You're whole understanding of electronics is bogus. Vce(sat) is given on the data sheet as .25 volts, and that's consistant with pretty much all transistors. You will NEVER find a datasheet that lists says VCE is limited to 1.5.
No matter how low vce goes, the power that is dissapated increases with the current, and as every 10th grade physics student knows, power it I^2*R, I is current and R is the saturation resistance. You really should spend some time learning basic electronics before you try to appear to know what you're talking about, because you clearly don't.
You're whole understanding of electronics is bogus. You really should spend some time learning basic electronics before you try to appear to know what you're talking about, because you clearly don't.
I never said that the data sheet limits Vce to 1.5 volts. I said that the 2 diodes across the transistor limit the voltage to 1.5. Didn't you understand I was talking about the circuit? You should read the posts better. The transistor does not get hit with 3 volts like you calculated earlier.
Yes, but what is the current? What is going to supply the excessive current you imply is going to exist? How long is the current going to last?
The General Semiconductor has a misprint for VCEsat. That number is impossible.
I use the fairchild statsheet https://www.fairchildsemi.com/ds/2N%2F2N4401.pdf. It gives vcesat = .25V
That's so naive, the voltage across the transistor can be anywhere from vcesat to VCC, whatever that is. There is no limit to vce aside from vcesat. You're understand of basc electronics is sorely lacking.
Strange, I get read the same misprint in the link you presented. We need a third party opinion. Anyone?
But in this circuit, it is limited to two forward biased diode drops.
The diodes DO NOT limit vce to 1.5V. That's more fantasy crap to cover your incorrect statements about the circuit.
I already showed the current to be 1.8A. It comes from the supply. It's going to last until something blows up (either the transistor or the battery). You have to understand how transistors truly operate in saturation. This is really, really basic.
It won't matter that AA batteries may not supply 1.8A. The circuit should be operated from any 3V supply, regardless of the current it can supply. So my earlier caluclation of 1.8A for current and ~5W for power is correct. No fantasy voltage limit changes that.
It is strange. The chart says .75 volts, and the graph below says .25V. Oh well.
No transistor is ever limited to two forward biased diode drops. The power is 5W, as I've previously shown, regardless of what vce is limited to.
Like I said guys...he is baiting.
Until MrAl shows me differently, I will stick with my calculation of 0.5*0.75 = 0.38 watts. But consider this. Why don't you also say that the two diodes will fry from the 3 volts source? Evidently there must be some resistance from the coil that limits the current.
The circuit limits the voltage across the transistor. The voltage across the transistor without the diodes can be much higher. Or are you talking about how low the voltage across the transistor can be?
Well, that explains the descrepancy. My data sheet does not even have charts. Using the chart figures, my calculations for the power dissipated during saturation are 0.5*0.25 = 0.13 watts.
Why don't you also say that the two diodes will fry from the 3 volts source?
MrAl,
This is turning in to philosophical argument instead of a technical one. Read further on.
Thanks, I thought it might be that, but I was not sure.
Good, a solid explanation. The time constant is too long to see the curvature during the time span. Actually, the rate of current increase of a LR circuit is greatest at the beginning and tapers off to zero at the end. I don't need a $2000 circuit simulator or a $50000 Cadence to know what should happen. I did investigate by asking you.
The saturation condition of a BJT is a stable condition. There is a large range of current values that a saturated transistor can support. The coil goes through a range of current values during the cycle. We can substitute a resister and a variable B+ supply for the coil, and duplicate the current range of the coil while still keeping the transistor in saturation. There is only one voltage point where the transistion from saturated to active occurs, and that is when the c-b voltage becomes reversed biased. It is not important what the current through the coil is. The current can be a lot of different values. It is the what voltage the the coil allows to the collector that determines whether the transistor is in saturation or not.
Not for a long time. I will have to get something a lot cheaper than $2000.
Ratch
There is only the minute resistance of the wires, and the minute resistance of the coil. Neither can limit the current to a safe level. The transistor is toast, as you describe it's operation.
It doesn't matter what vce sat is anyway. The current is 1.8A, giving I^2*R = 5W. To believe that only vce from the datasheet can be used demonstrates a lack of understanding of transistor operation.
There is a single LED. Who knows what it's forward voltage is? It's not given, so I'm not concerned about it. VCEsat is going to be much less that the forward voltage anyway, that's why the diode is turned off while the transistor is saturated.
The purpose of the circtuit is most likely to step up the 3V to overcome the LED's forward voltage anyway.
How do you know what the resistance of the coil is? And why don't you also wonder about the current through the diode(s) is/are? They are in series with the power supply and the low resistance coil.
It's not really philosophical though, the supply voltage was always there, it just took time for the current to build up to a high enough value to force (ok) Vcb up to a high enough value. Vcb doesnt occur there and then the coil current follows, it's the other way around and the simulation shows this. It shows "everything is just fine" until the current gets too high, and then the transistor is forced out of saturation. Now you can measure Vcb if you like and see that it does become reversed, but that's after the fact. Vcb does not force Ic, Ic forces Vcb...that's the point.
If we look at a child's see-saw, when one child pushes down on the ground with the feet their end goes up. The other child's end goes down because the beam is rigid. It's not gravity that pulls the other child's end down, it's the other child pushing up that really does it, as both children can weight the same and the beam (theoretically) can be very long.
You can not say that the other child's end goes down because of gravity or any other reason, the real forcing function comes from the child that pushed down with the feet (and thus rose up).
Saying that the Vcb somehow made Ic go up is just like trying to say that the other child's gravity pulled it down. It's really Ic doing the work. Charges accumulate.
Ratchit, please excuse me for asking what your career covered. I know you earned a BSEE, but after that, what was the nature of your work? If it's something you'd rather not discuss, I understand.
I'm just curious as to what your info is based on. Disagreeing with me is not a problem if you can show me peer-reviewed references, herein called "PRR", that support your position. You have cited various PRR, but they don't state what you want to interpret them as stating. Although your cited PRR give valid info, you draw conclusions that are not supported by said PRR.
Have you designed a LOT of bjt networks that sold in the market place?
I have, & from what I'm reading Mr. Al, & Brownout have a good background in electronics as applied to device behavior.
None of us are infallible, but I don't enjoy telling you that your comprehension of electronics is not as deep as you believe.
You have some knowledge, but not enough to tell veterans what's what. To stand on a soapbox & lecture the EE/physics community telling them that the OEMs & unis have "myths, misconceptions, misunderstandings", then proceed to lecture us on "this is how it really works" is a bit over your head.
You don't have the technical chops to debate tried & proven principles employed by EE divas such as myself & the others just mentioned.
You impress me as an intelligent person w/ an ability to practice science. But raw ability is only a start. Much post graduate study & decades of applying theory into practice is needed as well. I'm not saying that you cannot practice EE, but only that you have not yet arrived.
do not wish to WIN an argument. My motive is to let newbies considering entering science, know that these issues have been thoroughly examined & there are good reasons why things are taught the way they are. I cannot take credit for the CC external bjt model, nor the QC internal model. These are the work of great thinkers before my time. I have 2 published papers, & 3 patents + 2 pending. I've made contributions to electronics state of the art, but I am no superstar by any stretch. I never claimed to be anything more than plain old Claude.
If what the PRR say was wrong, the EE canon would have been corrected decades ago. You're literally telling the EE community, "no, you're wrong, this is how it really is!" Indeed! Nothing personal, but if you know more than TI, Natl Semi, etc., why are you not the one designing IC's, or writing tech briefs? The EE community would benefit immensely from a superstar who could purge all the myths & set the record straight. I'd be grateful.
Coils have low resistance since they are made of copper wire. Proper design would NEVER assume the wire has enough resistance to protect the device. Again, really basic stuff.
There aren't diodes, there is a single LED.
Yes, I already said previously that transistor saturation is what turns the diode(s) off.It will always have a higher voltage than the bjt's sat voltage, or else the circuit doesn't work.
The purpose of the circuit it to step the voltage higher than the LEDs forward voltage, as I've already pointed out.
The fact that the LED is in series with the coil is a valid point. If it were my design, I would probably insert a low value resistor in series with the LED for protection
Brownout is more of a complainer than explainer.
I had a long post going here....
Then deleted it. Quickly.
Good luck Ratch.
You ain't making friends here though.
Small bit of advice Ratch....you are trying to be a smart ass. And that never works. Without factual proof.
Done and dusted.
BTW, how old are you?
And what are your qualifications???
Cheers and sleep well.
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