SpotTheMistake Critiques Roman Black's Two Transistor Switcher Design

[colin55]

“When C1 discharges from 5.6v to 3v through C2 it releases stored energy, and delivers current into point A. “
Because you have never drawn the high-speed diode “flipped-over” you don’t understand that point A is actually at a potential of -0.6v during this part of the cycle.
”there is a RC time delay formed by RZ and C1”
How long do you think it will take to charge C1 via a 10k and provide enough energy to turn on the transistor????? (as well as the leakage into the zener.)
The electo on the output does this via the inductor and 1n.
You have a completely wrong idea of how the circuit works.

 

[ericgibbs]

hi Roman,
LTSpice simulations of your circuit.

During Period A to B, Q1 is supplying current to the Load via the L1 inductor.
This is due to the forward Vbe voltage on Q2 , relative to Vout.

At time B, Vout rises so that Q2 stops conducting, as it does so positive feedback via C2 drives it off hard and fast.

At time B the current flowing thru the L1 inductor is suddenly cut off, this causes the magnetic field around the inductor to collapse back into the inductor , this results in a current thru the D1 diode , L1 and the Load. So the collapsing field of the inductor is the source of this current.

At time C, the Vout has fallen enough so that Q2 starts to conduct, as it does so positive feedback via C2 drives it hard and fast on.

So the major current sources for the load during time A to B is Q1 and from time B to C is the current loop around the L1, Load and D1 due to the collapsing magnetic field of the inductor.

The function of C2 is positive feedback at the switching times of Q2, so that Q2 and Q1 switch hard and fast.

Regards
Eric

 

[colin55]

During Period A to B, Q1 is supplying current to the Load via the L1 inductor.
This is due to the forward Vbe voltage on Q2 , relative to Vout.

At time B, Vout rises so that Q2 stops conducting, as it does so positive (negative) feedback via C2 drives it off hard and fast.

At time B the current flowing thru the L1 inductor is suddenly cut off, this causes the magnetic field around the inductor to collapse back into the inductor , this results in a current thru the D1 diode , L1 and the Load. So the collapsing field of the inductor is the source of this current.

At time C, the Vout (actually the voltage produced by the inductor - not Vout) has fallen enough so that Q2 starts to conduct, as it does so positive feedback via C2 drives it hard and fast on.

So the major current sources for the load during time A to B is Q1 and from time B to C is the current loop around the L1, Load and D1 due to the collapsing magnetic field of the inductor.

The function of C2 is positive feedback at the switching times of Q2, so that Q2 and Q1 switch hard and fast.

This is what I have been saying all along. I have made some corrections to the wording above.

 

[ericgibbs]

Colin,

Of course your supposed corrections are incorrect, as usual.

You do not seem able to grasp the concept of positive feedback when applied to an oscillator.

POSITIVE feedback is used when it required to enhance the switching rate ie: effectively increase its INSTABILITY in order to change its state.

NEGATIVE feedback is used to stabilise the operation of a circuit.

From what I have read of your posted descriptions, how you can claim this is what I have been saying all along is amusing .

Its sad you had to copy my text rather than explain in your own words how you think the circuit works.

 

[colin55]

You are entirely incorrect ericgibbs.
We are NOT talking about positive or negative feedback as applied to an oscillator.
We are talking about positive or negative voltage on the base of one of the transistors.

 

[colin55]

If you read my full description on my website, you will find that the description in #22 is what I have been saying all along.
C2 provides a negative voltage (a reduced voltage ) on the base of Q2 to turn it OFF and a positive voltage to turn it on harder.
You can also call these FEEDBACK VOLTAGES.

 

[ericgibbs]

You are entirely incorrect ericgibbs.
We are NOT talking about positive or negative feedback as applied to an oscillator.
We are talking about positive or negative voltage on the base of one of the transistors.

Your above reply displays your complete lack of understanding regarding electronics fundamentals.

I can see no further point in trying to explain this topic to you, as are not prepared to listen and learn.

 

[colin55]

From what I have read of your posted descriptions, how you can claim this is what I have been saying all along is amusing

You remind me of the three blind men and the elephant.

The thread starts with: page 13 of Talking Electronics

 

[RichTheDude]

If you read my full description on my website, you will find that the description in #22 is what I have been saying all along.
C2 provides a negative voltage (a reduced voltage ) on the base of Q2 to turn it OFF and a positive voltage to turn it on harder.
You can also call these FEEDBACK VOLTAGES.

Just wondering if you would care to share your analysis and simulation you are basing your textual description of the circuit from.

 

[colin55]

Your above reply displays your complete lack of understanding regarding electronics fundamentals.

I can see no further point in trying to explain this topic to you, as are not prepared to listen and learn.

No. You have got it wrong.
The overall feedback is called POSITIVE FEEDBACK as it keeps the circuit oscillating. But we are now talking about the individual positive and negative part of the feedback signal (actually the CONTROL SIGNAL - because it is not directly connected from the output to the input) and its effect on turning the transistor ON or OFF.

 

[colin55]

Just wondering if you would care to share your analysis and simulation you are basing your textual description of the circuit from.

What do you mean?

 

[ericgibbs]

No. You have got it wrong.
The overall feedback is called POSITIVE FEEDBACK as it keeps the circuit oscillating. But we are now talking about the individual positive and negative part of the feedback signal (actually the CONTROL SIGNAL - because it is not directly connected from the output to the input) and its effect on turning the transistor ON or OFF.

Pleased to see that you have finally figured out it is positive feedback.

Are you saying that C2 is not the positive feedback path.??

because it is not directly connected from the output to the input)

Is not connected to the voltage output because that would not work.! its connected to collector of the driver transistor where its supposed to be, as thats the switching output, not the Vout which is the DC output of the circuit.

The DC level sensing is via the emitter of Q2 which is connected to the Vout.

Are there other points that you need explaining.?

 

[Mr RB]

hi Roman,
LTSpice simulations of your circuit.
...

Thank you very much! It's also nice to see your simulated voltage waveforms match so well with my real world 'scope waveforms, especially the voltage at C1, and the "RC delay" time where C1 is charged in a neat ramp via RZ.

Your description of the circuit is correct of course and you describe a buck converter with C2 being the very obvious positive feedback, no argument from me.

However there are two specific points I am arguing with Colin that you did not comment on;

1. During time B-C there is an RC delay where the transistors cannot turn on, a delay caused by RZ and C1. This delay is roughly "a fixed delay" and I designed this to mimic the actions of SMPS ICs (which have a fixed OFF delay before allowing re-turnon) to improve efficiency for all the same reasons as used in SMPS ICs. I spent a bit of time deliberately designing that "delay" and fine tuning it, and examining its effects on the circuit. Colin argues there is no delay in my circuit.

2. I am arguing to defend the arrow on my diagram showing that when energy is dumped from C1 during your point B, this energy is supplied to the load. Colin argues this energy is "wasted".

 

[Mr RB]

(Specifically on the argument of whether C1 current goes to the load or "is wasted", I have made the current processes more clear so Colin may hopefully understand it);

In a buck converter, at the time when the transistor turns off the inductor continues to supply the same current (in this case 1A) so for simplicity it can be drawn as a 1A constant current source.

**broken link removed**

Now during the start of the buck cycle, the capacitor C1 in my SMPS discharges from 5.6v to 3v, this is releasing energy from the capacitor, so the capacitor acts as a current source and is supplying energy into the circuit (shown as 0.5A current into point A, see below).

**broken link removed**

This current into point A contributes to L1 current, so there are now TWO power sources contributing current in the same direction. (For the C1 power to be "wasted" as Colin argues, it would have to be dissipated as heat elsewhere in the circuit like the diode and this is not happening).

This additional current from C1 fed into point A means that there is an additional 0.3W of power supplied INTO the buck loop.

The end result is that the inductor now is supplying 0.6W less, for the load to recieve the same power.

So not only is the energy from C1 contributing directly to "supply the load" but since it does not need to go through the diode to supply the load it drives the load with a higher efficiency than the inductor "buck" can.

 

[ericgibbs]

hi Roman,
I would agree with your latest Fig #1, but not with Fig #2 analysis.

Look at this image showing the simplified circuit for the feedback cap, it shows the 'sense' of the C2 pulse is positive regenerative feedback to the switching transistors

I will check the circuit in more detail regarding the function of C1, let you know.

Eric

 

[ericgibbs]

hi Roman,

Two circuits, left side is original and right side has no C1 [4n7], the pulses are a little irregular, but looking at the Upper Plot for C1 you can see that the current spike is less than 50mA with a duration of approx 1uSec.

The Middle plot shows the Schottky diode currents for circuit #1 and #2. you can see the load current is all from the inductor field collapse. In the D2 circuit, there is no C1 [4n7] yet the two circuits currents are almost identical.

The Lower plot shows the current thru the inductors, again almost identical.

The purpose of C1 is to store the positive current feedback from C2, so that the feedback acts on Q2 for a slightly longer period, thus ensuring a defined switching action.
As you can see without the C1 cap the switching is erratic, but it works.

If you would like me to run more checks, just ask.

Eric

 

[MrAl]

Hi guys,

Interesting discussion here

The only question i have is why all the fuss about the energy from C1? C1 is in parallel with the zener so some of the energy would make it through the base of Q2 and some through the zener. The average power is going to be less than 1mw anyway so i dont see why all the concern, except maybe for the general understanding of the circuit.

To understand better it makes sense to look at the waveforms at the switching times as well as just a few microseconds just before each switching time.
When Q1 turns on it quickly charges up C2 and that being in series with C1 it charges up C1 a little more too. C1's extra energy would quickly dissipate through the zener (having it's voltage now above the zener value) and through the base of Q2 which of course turns Q2 on harder (your positive feedback). Not all of the energy in C1 dissipates, just the extra energy. This might take around 1/20 th of the cycle or so (maybe a microsecond or two).
Now that Q1 is turned on, it stays on and the voltage across L1 causes more current to flow to the output thus eventually raising the output voltage more and more. Meanwhile, since the extra energy in C1 is gone Q2 goes into a more relaxed mode and as the output reaches a certain voltage level Q2 base current gets lower and thus it's collector current gets lower and thus Q1 will find it easier to pull out of sat as the inductor current in L1 rises as well.
Finally the tripover point is reached where the base drive in Q1 and the inductor current forces Q1 to start to turn off. This small change causes the collector voltage of Q1 to fall slightly too because of the load from L1. This small change is immediately felt at the base of Q2 because of the coupling from C2, which pulls the base down slightly. Thus some of the energy from C1 is pulled through C2 now and the current goes through the inductor to the output because the inductor provides the path to the output load to ground. Thus some of the energy from C1 gets to the output during this short time when Q1 starts to turn off.

Q1 isnt off yet, but is turning off. The collector voltage has fallen slightly and the diode is still off. The path of current is up through C1 and down through R3 and through C2 and through the inductor and through the load to ground. There is still some current flowing through the base emitter of Q2 but it is decreasing.

The decreasing current in the base of Q2 causes it to turn off more, which in turn causes Q1 to turn off even more now (your positive feedback again even though it is turning off). This in turn causes the collector voltage of Q1 to go even lower, which causes more current flow in C2 and C1 and so more energy comes from C1 now and makes it to the load. The diode is still not conducting because the voltage at the cathode is still positive even though it is decreasing.

Now if we stop here and take a look at the circuit we see that the energy from C1 is making it to the output, but this period will not last very long relative to the whole cycle and this isnt the only time we have had to look at to completely determine the energy behavior of C1.

Eventually the collector voltage of C1 falls low enough to turn on the diode and after that C2 voltage no longer changes the same way so the inductor can only get it's current up through the diode, so now only the inductor supplies the energy to the output.
In the mean time, C1 starts to charge back up through R3 (the 10k resistor). C2 starts to charge through R3 and the diode (base of Q2 positive and right side of C2 negative).

So i think the main point is that the energy from/to either cap is quite small compared to the total output energy, and that some energy from C1 gets to the output and some through the zener.

Also, because the cap C2 causes a snap action of Q1 in either direction (on or off) C2 is responsible for the positive feedback. Because the emitter of Q2 causes less current to flow in Q1, the emitter of Q2 is responsible for the negative feedback similar to the function in a linear regulator.

 

[colin55]

Finally the tripover point is reached where the base drive in Q1 and the inductor current forces Q1 to start to turn off. This small change causes the collector voltage of Q1 to fall slightly too because of the load from L1. This small change is immediately felt at the base of Q2 because of the coupling from C2, which pulls the base down slightly. Thus some of the energy from C1 is pulled through C2 now and the current goes through the inductor to the output because the inductor provides the path to the output load to ground. Thus some of the energy from C1 gets to the output during this short time when Q1 starts to turn off.

The above needs to expanded.

Finally the tripover point is reached because the current through the inductor is not increasing and thus the flux is not expanding. Thus the voltage produced across it decreases and this reduced voltage is passed through C2 to lower the voltage on the base of Q2.
This action charges C2 but the energy from C2 does not get passed to the load though the inductor.
The inductor has a much-greater effect on charging the capacitor during this time. The capacitors C1/C2 have no “strength” or “ability” to pass their energy to the load.

 

[MrAl]

Hi Collin,

Im not sure where you got the idea that: "Finally the tripover point is reached BECAUSE the current through the inductor is not increasing..."

That statement suggests that the inductor current somehow magically stops increasing just before the transistor turns off which would be quite a trick unless there was some sort of resonance built in that was significant enough to work with the inductor of that size in the short time that we would have for this sort of resonance. There is a bit of resonance in all these circuits but the cycle time is too long to act independently enough to control the whole feedback process.

What controls the feedback trip point is the fact that the inductor current is increasing. This is very typical in these forced out of sat circuits. The inductor current continually increases because there is nothing to stop it when the output voltage is lower than the collector voltage of the drive transistor. 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. It would never stop until the output voltage rose as high as the collector voltage were it not for the drive transistor coming out of sat. It's not very common to see the collector of a transistor being used as the control pin of that transistor except in these kind of circuits. Usually either the base or the emitter is used as the control pin. In this circuit we have both the base and the collector controlling the action of Q1.

The time when C2 passes energy through the inductor is the time just after the Q1 transistor starts to turn off up until the time when the diode starts to conduct. Remember the collector of the transistor is not a current sink, it is a current source, whether it is all the way on or just partially on. The only sink then is the inductor (through the load resistance). That's the only thing that can pass or absorb the energy from C2 until the diode conducts. Think of what would happen if the inductor were suddenly removed from the circuit with Q1 somehow still turning off...the cap would stay charged up plus on the right negative on the left, with nothing to discharge it but it's own self discharge and perhaps a small leakage current of Q1.

I am certainly not arguing that the energy from either small cap is very significant, it is not significant to this circuit with the typical load, but some of that energy although small as it is does get passed to the output. If we were to somehow block this energy from getting to the output the efficiency would probably only change by a tenth of one percent, not worth mentioning in most cases, but it is in fact there.

 

[ericgibbs]

The above needs to expanded.

Finally the tripover point is reached because the current through the inductor is not increasing and thus the flux is not expanding. Thus the voltage produced across it decreases and this reduced voltage is passed through C2 to lower the voltage on the base of Q2.
This action charges C2 but the energy from C2 does not get passed to the load though the inductor.
The inductor has a much-greater effect on charging the capacitor during this time. The capacitors C1/C2 have no “strength” or “ability” to pass their energy to the load.

The above statement is incorrect, the current thru the inductor is increasing, so the Vout voltage is increasing.

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.

Its the Q2 emitter DC connection to the Vout that determines the switching point of Q2, not the current thru the inductor.

The capacitor C2, provides the positive feedback required when the Q2 switching occurs due to the emitter connection to the Vout.


hi Al,
You beat me too it while I was typing..
E