How does a transistor amplify current or voltage?

[Ratchit]

Jony130,

Well I think that it is another way around. When Q2 is first "ON" is in saturation and with time during the increase of current in Q2. Q2 "switch" from saturation to active region. And thanks to C1 an this "switch" to active region Q1 and Q2 start to cut-off.
So it is the increases of a collector current that cause BJT to cut-off.

I am sorry, but I cannot understand what you are trying to say. It just does not make sense to me. I cannot comprehend your reasoning and conclusion. However, Brownout appears to understand what you are saying, so mabe he can explain your thoughts.

Ratch

 

[MrAl]

MrAl,



That is an easy one to answer. When a transistor goes into saturation, it does not short. It has a small saturation resistance. According to the data sheet for the 2N4401, its saturation resistance is around 1.5 ohms at Ib = 500 ma. With a 3 volt supply, you are just not going to get an "infinite" increase in saturation current. So when the saturation does level off, as certainly will, a positive voltage will be applied to the base of Q1, and turn off both Q1 and Q2. By the way, what is Resr?

Next you propose three "experiments" for whatever reason. In order to keep disciplined and stay focused on the original question of how your oscillator works, I am going to ignore them, and reply only to your answer about what I said in regard to saturation resistance, or any mistakes I made. Since I was the first to explain in detail how your oscillator works, we should address my explanation first before going off on any other divergent tangents. So I expect your next posting to point out any mistakes in my analysis.



It is a comparative illustration with no meaning. The two equations are simply inverses of each other. No one tries to control Vbe with Ic.

Ratch

Hello again,


Well guess what? That's what we are doing indirectly. But to make the point clearer, we will now control the transistor directly with the collector current. This was the objective of the other experiments i pointed out to you. Why do you think i mentioned those experiments, just for the heck of it?

Resr was the inductor esr, series resistance.

Here's an illustration of how the transistor works in that oscillator boost circuit. In this circuit we test the theory by increasing collector current until something gives. Note we dont even have to think about the oscillator circuit anymore, just pay attention to what is happening to the transistor while the collector current is rising.

Note that blue is the collector emitter voltage, red is base emitter voltage (constant) violet is the current through the collector resistor R3, green is simply the supply voltage.
We increase the supply voltage and watch what happens as the collector voltage rises.

At first, Vce stays low, a few tenths of a volt, which is typical for a bipolar in saturation as the collector current rises a little. Vce rises too but only a little before the knee.
Now as we approach the knee, the Vce starts to rise more sharply, until it rises quite a bit and this is where it comes out of saturation. Note after the knee Vce rises following V2 (the supply voltage). Note this change is gradual, however with an inductor in place of R3 this change is more abrupt, and so the voltage rises more sharply.

So it's the collector current that causes the transistor to pull out of saturation and that circuit shows that this is of course what happens.

With the oscillator/boost circuit as i said before, increase the inductor series resistance and see how the circuit stops oscillating.

 

[BrownOut]

We seem to have a definition problem here. What you call the barrier voltage I would call the junction voltage. You don't think the Vbi is the barrier voltage, and you wish the term "junction voltage" would just go away. No one will ever convince you otherwise, but I think you do have the empirical relationship down correctly.

How do you know what I wish? Now you’re psychic too? I don't care what you call it; call it anything you wish, I'll stay with the vocabulary in Sedra and Smith, and the proven stated fact that barrier voltage is reduce by the excitation source.

According to the data sheet for the 2N4401, its saturation resistance is around 1.5 ohms at Ib = 500 ma. With a 3 volt supply, you are just not going to get an "infinite" increase in saturation current. So when the saturation does level off, as certainly will, a positive voltage will be applied to the base of Q1, and turn off both Q1 and Q2

That means the transistor is heating with power V^2/R, or using your own numbers, 9/1.5 = 6Watts. Before the current "levels off" the transistor becomes a little pile of ash. Something better happen to limit the current before then.

Or you could cite the series resistance of the AA cells, but that's not how it works either. Oscillators never depend on cell resistance for their operation.

 

[MrAl]

Hi again,


Brownout:
Yeah he he. So we see the transistor turn into a fuse after all

Ratchit:
This next circuit is the same as the circuit in my previous post, only this time i've replaced the collector resistor with an inductor with series resistance equal to 0.1 ohms (R3) as shown in the attached diagram. The source voltage (V2) is now fixed at +5vdc.
We concentrate on the region between the two red arrows as on the diagram. Note how the current levels off AFTER the transistor comes out of saturation (see dashed violet line indicating time sync), and note how fast the transistor comes out of sat using an inductor with low esr, and note that that speed happens even with a linear inductor (L1) while in practice it might be even faster as the inductor goes into saturation. Note however that the inductor doesnt really even have to enter saturation, it just has to have low esr.
Also note that again we are driving the transistor with a fixed Vbe.

The sequence of events for this experiment is as follows:
Turn V1 on and wait for it to stabilize, then later turn on V2. Note the time it takes for the transistor to pull out of saturation after V2 is turned on (about 2.2us).
This clearly illustrates the control of the transistor using the collector current, not Vbe or related.

 

[Ratchit]

MrAl,

Well guess what? That's what we are doing indirectly. But to make the point clearer, we will now control the transistor directly with the collector current. This was the objective of the other experiments i pointed out to you. Why do you think i mentioned those experiments, just for the heck of it?

Experiments are only as good as the conclusions. Read on.

Resr was the inductor esr, series resistance.

It would have been clearer if you designated it Rl and defined it in the first place. I assumed it was close to zero.

Here's an illustration of how the transistor works in that oscillator boost circuit. In this circuit we test the theory by increasing collector current until something gives. Note we dont even have to think about the oscillator circuit anymore, just pay attention to what is happening to the transistor while the collector current is rising.

Note that blue is the collector emitter voltage, red is base emitter voltage (constant) violet is the current through the collector resistor R3, green is simply the supply voltage.
We increase the supply voltage and watch what happens as the collector voltage rises.

At first, Vce stays low, a few tenths of a volt, which is typical for a bipolar in saturation as the collector current rises a little. Vce rises too but only a little before the knee.
Now as we approach the knee, the Vce starts to rise more sharply, until it rises quite a bit and this is where it comes out of saturation. Note after the knee Vce rises following V2 (the supply voltage). Note this change is gradual, however with an inductor in place of R3 this change is more abrupt, and so the voltage rises more sharply.

The plot had me scratching my head because the x and y axes were not labeled. However, I understand what you are try to aver.

So it's the collector current that causes the transistor to pull out of saturation and that circuit shows that this is of course what happens.

Here is an example of doing an "experiment" and coming to the wrong conclusion. I said back in post #283 of this thread that the only way to bring a BJT out of saturation was to cut off its base current or increase the reverse collector-base voltage. You are doing the latter, and concluding that it is the collector current that is the reason for coming out of saturation. It is really the increase in the c-b voltage that brings it out of saturation. From what I see of your plot, the current of R3 remains constant throughout.

Anyway, this does not apply to your oscillator circuit. The two diodes across Q2 will limit any increase of c-b reverse bias voltage, so that base current cut off is your only option. So I stand by my statement that it is the "inductive kick" of the coil that puts a positive voltage on Q1 and thereby cuts off both Q1 and Q2.

With the oscillator/boost circuit as i said before, increase the inductor series resistance and see how the circuit stops oscillating.

Not surprising, increasing the resistance in series with the coil will reduce the positive voltage it delivers.

Ratch

 

[Ratchit]

Brownout,

That means the transistor is heating with power V^2/R, or using your own numbers, 9/1.5 = 6Watts. Before the current "levels off" the transistor becomes a little pile of ash. Something better happen to limit the current before then.

Or you could cite the series resistance of the AA cells, but that's not how it works either. Oscillators never depend on cell resistance for their operation.

Tsk, tsk. Too bad you don't look at the spec sheets once in a while. Its saturation voltage is listed as 0.75 volts, giving it a power dissipation of 0.38 watts, which is well within what it can handle.

Ratch

 

[MrAl]

MrAl,

Here is an example of doing an "experiment" and coming to the wrong conclusion. I said back in post #283 of this thread that the only way to bring a BJT out of saturation was to cut off its base current or increase the reverse collector-base voltage. You are doing the latter, and concluding that it is the collector current that is the reason for coming out of saturation. It is really the increase in the c-b voltage that brings it out of saturation. From what I see of your plot, the current of R3 remains constant throughout.

You can say that the cb voltage indicates that the transistor has come out of saturation, but how do you think that cb voltage increased? It increased because Ic increased. That's the point.

Anyway, this does not apply to your oscillator circuit. The two diodes across Q2 will limit any increase of c-b reverse bias voltage, so that base current cut off is your only option. So I stand by my statement that it is the "inductive kick" of the coil that puts a positive voltage on Q1 and thereby cuts off both Q1 and Q2.
Ratch

It does apply, and we dont really need an inductive kick we just need an increase in voltage. That voltage gets coupled through the cap into the base of Q1.
You can see in the third (last) experiment posted that the inductor voltage does not 'kick', it just stalls. The 'kick' comes in only AFTER the cap couples the signal to Q1 and turns it off and Q1 turns off Q2. When Q2 turns off, then you get the inductive kick, and that's the only way. It's the Q2 cutting off that changes the current and that's how the kick happens. That's how it happens with any inductor even with a switch. When you turn the switch 'off' the kick occurs, and not before then. Look at the waveform in that diagram and see that since Q2 always has a bias the inductor does not kick. In the osc circuit Q2 bias is removed after the collector voltage rises so that's how the inductor gets to kick and give us the required boost in voltage. But that can only happen after Q1 turns off Q2, and Q1 turns off when Q2 collector voltage RISES sharply as it pulls out of saturation, but at that point Q2 still has a strong bias so it is still conducting.
Note also that there is no reason why Q2 can not conduct with either a low Vce or high Vce.

Again, the sequence of events is as follows:
1. Q2 turns on.
2. Current rises in inductor
3. Current exceeds that which Q2 can hold with it's current bias
4. Vce voltage starts to rise
5. Vcb gets higher, transistor comes out of sat.
6. Voltage rises Vce
7. Rising voltage couples through cap to Q1 base, turning it off quick
8. Q1 turning off turns off Q2
9. Q2 turning off gives rise to inductive kick back and high boosted voltage Vce

If you dont agree with that sequence of events then list your own idea of how you think it happens. You really should do a circuit simulation first though because it looks like you havent done that yet.

 

[Jony130]

So I stand by my statement that it is the "inductive kick" of the coil that puts a positive voltage on Q1 and thereby cuts off both Q1 and Q2.
Ratch

But this circuit can still oscillate if we replace the coil with resistor.

 

[Ratchit]

MrAl,

This next circuit is the same as the circuit in my previous post, only this time i've replaced the collector resistor with an inductor with series resistance equal to 0.1 ohms (R3) as shown in the attached diagram. The source voltage (V2) is now fixed at +5vdc.
We concentrate on the region between the two red arrows as on the diagram. Note how the current levels off AFTER the transistor comes out of saturation (see dashed violet line indicating time sync), and note how fast the transistor comes out of sat using an inductor with low esr, and note that that speed happens even with a linear inductor (L1) while in practice it might be even faster as the inductor goes into saturation. Note however that the inductor doesnt really even have to enter saturation, it just has to have low esr.
Also note that again we are driving the transistor with a fixed Vbe.

The sequence of events for this experiment is as follows:
Turn V1 on and wait for it to stabilize, then later turn on V2. Note the time it takes for the transistor to pull out of saturation after V2 is turned on (about 2.2us).
This clearly illustrates the control of the transistor using the collector current, not Vbe or related.

Again, the back voltage of the coil is preventing the V2 voltage from reaching the collector, so the transistor stays in saturation because it has no c-b reverse bias. When the coil's back voltage lowers enough, more voltage is applied to the collector and brings the transistor out of saturation. The increase in current did not break the saturation, the establishment of a c-b reverse bias did.

The y-axis are not labeled, and the numbers in the y-axis are cut off. Does 30.000m mean milliamps or microamps? Does u mean microseconds in the x-axis? Why is the i(R3) curve so straight at the beginning? It should show a exponential charge curve with a L/R time constant. Better check your plotting program.

Ratch

 

[MrAl]

MrAl,



Again, the back voltage of the coil is preventing the V2 voltage from reaching the collector, so the transistor stays in saturation because it has no c-b reverse bias. When the coil's back voltage lowers enough, more voltage is applied to the collector and brings the transistor out of saturation. The increase in current did not break the saturation, the establishment of a c-b reverse bias did.

The y-axis are not labeled, and the numbers in the y-axis are cut off. Does 30.000m mean milliamps or microamps? Does u mean microseconds in the x-axis? Why is the i(R3) curve so straight at the beginning? It should show a expontential charge curve with a L/R time constant. Better check your plotting program.

Ratch

Hello again,

You can not get an increase in Vcb without more current through the coil!

30.000m means milliamps, 3u means microunits, and these are electrical standards.
3u on the plot x axis (time) stands for 3 microseconds.

i(R3) is exponential but if you knew what was happening here you would realize that
'exponential' has a relatively constant increase toward the beginning. You seem to
put your own 'knowledge' higher than the $2000 dollar circuit simulator. At the
very least, you should investigate.

The back voltage of the coil decreases but only as the transistor comes out of saturation.
You need to look at the simulation to see that. The voltage across Vce does not simply
appear for no reason at all, and the coil voltage does not fall for no reason at all, and we
can keep the coil voltage higher if we bias the transistor harder. That's how we can get
more output too.

When the current increases then and only then does the Vce rise.

You need to do a few simulations. Have you ever even done a single simulation?

 

[BrownOut]

Tsk, tsk. Too bad you don't look at the spec sheets once in a while. Its saturation voltage is listed as 0.75 volts, giving it a power dissipation of 0.38 watts, which is well within what it can handle.

Too bad you never learned basic electronics. The spec sheet gives resistance for the transistor as 1.5 ohm, according to you. When the coil current "levels off" the full 3V is across the transistor. The current is:

(3V - 0.75V)/1.5 = 1.5A. Power is (1.5)^2*1.5 ohms = 3.375Watts, and that's best case for IB=50ma, which is not guaranteed. That's still many times more then it's designed for.

You should read a book on basic electronics someday. Then you might know this stuff.

 

[Ratchit]

MrAl,

You can say that the cb voltage indicates that the transistor has come out of saturation, but how do you think that cb voltage increased? It increased because Ic increased. That's the point.

It increased because the back voltage of the coil decreased, allowing the supply voltage to increase the reverse c-b voltage. The current increase was incidental.

It does apply, and we dont really need an inductive kick we just need an increase in voltage. That voltage gets coupled through the cap into the base of Q1.

You are right about the inductive kick. It does not happen unless the current through the coil is interrupted. But the back voltage does not last forever, and when it decreases enough, the reverse c-b bias reestablishes itself, and the increase is coupled to Q1 so as to turn off the base current of Q2 and hasten its departure from saturation.

You can see in the third (last) experiment posted that the inductor voltage does not 'kick', it just stalls. The 'kick' comes in only AFTER the cap couples the signal to Q1 and turns it off and Q1 turns off Q2. When Q2 turns off, then you get the inductive kick, and that's the only way. It's the Q2 cutting off that changes the current and that's how the kick happens. That's how it happens with any inductor even with a switch. When you turn the switch 'off' the kick occurs, and not before then. Look at the waveform in that diagram and see that since Q2 always has a bias the inductor does not kick. In the osc circuit Q2 bias is removed after the collector voltage rises so that's how the inductor gets to kick and give us the required boost in voltage. But that can only happen after Q1 turns off Q2, and Q1 turns off when Q2 collector voltage RISES sharply as it pulls out of saturation, but at that point Q2 still has a strong bias so it is still conducting.
Note also that there is no reason why Q2 can not conduct with either a low Vce or high Vce.

I said I was not going to look at another experiment until by explanation has been picked to death. I broke my word twice already. I will not do it again.

Again, the sequence of events is as follows:
1. Q2 turns on.
2. Current rises in inductor
3. Current exceeds that which Q2 can hold with it's current bias

Yes, that sends Q2 into saturation.

4. Vce voltage starts to rise

Tilt! Vbe is a fraction of a volt in saturation. Saturation lowers Vce greatly. Even in the active region, the Vce lowers when current increases.

If you dont agree with that sequence of events then list your own idea of how you think it happens. You really should do a circuit simulation first though because it looks like you havent done that yet.

I already did so in post #289 and #295 of this thread. You are right about a circuit simulation, I have not done one for a long time. It will take some time to get up to speed on it.

Ratch

 

[Ratchit]

Jony130,

But this circuit can still oscillate if we replace the coil with resistor.

Can you supply a description on why that happens?

Ratch

 

[BrownOut]

Ps, the number rachit gave for saturation voltage was in fact, BE saturation voltage. Too bad he can't read a datasheet. The real number for CE sat voltage is around .25v for for IC=500ma. So I change my math to account for the correct VCE saturation voltage:

(3V - 0.25V)/1.5 = 1.8A. Power is (1.8)^2*1.5 ohms ~= 5Watts, which is closer to my original power calculation. The transistor becomes just a pile of ash at that power level.

 

[Ratchit]

MrAl,

You can not get an increase in Vcb without more current through the coil!

This is turning in to philosophical argument instead of a technical one. Read further on.

30.000m means milliamps, 3u means microunits, and these are electrical standards.
3u on the plot x axis (time) stands for 3 microseconds.

Thanks, I thought it might be that, but I was not sure.

i(R3) is exponential but if you knew what was happening here you would realize that 'exponential' has a relatively constant increase toward the beginning. You seem to
put your own 'knowledge' higher than the $2000 dollar circuit simulator. At the
very least, you should investigate.

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 back voltage of the coil decreases but only as the transistor comes out of saturation.
You need to look at the simulation to see that. The voltage across Vce does not simply
appear for no reason at all, and the coil voltage does not fall for no reason at all, and we
can keep the coil voltage higher if we bias the transistor harder. That's how we can get
more output too.

When the current increases then and only then does the Vce rise.

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.

You need to do a few simulations. Have you ever even done a single simulation?

Not for a long time. I will have to get something a lot cheaper than $2000.

Ratch

 

[Ratchit]

Brownout,

The spec sheet gives resistance for the transistor as 1.5 ohm, according to you. When the coil current "levels off" the full 3V is across the transistor. The current is:

(3V - 0.75V)/1.5 = 1.5A. Power is (1.5)^2*1.5 ohms = 3.375Watts, and that's best case for IB=50ma, which is not guaranteed. That's still many times more then it's designed for.

Your calculations are bogus. In the first place, the voltage across the transistor is limited by the two diodes to 1.5 volts. In the second place, Vce is lowered while in saturation due to the back voltage of the coil. When the coil back voltage decreases enough, the transistor comes out of saturation for the reasons I explained before. So what's the problem?

Too bad you never learned basic electronics.
You should read a book on basic electronics someday. Then you might know this stuff.

Do really believe all the agitpap you put out?

Ratch

 

[Ratchit]

Brownout,

Ps, the number rachit gave for saturation voltage was in fact, BE saturation voltage. Too bad he can't read a datasheet. The real number for CE sat voltage is around .25v for for IC=500ma. So I change my math to account for the correct VCE saturation voltage:

(3V - 0.25V)/1.5 = 1.8A. Power is (1.8)^2*1.5 ohms ~= 5Watts, which is closer to my original power calculation. The transistor becomes just a pile of ash at that power level.

I used the data sheet for the 2N4401 by General Semiconductor. See for yourself. http://www.datasheetcatalog.org/datasheet/GeneralSemiconductor/mXvruwu.pdf

You equations are bogus anyway. The voltage across Vce is limited to 1.5 volts by the diodes, the coil protects it from the B+ voltage while in saturation, and the transistor comes out of saturation before it can get hit by those power levels.

Ratch

 

[BrownOut]

Your calculations are bogus. In the first place, the voltage across the transistor is limited by the two diodes to 1.5 volts. In the second place, Vce is lowered while in saturation due to the back voltage of the coil.

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. That's lame beyond all belief. No matter how low vce goes, the power that is dissapated increases with the current, and as every 10th grade physics student knows, power is 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.

 

[tvtech]

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 vcesat at 1.5V. That's lame beyone all belief. 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.

Like I said guys...he is baiting.

Cheers

 

[BrownOut]

Brownout,



I used the data sheet for the 2N4401 by General Semiconductor. See for yourself. http://www.datasheetcatalog.org/datasheet/GeneralSemiconductor/mXvruwu.pdf

The General Semiconductor has a misprint for VCEsat. That number is impossible.
I use the fairchild statsheet http://www.fairchildsemi.com/ds/2N/2N4401.pdf. It gives vcesat = .25V

You equations are bogus anyway. The voltage across Vce is limited to 1.5 volts by the diodes, the coil protects it from the B+ voltage while in saturation, and the transistor comes out of saturation before it can get hit by those power levels.

Ratch

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.