I think it would because with no feedback if the resistor is of a low enough value the transistor will come out of sat. So it may depend on the value of the resistor.
You have to realize though that you can not make a statement like that when you talk about saturation, ie collector current 'only'. Yes, 100ma is within the 'range of saturation' but then so is 10ma, and 1ma, but without the proper drive we still cant obtain (nor maintain) saturation.
In other words, we have to specify collector current and also base current to tell if the device is in sat.
Example:
100ma collector current and 10ma base current, transistor is (most likely) in saturation.
100ma collector current and 1ma base current, transistor might be in sat.
100ma collector current and 100ua base current, transistor probably is not in sat.
100ma collector current and 10ua base current, transistor is most definitely not in sat.
100ma collector current and 0.00 base current, transistor is not in sat.
So it's a matter of BOTH Ic and Ib, not just Ic.
I think what Jony was saying is that for an ideal coil (as in simulation) the coil can support any voltage you throw at it so there's no reason for that coil to force a zero voltage (across it) condition. Also, you have to keep in mind that even without that feedback circuit (Q1) the transistor comes out of sat, and that's with no feedback whatsoever.
We assume that at time 0 Q1 and Q2 are full "ON".
And when Ic_Q2 reaches the point 2 Q2 will start to come out of saturation.
So the primary cause for all BJT start to cut-off is Q2 that leaves saturation region and enter into active region.
MrAl,
You are making my point for me by listing all those currents. Neither Ic or Ib determine saturation or not. When Vcb is zero or less, and the transistor loses its reverse bias, it will go into saturation. Ib and Ic don't affect saturation unless they pull down the reverse bias. Do a differential voltage measurement between the base and collector. You will find its zero point syncs perfectly with whether the transistor is saturated or not.
Why are we talking about ideal coils? The coil has resistance, as does the voltage source and the transistor. That establishes a current limit. The coil's back voltage eventually becomes close to zero as it approaches its current limit.
By disconnecting the capacitor, I see that the transistor operates permanently in the active region with no excursions into saturation. I surmise that on startup, the capacitor momentarily sends Q2 into saturation by making its collector the same voltage as the base of Q1 (0.7 volts). Then it latches and stays in saturation until the back voltage across the coil decreases.
Ratch
Q2 will come out of saturation when its c-b reverse bias is restored regardless of what Ic is.
Q2 is always either in the cutoff or saturation region.
Yeah, dont try to turn it around to make your own point.
You keep saying Vcb and i already agreed with you, but it takes Ic to get to Vcb that you talk about. You just dont see it and you never will.
We talk about ideal coils when the resistance in the coil is low enough to be considered with no resistance. The theory goes with either some resistance or no resistance and you are the only one saying now you dont want to work with zero resistance. Look at it this way, 1.4 ohms would draw 3/1.4 amps at 3 volts which is at least 2 amps. The transistor wont allow that, and it is not the inductor series resistance that limits the current! If you think so, lower the resistance more then to 0.01 ohms. That's 3/0.01 amps, which the cells would not be able to supply, yet their minuscule series resistance (0.2 ohms) wont limit it either: 3/0.2 is 15 amps. That's not limiting.
The coils voltage does not really have to go to zero it just has to conduct more current, and 2 amps is pretty high already, yet it never gets there. Why? If that was the current limit then the transistor would see 2 amps, yet it doesnt. Why?
By disconnecting and operating as you say, you are working with a different setup then. You have to stick to what the transistor actually gets no matter what circuit you use. By using the correct circuit(s) you will see the transistor behave no matter what circuit you choose. By chosing the wrong circuit or different operation, you wont see it happen the same way. The circuit has to start from low current as the real circuit does if you are to use a different circuit, not from full supply voltage.
You just wont see this no matter what anyone says because you are only looking for ways to proved whatever it is you want to try to prove, which im not even sure what that is.
I already agreed with you about Vcb yet you keep bringing that up again. Without Ic Vcb can never rise. Try it, try to get Vcb to rise without increasing Ic at the point of operation where it matters.
Also, run the osc circuit as it should run and look at the collector current and bias on the base of Q2. The collector current increases, it's Vce rises, the transistor comes out of sat, and that means Vce rises more, fed back through the cap Q1 turns off, that turns off Q2. Look at the simulation you run and look at the one i posted. It's pretty clear.
Look at Jony's post! That plot shows Vce vs Ice. Now try to get out of that one
If you increase Ice (as the graph y axis) you see the sat voltage rise and then break out of sat at the knee(s).
Try to argue against that one.
This is exactly what i had been talking about here, and this shows how the circuit forces the transistor out of sat.
Like i said, if you want to 'detect' Q2 coming out of sat with a measurement of Vcb, that's fine with me, but it will take increaseing Ice as in that plot to get it there. You're going to have to acknowledge this sooner or later.
Your Point 1:
The series resistance of the coil limits the current.
Counter point proven:
No it doesnt, because its resistance is too low, 1.4 ohms, which is 2 amps and the transistor never gets there.
Your Point 2:
The series resistance of the batteries limits the current.
Counter point proven:
No it doesnt, because their resistance is too low, 0.2 ohms, and that would be 15 amps and the transistor never gets there.
Your Point 3:
The series resistance of the cells AND the coil limits the current.
Counter point proven:
No they dont, because their resistance together is 1.6 ohms, which is roughly still close to 2 amps and the transistor never gets there.
Your point 4:
The Vcb is an indicator of the saturation state of the transistor.
Counter point:
None, agreed.
Your point 5:
The Vcb takes the transistor out of saturation itself.
Counter point:
Not really, but partially agreed because it's too hard to prove otherwise, but in any case Vcb can only get to the required value by an increase in current Ice.
Your point 6:
It's Vcb that takes the transistor out of saturation itself, not Ice.
Counter point:
You can not EVER reach Vcb required unless you increase Ice.
Your point 7:
You dont have to increase Ice because of the feedback which cuts off Q1 and hence Q2.
Counter point:
Q2 comes out of saturation before the feedback arrives at Q1, as proven by simulation. In fact, if you look closely at the simulation you can see the transistor behaving exactly as Jony's plot of Ice vs Vce indicates it should. You can see the Vce start to rise BEFORE the feedback arrives at Q1.
Your point 8:
You dont agree with Jony's plot of Vce vs Ice.
Counter point:
It's a simple two dimensional plot that the manufacturer supplies to show typical operation of the transistor.
Your point 9:
The transistor region of saturation includes Ic of 100ma and even higher, so the transistor would stay in saturation.
Counter point:
You can not conclude that the transistor stays in sat unless you know Ib also, and given that Q1 is fully turned on, Ib will be constant and will not change until the feedback arrives. The transistor operates at constant Ib and it's only Ic that changes.
point 10:
The general operation of the osc circuit is that something other than Ic causes the rise in collector voltage and therefore the rise in Vcb.
Counter point PROVEN:
It is actually Ic that causes the rise in Vce which causes the rise in Vcb. To prove this, we only have to look at the simulation and take it one microsecond at a time. The simulation clearly shows that the transistor Vce starts to rise BEFORE the feedback gets high enough to cut off Q1. To make this more clear, we can use another circuit that has no feedback, or even simply short out Q1 collector to emitter to keep Q2 turned on constantly. What we see is as we turn the circuit on is that the collector current rises as the inductor allows with time, and it continues to rise until the transistor starts to come out of saturation. Lo and Behold, the transistor comes out of saturation even with no feedback...what a surprise
The simulation clearly shows Q2 coming out of saturation when IC=120mA.
That's incorrect, as the simulation clearly shows. Q2 leaving the saturation region is the only thing that stops the current from rising past 120mA.
A saturated bjt can't do it.
No matter how much you distort the basic laws of circuit voltage, current and resistance, they prove you wrong each time you make such a mis-statement.
Brownout,
130 ma in my simulation.
Certainly that is correct. Who said it was not?
Can't do what?
Whatever are you talking about? Let's start over. I said that Q2 is either in saturation or cutoff when oscillating. Your turn.
Ratch
Here are two simulation runs for the Brinkmann circuit.
Fig 1 shows Q1 cutting off, but only after Vce starts to rise. The point where Q1 starts to cut off is after Vce rose about 0.6v, which means it's hard to tell what happens first (however it did actually rise not stay the same), so in Fig 2 the capacitor was decreased to 380pf and now we can see that Q1 does NOT start to cut off until Vce rises to nearly 1v (0.99v).
Now in the Fig 1 we might be able to argue that Vce only had to rise a little, so it 'might not' have come out of sat yet because Vce only equals 0.6v and so the feedback couples that to Q1 base. But, in Fig 2, Vce rises clearly to 0.99v before any action occurs in Q1, and so if we argue that in Fig 1 the transistor came out of sat at 0.6v due to the feedback then we can not also argue that the transistor came out of sat at a different voltage (0.39 volts higher than 0.6v) too because the base drive is the same in both cases (Q1 is conducting).
Thus, at the very least in Fig 2 the transistor Q2 comes out of saturation due to some reason other than the feedback and i dont see any way around this, and the only reason left is Ic increasing.
NO saturated bjt can limit the current to 120 mA in the circuit.
No cutoff bjt can limit the current to 120mA
Only a bjt in active mode can.
The simulation of the bjt alone shows that only in active mode can the current limit at 120mA. That's shouldn't be so hard to understand.
That 's right, But the saturated BJT cuts off before the current rises further, for reasons I articulated to MrAl.
That is right. But the current of the transistor does go to zero as you can see in the simulation. The current of the coil then switches to the diode until the transistor saturates again.
That does not happen in this circuit.
Look at either the Ic or Ie of the simulation again. You will see the current rise from 40 ma to 130 ma during the saturation period, and fall to zero during the cutoff period.
MrAl,
I agree that Vce rises very shortly before feedback starts. The data sheet shows Vce rising slightly with increased current. How are you getting such humongous Vcesat values when the data sheets show a much lower value? What is causing those values to be so high? Are they really transitional values of Q2 coming out of saturation? What caused it to come out of saturation?
Ratch
From ONsemi datasheet it's clearly shows that if IB is equal 600uA then BJT is comes out of saturation for Ic > 100mAAccording to the Fairchild data sheets, the transistor can easily exist in saturation at 500 ma, so just increasing the current to 120 ma is not going to hack it.
And here you have the photo from very old Polish book writes by professor Jerzy Baranowski in 1976r.You use the familiar Ic/Vce curves to do this. I don't think that is valid because those curves are predicated on a reverse bias being present on the c-b segments of the BJT. If you were jumping from one point to another in the active region, then I would say you would be correct. But you start in the saturation on the left of the base current line, where the base current no longer controls the transistor, and the Vce and Ic can have a wide range of values.
Why current stop in 120mA. Well the answer is simple.You submitted two simulations to show a transistor comes out of saturation by its current, the first one at 120 ma. But why did it stop at 120 ma? It could still be feeding the coil during saturation up to at least 500 ma. I notice again that the exact point where the current increase stopped was when the transistor became reversed bias. If the Vce increases by 0.69 volts, then it will become reversed bias and break out of saturation at 120 ma
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