Electricity characteristic

[RivaultUser]

Hellooo againnn friends, Can you explain to me some electricity characteristic in a circuit?

Like in this picture :
 Why in the left picture the base of the right transistor doesn't get any current? ( I've tried this circuit )


Is this have something to do with electricity characteristic ?

[CarlG]

In order to not make it to easy for you: what voltage do you have at the collector of Q1?

//C

[lewis]

It's to do with the characteristics of transistor operation. Transistors need about 0.6V between the base terminal and the emitter to turn them on.

So in the RIGHT hand circuit, Q1's base is connected to its emitter through the switch and R2, so there is no voltage between the base and emitter, and therefore the transistor is off. That means no current flows between collector and emitter. This allows R3 to pull up the base of Q2 - current flows from the positive supply, through R3, through the base of Q2 and out the emitter. This means the base voltage will be slightly above 0.6V, and Q2 turns on allowing current to flow through the collector and emitter, and through the lamp. Q2 is fully on and is said to be 'saturated', conceptually the collector and emitter now shorted together (with a small voltage drop - it's not a 'perfect' short, but assume it is for the moment).

The important thing to remember is that when one transistor is off, the other one is on.

In the left hand circuit, the operation is reversed. Q1 is turned on fully (saturated) because current flows through R1, through the switch, into the base and out the emitter causing the base-emitter voltage to be a bit higher than 0.6V turning it on fully. That means that the collector and emitter are 'shorted together'. Like the switch in the right hand circuit, this means that there can be no base-emitter voltage on Q2 because Q1 has essentially shorted Q2's base-emitter junction, so Q2 remains off and does not light the lamp. Current flows through R3 and Q1, but there's not enough voltage at Q2's base to turn it on.

There are some good examples here of the use of the weird parameters you'll see in the datasheet for any transistor. The 0.6V I keep mentioning is V_be, and it varies hugely with temperature, voltage, current, time, all kinds of things. Current into the base of a transistor is I_b, the voltage drop across the emitter-collector terminals when the transistor is saturated is V_ce(sat), and current through the collector-emitter junction (eg the current through the lamp) is I_c.

[RivaultUser]

It's to do with the characteristics of transistor operation. Transistors need about 0.6V between the base terminal and the emitter to turn them on.

So in the RIGHT hand circuit, Q1's base is connected to its emitter through the switch and R2, so there is no voltage between the base and emitter, and therefore the transistor is off. That means no current flows between collector and emitter. This allows R3 to pull up the base of Q2 - current flows from the positive supply, through R3, through the base of Q2 and out the emitter. This means the base voltage will be slightly above 0.6V, and Q2 turns on allowing current to flow through the collector and emitter, and through the lamp. Q2 is fully on and is said to be 'saturated', conceptually the collector and emitter now shorted together (with a small voltage drop - it's not a 'perfect' short, but assume it is for the moment).

The important thing to remember is that when one transistor is off, the other one is on.

In the left hand circuit, the operation is reversed. Q1 is turned on fully (saturated) because current flows through R1, through the switch, into the base and out the emitter causing the base-emitter voltage to be a bit higher than 0.6V turning it on fully. That means that the collector and emitter are 'shorted together'. Like the switch in the right hand circuit, this means that there can be no base-emitter voltage on Q2 because Q1 has essentially shorted Q2's base-emitter junction, so Q2 remains off and does not light the lamp. Current flows through R3 and Q1, but there's not enough voltage at Q2's base to turn it on.

There are some good examples here of the use of the weird parameters you'll see in the datasheet for any transistor. The 0.6V I keep mentioning is V_be, and it varies hugely with temperature, voltage, current, time, all kinds of things. Current into the base of a transistor is I_b, the voltage drop across the emitter-collector terminals when the transistor is saturated is V_ce(sat), and current through the collector-emitter junction (eg the current through the lamp) is I_c.

Thx for your reply man! I slowly getting the hang of it.

What confused me is on the LEFT picture, when Ic is flowing through R3, according to Kirchoff law isn't the current should be divided ? ( some of the Ic current go to the Q1 & some of the Ic currrent go to the base of the Q2)

[RivaultUser]

In order to not make it to easy for you: what voltage do you have at the collector of Q1?

//C

That would be the Vs - the voltage drop that cause by R3 Am I wrong??

[CarlG]

In order to not make it to easy for you: what voltage do you have at the collector of Q1?

//C

That would be the Vs - the voltage drop that cause by R3 Am I wrong??

No, you're right. But how much is that? I thought you had the circuit up and running, so you could measure it

Ok, it's not so easy to figure out without component values and transistor datasheet (except for just "knowing" what it must be). But as explained above by Lewis, Q1 is saturated in the figure on the left, leading to very low V_CE drop, let's say 100 mV or less. It's not saturated just like that: the values of R1 and R3 are chosen in order to ensure that the actual collector current is much less than the current gain (B) times the base current:

I_C << B*I_B  (this is the definition of saturation)

as opposed to a transistor operating in its linear region, where I_C = B*I_B.

Knowing the current gain B, you can then calculate R1 and R3 at the given V_S.

//C

[RivaultUser]

In order to not make it to easy for you: what voltage do you have at the collector of Q1?

//C

That would be the Vs - the voltage drop that cause by R3 Am I wrong??

No, you're right. But how much is that? I thought you had the circuit up and running, so you could measure it

Ok, it's not so easy to figure out without component values and transistor datasheet (except for just "knowing" what it must be). But as explained above by Lewis, Q1 is saturated in the figure on the left, leading to very low V_CE drop, let's say 100 mV or less. It's not saturated just like that: the values of R1 and R3 are chosen in order to ensure that the actual collector current is much less than the current gain (B) times the base current:

I_C << B*I_B  (this is the definition of saturation)

as opposed to a transistor operating in its linear region, where I_C = B*I_B.

Knowing the current gain B, you can then calculate R1 and R3 at the given V_S.

//C

okay dude, I'm slowwwwlyyyyyy got the hang of it.
But still, What confused me is on the LEFT picture, when Ic is flowing through R3, according to Kirchoff law isn't the current should be divided ( some of the Ic current go to the Q1 & some of the Ic currrent go to the base of the Q2)

This is the thing that confused me the most

[Shuggsy]

okay dude, I'm slowwwwlyyyyyy got the hang of it.
But still, What confused me is on the LEFT picture, when Ic is flowing through R3, according to Kirchoff law isn't the current should be divided ( some of the Ic current go to the Q1 & some of the Ic currrent go to the base of the Q2)

This is the thing that confused me the most

Look at it this way: the BJTs are used as switches (either they are fully on [saturated] or fully off [cutoff], no operation in the linear region [region with linear gain based essentially on input current to the base]}. Now, with the assumption that the BJTs are switches, the left picture is simplified somewhat. In that instance, Q1 is turned on and is acting like a closed switch. Q2's base is shorted to ground by the "switch" of Q1. The emitter of Q2 is also at ground, so the base-emitter voltage (Vbe) of Q2 is 0. Since BJTs need ~0.6V to turn on, Q2 is kept in the cutoff region and acts like an open switch. With Q2 in this open state, no current flows through the lamp. No current is drawn through the base of Q2 simply because the voltage on the base is not high enough to begin to draw current.

Now, in reality, transistors are not ideal switches. There is some equivalent resistance through the collector-emitter junction of a BJT. This gets to what CarlG was talking about in his earlier post. The equivalent resistance of the "closed switch" of Q1 is so small that, unless you have a massive supply voltage and/or R3 was incredibly small, the voltage drop across Q1's collector-emitter junction (Vce) is well below the ~0.6V [~600mV] required to "turn-on" a BJT. In reality, the amount of current required to drop that much voltage across Q1's collector-emitter junction would likely blow up Q1 unless it was a very powerful transistor with a good heatsink and enough airflow to dissipate all the power.

So, what that tells us is that, in reality, there may be an extremely small current flowing into the base of Q2 in the left picture, but it is not enough to turn Q2 on or even get it into the linear region. Put another way, the voltage across Q1's collector-emitter junction (Vce) is so small when Q1 is on that Q2's base does not get to the ~0.6V required to turn on Q2.

All of this is essentially captured in Ohm's Law: V = I * R

Since the base-emitter voltage of Q2 is so small, Q2's base acts like a very high resistance. Or you could say that Q2's base has a very high resistance and thus the current going into it will be very small for a small voltage across Q1's collector-emitter junction. Everything fits together, it just depends on how you look at it! If you knew what transistors were being used for Q1 and Q2, knew the values of R1, R2, and R3, knew the state of the switch, and knew the voltage Vs then you could do a full analysis of the circuit to come up with very, very close values for what the currents and voltages would be at each point in the circuit. However, intuitively, many of the people here have described the operation of the circuit without knowing any of these values based on what the given components would do in general. That is, knowing that transistors operated in this way would act like open and closed switches allows people here to describe the operation of the circuit. This knowledge comes with time and effort in looking at circuits to understand what they do and why.

[RivaultUser]

okay dude, I'm slowwwwlyyyyyy got the hang of it.
But still, What confused me is on the LEFT picture, when Ic is flowing through R3, according to Kirchoff law isn't the current should be divided ( some of the Ic current go to the Q1 & some of the Ic currrent go to the base of the Q2)

This is the thing that confused me the most

Look at it this way: the BJTs are used as switches (either they are fully on [saturated] or fully off [cutoff], no operation in the linear region [region with linear gain based essentially on input current to the base]}. Now, with the assumption that the BJTs are switches, the left picture is simplified somewhat. In that instance, Q1 is turned on and is acting like a closed switch. Q2's base is shorted to ground by the "switch" of Q1. The emitter of Q2 is also at ground, so the base-emitter voltage (Vbe) of Q2 is 0. Since BJTs need ~0.6V to turn on, Q2 is kept in the cutoff region and acts like an open switch. With Q2 in this open state, no current flows through the lamp. No current is drawn through the base of Q2 simply because the voltage on the base is not high enough to begin to draw current.

Now, in reality, transistors are not ideal switches. There is some equivalent resistance through the collector-emitter junction of a BJT. This gets to what CarlG was talking about in his earlier post. The equivalent resistance of the "closed switch" of Q1 is so small that, unless you have a massive supply voltage and/or R3 was incredibly small, the voltage drop across Q1's collector-emitter junction (Vce) is well below the ~0.6V [~600mV] required to "turn-on" a BJT. In reality, the amount of current required to drop that much voltage across Q1's collector-emitter junction would likely blow up Q1 unless it was a very powerful transistor with a good heatsink and enough airflow to dissipate all the power.

So, what that tells us is that, in reality, there may be an extremely small current flowing into the base of Q2 in the left picture, but it is not enough to turn Q2 on or even get it into the linear region. Put another way, the voltage across Q1's collector-emitter junction (Vce) is so small when Q1 is on that Q2's base does not get to the ~0.6V required to turn on Q2.

All of this is essentially captured in Ohm's Law: V = I * R

Since the base-emitter voltage of Q2 is so small, Q2's base acts like a very high resistance. Or you could say that Q2's base has a very high resistance and thus the current going into it will be very small for a small voltage across Q1's collector-emitter junction. Everything fits together, it just depends on how you look at it! If you knew what transistors were being used for Q1 and Q2, knew the values of R1, R2, and R3, knew the state of the switch, and knew the voltage Vs then you could do a full analysis of the circuit to come up with very, very close values for what the currents and voltages would be at each point in the circuit. However, intuitively, many of the people here have described the operation of the circuit without knowing any of these values based on what the given components would do in general. That is, knowing that transistors operated in this way would act like open and closed switches allows people here to describe the operation of the circuit. This knowledge comes with time and effort in looking at circuits to understand what they do and why.

Thx a lot for your help man, I'm starting to know why......

[nadona]

I underlined the answer in the page.

Excellent book by the way.

[olsenn]

The easiest way to analyze circuits like these is to make assumptions and solve it at a glance.

In the case of the left photo you posted, first assume that IB1 is zero; with no current passing through R1 there is no voltage drop accross R1, in which case the voltage accross the base-emitter junction of Q1 is Vs. If we assume that Vs is greater than the voltage needed to forward bias this junction then we know that our first assumption is wrong, in which case IB1 is NOT zero and therefore neither is IC1.

Next, assume that the value of R3 was chosen such that Q1 is saturated. In this case, the voltage drop accross R3 (IC1 * R3) is approx Vs, and there is not enough voltage left at the base of Q2 to forward-bias its base-emitter junction. With no current flowing through the lamp, it is off!

As a side note, R2 in this circuit is useless, even when the switch is flipped.