A Perspective On The B Type EESD - Robert Murray-Smith - Any issues?

[MileHigh]

MH

You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit.
The battery is also a resistor,and as the voltage drops,the internal resistance grows,and so do the I/R losses associated with that battery.

In the second circuit,these losses are only had when the transistor is on.
In the first circuit(your circuit) these battery I/R losses are not only included when the transistor is on,but they are also included when the transistor is off.

During the off time of the transistor in circuit 2,the current loop excludes the battery,and thus the losses associated with the batteries internal resistance.
In the first circuit(your circuit) the battery becomes part of the current loop during the off time of the transistor,and so also includes the I/R losses associated with the battery.

Why you find this so hard to understand-i guess we will never know.\

Brad

Okay Brad, here we go.

For starters, you are just playing a game when you say, "You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit."  It's all complete crap and you know it and everybody knows it.  It's a defensive measure by mounting a fake offense, nothing more than that.  It's you making a shameless spectacle of yourself.

Secondly, it's the old cliche at this point.  What I said to you passed right though you like you weren't even there.  That puts you and your understanding in question, again.  It's like you have an aerogel brain, and an idea can travel for 14 light-years though your head before it strikes a neuron.

In the second circuit,these losses are only had when the transistor is on.
In the first circuit(your circuit) these battery I/R losses are not only included when the transistor is on,but they are also included when the transistor is off.

Yes, but if the two circuits have the same average current draw, that means by definition that there must be a higher current draw when the transistor is ON in the second circuit.  Higher current draw equals higher internal losses when the transistor is ON.

So you have a case of lower internal losses with continuous current flow and higher internal losses when the current flow toggles ON and OFF.  They balance out.  It's like six of one and half a dozen of the other.  How many light-years will this concept have to travel in Brad's brain before it hits a neuron?

Why you find this so hard to understand-i guess we will never know.

I think we know why you find this so hard to understand.  You have to work on improving your ability to analyze and think beyond just your first impressions.  You have to work on applying the knowledge that you already possess.   You have to work on reading and understanding what people say to you.  And you most certainly have to work on your language skills so you can express yourself properly.

MileHigh

[tinman]

Okay Brad, here we go.

For starters, you are just playing a game when you say, "You really are lost,and it is becoming more apparent each day as to how little you understand the simple JT circuit."  It's all complete crap and you know it and everybody knows it.  It's a defensive measure by mounting a fake offense, nothing more than that.  It's you making a shameless spectacle of yourself.

Secondly, it's the old cliche at this point.  What I said to you passed right though you like you weren't even there.  That puts you and your understanding in question, again.  It's like you have an aerogel brain, and an idea can travel for 14 light-years though your head before it strikes a neuron.







MileHigh

Yes, but if the two circuits have the same average current draw, that means by definition that there must be a higher current draw when the transistor is ON in the second circuit.  Higher current draw equals higher internal losses when the transistor is ON.

Only not

Quote TK:==PSU
So, Circuit 1 ran at an average input power of 90 mW and produced 63.9 lux at the sensor
Circuit 2 ran at an average input power of 40 mW and produced 30.0 lux at the sensor

Quote TK:==battery
Circuit 1 gave 49.3 Lux at an average input power of 54.6 mW for an efficiency of 903 Lux/Watt.
Circuit 2 gave 26.1 Lux at an average input power of 28 mW for an efficiency of 932 Lux/Watt.

Now why do you suppose that is MH? 

So you have a case of lower internal losses with continuous current flow and higher internal losses when the current flow toggles ON and OFF.  They balance out.  It's like six of one and half a dozen of the other.  How many light-years will this concept have to travel in Brad's brain before it hits a neuron?

Yep--you are lost.
In fact,i cant even believe i am reading this from you,it is like you went from some sort of reality,to a land of MH dreams. Do you have any idea as to how high the internal resistance go's in a AA battery when the voltage is down around .9 volts MH ?--lets see your mighty pen work that one out without the bench. Well i do know the answer to that MH,and i know it because i can measure it on the bench. So give it a shot MH,show us how mighty your pen is without the bench,and give us a round about internal resistance value of a AA battery at around 900mV--lets not get to carried away with different types of batteries ATM,just use an eveready heavy duty battery as an example.

I think we know why you find this so hard to understand.  You have to work on improving your ability to analyze and think beyond just your first impressions.  You have to work on applying the knowledge that you already possess.   You have to work on reading and understanding what people say to you.  And you most certainly have to work on your language skills so you can express yourself properly.

My language skills are just fine that you MH,and the fact that you keep bringing that up,is just a sign of weakness.
It is also true that it is you that dose not listen to others--you know,the guys on the bench.
It is also becoming very apparent that i could indeed run rings around you on the bench.

MH,i have made this very easy for you to follow and understand.
Below are the two circuits,where in both cases the same amount of energy will be delivered to the inductor--lets say 100mJ of energy. I have included the batteries internal resistance that i measured !on the bench! of a normal alkaline battery with 900mV across it,and subtracted 10% of that calculated value for error margin.
Now we have the inductor in both cases storing the same amount of energy.
Can you please tell everyone here,which circuit will deliver the greatest amount of that stored energy to the LED?.

It is extremely clear MH that you throw all common sense out the window,when you need to be right outways your intelligence.
I can also see now,why you will not take me on in a simple little JT challenge--even when some one else was going to do the building for you. All you had to do,was put your !claimed! knowledge to paper-->you know,that mighty pen

So MH,i am afraid that i have clearly shown that the bench just kicked your ass ,and that my learning's on the bench just showed how your pen was not even in the race.
I am still amazed at how you made such a blunder,and as to how you came up with circuit 1 being more efficient(in any way) than circuit 2.

For me, there are only two outstanding issues and I will mention them again and I will put them in a better sequence this time:

1.  MH gets up the learning curve and understands as to how the JT circuit work's,and can be made more efficient,and clearly demonstrates that he understands what he is doing.
2.  MH admits that he is wrong when he stated that circuit 1 is more efficient than circuit 2.


Brad

[tinman]

author=MileHigh link=topic=16225.msg488167#msg488167 date=1468116506]

I seriously doubt that TK made a mistake in his power measurements.  It doesn't make any sense.  All that he had to do was scope the voltage and current from the power supply for both circuits.

Indeed,and when he carried out the power measurements while using the power supply,circuit 2 was more efficient at various voltages.
BUT-when he carried out those same measurements while using a battery,the outcome was the reverse
Now,why do you suppose that was MH?
I wonder if TK took into account as to how,or what the current was traveling through during the off time in each test in circuit 1 
If the two test were carried out where the supply voltages where the same,then what changed between the PSU and battery?
You say i should use my brain a bit better?,well perhaps you should use yours a little better,and think before plastering such rubbish all over this thread.

your logic is flawed and makes no sense.

My logic is just fine,but seems your has gone missing.

I was not talking crap at all.  I tried desperately to explain to you what the true reason for having a base resistor was but you would have nothing to do with that, it was pure willful ignorance on your part.

And it is rubbish MH.
The less resistive losses you have in a circuit,the more efficient that circuit becomes.
I so wish you would have taken up the challenge presented to you MH,but know i see why you did not want to go there.

You are just talking stupid gratuitous foolish idiocy, you need to blow off some steam and thar she blows!

It would seem ATM MH,that your boiler has exploded.

My pen is absolutely fine.  You need to work on thinking straight and being able to string five sentences together that make sense.  Those are some of your biggest problems.

Only you have a problem understanding MH--no one else dose.


Brad

[MileHigh]

Brad:

Only you have a problem understanding MH--no one else dose.

You don't know that.  If nobody says anything you don't know one way or the the other.  Critical thinking skills, Brad, say something that makes logical sense.

Look at this mangled pathetic excuse for the English language:

As  i pointed out to MH,the second circuit eliminates the losses associated with charging the battery also,when a battery is used in place of your PSU. The battery will produce more waste heat,when being discharged and charged continuously,and also there are the internal resistive losses that grow as the battery voltage drops.

You've got problems.  You can't even get "charging" vs. "discharging" right in your head.  The second sentence is an illogical disaster.

I asked you how you would measure a supercapacitor on your bench and you had nothing to say except for one obtuse and near-ridiculous statement.  So that is telling me you can't conceptualize by yourself how to measure the size of a supercapacitor and any other possible parameters.  So much for your "bench smarts," you are coming up goose eggs for the capacitor question.

So, knowing that, please tell me how you measured the output resistance of your battery.  I take nothing for granted with you and you will have to state how you did it so I can verify that you got it right.  And there is another challenge for you.  Don't you dare rattle off one vague and ambiguous sentence.  You have to properly articulate how you go about measuring the output resistance of a battery.  You have to put a sequence of sentences together that actually make proper sense.

MileHigh

[MileHigh]

Okay Brad.

Let's just sample some approximate numbers and work out a very simple problem for illustrative purposes.

Let's say we have a source voltage of 1.5 volts and an output impedance of seven ohms.
Let's say that we have circuit #1 that draws 50 milliwatts from the power supply and draws continuous DC current.
Let's say that we have circuit #2 that draws 50 milliwatts from the power supply and draws current with an 80% ON time and a 20% OFF time.

Let's examine these two circuits.

Circuit #1:

The current is 0.050/1.5 = 33.3 milliamps
The power lost in the internal resistance of seven ohms is 0.0333^2 x 7 = 7.78 milliwatts

Circuit #2:

We know from above that the average current is 33.3 milliamps.
Therefore the ON current for 80% of the time is 0.0333 x 5/4 = 41.7 milliamps.
The power lost to the internal resistance of seven ohms is 0.0417^2 x 7 x 4/5 = 9.72 miliwatts

Well look at that Brad.  When you put the two circuits on an even playing field where they draw the same amount of power from the fixed 1.5 volt power supply, circuit #2 that has the 80% ON, 20% OFF duty cycle has more losses due to the internal resistance of seven ohms.

Brad, I have made this very easy for you to follow and understand.

MileHigh