor Switcher Design

[MrAl]

Hi Eric,

Yeah just a couple minutes

What i'd like to see is how Collin can explain why he thinks the inductor current magically stops increasing.

Colin:
Care to try to explain this?

 

[colin55]

As Vout increases it becomes high enough to start reverse biassing Q2 , via the emitter connection. This causes the voltage across D1 to fall, this fall is negative going and is coupled back to the Q2 base via C2 , which makes Q2 switch off faster.

The abvoe is entirely incorrect.

That statement suggests that the inductor current somehow magically stops increasing just before the transistor turns off

That's exactly what happens.
The current gets to a maximum due to the two transistors being fully turned ON. It is only the EXPANING FLUX that produces the "back voltage" that separates the high voltage on the emitter of Q1 and the 5v on the output of the circuit.
As soon as this expanding flux cannot be maintained, the voltage on the left terminal of the inductor reduces and this is passed to the base of Q2 via the 1n to turn the transistor OFF.

 

[colin55]

What i'd like to see is how Collin can explain why he thinks the inductor current magically stops increasing.

Because the two transistors are finally fully turned ON.

 

[colin55]

The problem is you cannot "see" the circuit working.
I have been investigating circuits for the past 40 years and can "see" how they work.
I wrote up the operation of the circuit weeks ago, long before ericgibbs produced a simulation and an explanation.
However ericgibbs and Mr Al does not understand the intricacies of the operation of the circuit and glosses over the points such as the vital fact that starts the “turn-off” portion of the cycle.
” It's basic inductor circuit operation where we have a (relatively) constant voltage across an inductor and that causes a continual increase (ramp) in current until something else occurs to stop that increase.”
This is entirely incorrect.
How can we have a constant voltage across an inductor and an increasing current through it???

 

[colin55]

We cannot progress to the next point in the discussion until you can see that the "back-voltage" produced by the inductor decreases and this is passed to the base of Q2 via the 1n.

 

[ericgibbs]

That's exactly what happens.
The current gets to a maximum due to the two transistors being fully turned ON. It is only the EXPANING FLUX that produces the "back voltage" that separates the high voltage on the emitter of Q1 and the 5v on the output of the circuit.
As soon as this expanding flux cannot be maintained, the voltage on the left terminal of the inductor reduces and this is passed to the base of Q2 via the 1n to turn the transistor OFF

Colin
As I have stated before, you have no idea of the fundamentals concepts of electronic circuits, the above quoted statement confirms your total lack of knowledge regarding this subject.

I can see no further point in continuing explaining this topic with you as you are unable to understand the explanations Al and I have given you.

 

[colin55]

ericgibbs You have absolutely no idea what you are talking about.
What do you think happens to the voltage produced by the inductor when the current reaches a maximum?
Obvious the “back voltage” reduces and this is passed to the base of Q2 via the 1n.

 

[3v0]

I expect at this point somebody may want to close this thread.

I hope that does not happen because the topic under debate is electronics and more then a few of us are finding it interesting.

 

[ericgibbs]

I expect at this point somebody may want to close this thread.

I hope that does not happen because the topic under debate is electronics and more then a few of us are finding it interesting.

hi H,
I must admit I have considered it, but I may be accused of bias.

I am hoping someone else will step in and try to get Colin to understand mine and MrAl's explanation on how the circuit actually works, I have expended as much effort as I am prepared to give to Colin.

Eric

 

[colin55]

We have already said the circuit is highly efficient.
This means transistor Q1 is fully turned on and the voltage across it is very small.
This means the voltage across the inductor is 12v – 5v = about 7v.
The right lead of the 1n is pulled HIGH and it charges to about 7v and the charge-current comes partially through the base emitter junction of Q2. This is what fully turns on Q2. However the current through the inductor ceases to be an increasing current and the voltage produced by it decreases slighty. This is passed to the base of Q2 to turn the transistor off slightly to start the turn-off process.
The voltage on the 1n changes from +7v (in one direction) to -6v, and this 13v differential is reflected in the 4n7 by changing its value by about 3v.
This has been proven by the waveform provided in one of the previous postings.

 

[colin55]

The problem is this:
Almost no-one understands the intricacies of the operation of the circuit.
One person has already dropped out of the discussion and the other two have incorrect interpretations.
ericgibbs has made statements that don’t make sense.
How can the rise in voltage from the output turn Q2 off.
It will be acting through a 100u electrolytic.
We are talking about a regenerative action, whereby the effect is started by a change in a condition and this is passed around very quickly to completely turn the circuit off.
We are talking about microseconds.
If you follow my discussion you will be able to see exactly how this part of the cycle is expedited.

 

[3v0]

We may have other people here with EE degrees that have not yet looked at this. Perhaps they can try their hand at it.

 

[MrAl]

The problem is you cannot "see" the circuit working.
I have been investigating circuits for the past 40 years and can "see" how they work.
I wrote up the operation of the circuit weeks ago, long before ericgibbs produced a simulation and an explanation.
However ericgibbs and Mr Al does not understand the intricacies of the operation of the circuit and glosses over the points such as the vital fact that starts the “turn-off” portion of the cycle.
” It's basic inductor circuit operation where we have a (relatively) constant voltage across an inductor and that causes a continual increase (ramp) in current until something else occurs to stop that increase.”
This is entirely incorrect.
How can we have a constant voltage across an inductor and an increasing current through it???

Hi again,

I'd hate to see this thread closed myself as well, and on the plus side colin is forcing us to explain this circuit in greater and greater detail, which isnt entirely bad if you look at it from the standpoint of more and more information getting out there
There may be others that have doubts like this too so we'll try to clear them up.

colin, when you posed the question, "How can we have a constant voltage across an inductor and an increasing current through it"
you seemed to have an air of doubt that this can not be possible. However, this is exactly how an inductor works in a switch mode power supply circuit.
The voltage stays relatively constant as the output increases slightly. I can say "slightly" because the collector voltage is constant at this point and the output voltage only changes by millivolts. The current is ramping because that is what happens in an inductor with a constant voltage across it.
For example, connect a 1.5v battery across an inductor. The current ramps up. If it were a perfect inductor it would ramp up in a straight line. If the battery could supply an infinite current the current would ramp up forever, with just 1.5v across that inductor. So all it really takes is a small voltage across an inductor to create a ramp current. Actually even 0.1v would do it, but usually a real life inductor has some series resistance that limits this activity. Inductors used in power supplies however have low resistance to keep efficiency high so they behave like near perfect inductors for the most part (not exactly, but close enough for basic theory).

So what i am saying is that there is no limit to the current the inductor would draw if the transistor never turned off, or unless the output rose quite a bit to near the power supply input voltage. Something eventually limits that current and that is the transistor turning off, and the reason the transistor turns off is because it's collector current starts to rise too high for it to maintain a very low voltage drop, so the drop increases which means the collector voltage falls, slowly at first.

You're talking about a detailed view of this circuit but then i think you should look at the operation near the turn off time of Q1 in increments no larger than 100ns each. What you should see is the transistor starting to turn off (well, it develops a larger voltage drop collector to emitter) and this would be a rounded looking waveform where it decreases a little bit, and if you look at the point where it starts to decrease faster than it was before (it decreases very very slightly the whole partial cycle because the inductor is conducting more and more current during the entire 'on' part of the cycle) you'll see the inductor current still rising. As i said though you need to look at 100ns intervals to see this happening, or even at shorter time slots.

For the inductor to turn off as you are saying it would the switching cycle would have to be synced with the resonate frequency of the output circuit (inductor and cap) and as i was saying before that cycle would be longer than the switching cycle observed so it can not be turning off for that reason. Following the reasoning given above, we can trace the circuit operation from one cycle to the next without any magic at all because one thing leads to the next and those things are always well explained. And then again if it really was synced in this way (which it is not) then that would explain it and we still wouldnt need any magic

There is a chance that an inductor being used in a power supply circuit like this has too high of an equivalent series resistance to allow proper operation. What would happen in that case though is the oscillations would stall because the inductor current would stop increasing on it's own just because the voltage across it isnt high enough (with the real life ESR) to force more current through it. In fact, circuits like that usually dont even start up because the transistor Q1 never turns off because the inductor current never reaches that critical point. But this is a case of using the wrong inductor for the circuit and replacing it with the proper lower ESR inductor makes the circuit work just fine.

Im sure you looked at a lot of circuits in the past 40 years no argument there, but perhaps just this one time you might take a fresh look at this one here.

ADDED LATER:
I almost forgot to mention the case where the inductor saturates. Obviously this would be the wrong inductor for the task, but if it did in fact saturate the current would go way up fast...it would never just level off unless again the series resistance could limit the current...but then again it would not oscillate either...unless of course the transistor heats up and the beta goes down low enough for it to start to pull out of sat.

 

[colin55]

How can we have a constant voltage across an inductor and an increasing current through it"

For example, connect a 1.5v battery across an inductor.

You cannot take the example of a 1.5v battery because the voltage is very low impedance.
In this circuit the impedance is much higher.

How can we have a constant voltage across an inductor and an increasing current through it"

The mere fact that the current is changing, means the "back-voltage" produced by the inductor will change.

 

[RichTheDude]

What do you mean?

I want to see the quantifiable formal proofs (rather than subjective qualitative text) you are using to base your argument off.

 

[tvtech]

Oh well, here we go....again.

Shakes head

This Forum is going NOWHERE unless we all pull together. It is not hard to do.

Please stop arguing amongst yourself Members. Common.

tvtech

 

[Mickster]

I expect at this point somebody may want to close this thread.

I hope that does not happen because the topic under debate is electronics and more then a few of us are finding it interesting.

I hope not too. A lot of this is going way over my head, but I'm interested nonetheless and picking some things up slowly.

We may have other people here with EE degrees that have not yet looked at this. Perhaps they can try their hand at it.

I'm definitely not in the above category, but did try my hand using a java applet found here:
https://www.falstad.com/circuit/index.html
Unfortunately, I couldn't find a polarised cap for 'CLOAD' in the app, so used a non-polarised cap & added a 50ohm resistor for simulation into a load. Whether or not that makes that much of a difference, I don't know, but you can slow the sim waaaay down and hover over different nodes, right-click wires and/or components & create 'scope plots etc. Useful to see the voltage & current paths under animation.

Someone previously asked Colin what he was using to analyse the circuit, but I don't seem to recall an answer being provided.

I'd like to raise the same question again and hope to see an answer so I can try to understand Colin's review and explanation.

Regards.

 

[skyhawk]

You cannot take the example of a 1.5v battery because the voltage is very low impedance.
In this circuit the impedance is much higher.

The mere fact that the current is changing, means the "back-voltage" produced by the inductor will change.

No!
VL = -Ldi/dt

A constant di/dt means a constant VL.

 

[JimB]

Has anybody considered building the circuit and doing a few tests to try and get a better understanding of what may or may not be happening?

JimB

 

[MrAl]

Oh well, here we go....again.

Shakes head

This Forum is going NOWHERE unless we all pull together. It is not hard to do.

Please stop arguing amongst yourself Members. Common.

tvtech

Hi,

Well this time we're just debating. It's bringing out some good points anyway