Hello,

I am new to electronics, trying to self-study, I amworking through the Arts of Electronics textbook. On page 67 it says:

"In an emitter follower the npn resistor can only "source" current".

What does "to source current" mean?

Here is the circuit the textbook refers to:

Also, it says that this circuit cannot go more negative than -5V output (-4.4V input), and that further negative swing at the input results in reverse biasing the base-emitter junction. How come? Why? It seems to me that there is no problem with a, say -6V input. There is still plenty of leeway to keep the base-emitter junction forward-biased...

Thank you.

 

-6V at the base will not have BE junction forward biased...
suppose transistor is off. then Re and Rload form voltage divider so Ve=-5V.
to get BE forward biased, you need Vb to be greater than Ve by 0.6V.
-5V + 0.6V = -4.4V

 

Look at the emitter resistor and the load resistor. With the transistor removed from the circuit,what would the voltage be at the node of the two resistors.

 

Look at the emitter resistor and the load resistor. With the transistor removed from the circuit,what would the voltage be at the node of the two resistors.

As panic-mode explained, with the transistor off we would have a simple voltage divider and the voltage at the emitter would be -5V. We would therefore need -5V + 0.6V = -4.4V at the base to get the base-emitter junction forward biased. However I fail to see how this implies that the output voltage cannot go lower than -5V ??

Thank you.

 

Common usage of the terms:




Obviously, Q1 could be a PNP, and Q2 could be an NPN... or FETs or whatever...

 

MikeML thanks for your schematic but I am not sure I get it

 

MikeML thanks for your schematic but I am not sure I get it

When Q1 turns on, conventional current flows downward from the positive supply rail Vcc, through R1 into the collector of Q1, and finally out Q1's emitter to the 0V point in this schematic, shown as the ground symbol. Q1 sinks current from the resistor R1 fed from the positive rail.

When Q2 turns on, conventional current flows downward from the positive supply rail Vcc, into the emitter of Q2, out of the collector of Q2 into the resistor R2, and finally out of the bottom end of R2 to the 0V point in this schematic, shown as the ground symbol. Q2 sources current into the grounded resistor R1.

 

As panic-mode explained, with the transistor off we would have a simple voltage divider and the voltage at the emitter would be -5V. We would therefore need -5V + 0.6V = -4.4V at the base to get the base-emitter junction forward biased. However I fail to see how this implies that the output voltage cannot go lower than -5V ??

Thank you.

In order to go lower than -5V, it would need to go to something like -6V. With the voltage divider made up of the emitter resistor and the load, the only way to have the node point go more negative would be to increase the ohmic value of the load resistor, thus re-defining the ratio of the voltage divider. The transistor will supply a positive pull, but cannot force a negative.

 

Hello,

When a source supplies current to a load we sometimes call it "sourcing" a current, but when a source takes current from a load we sometimes say it is "sinking" a current.

So if we have a resistor with the top terminal connected to +10 volts DC and the bottom terminal connected to ground, the power supply is 'sourcing" current and the ground is "sinking" current because the current flows INTO the upper terminal and OUT of the bottom terminal, using conventional current flow.

So in short, something that supplies current is a "source" and something that receives current is a "sink".

A better example is say we have two resistors connected as a voltage divider across a 10 volt DC source (like a battery). Say both resistors are the same value like 1k each. That puts +5v at the center tap.
Now if we connect a short circuit from the tap to ground, we are 'sinking' more current from the tap to ground, but the tap itself is "sourcing" more current.
But if we connect a short from the tap to the +10v source, then we are "sourcing" more current from the source to the tap, or the tap is sinking more current from the source.
So how we look at the nodes matters...
When we connect the short from the tap to ground we can say two things:
1. The tap is sourcing current to ground.
2. The short circuit wire is sinking current to ground (or the ground node is sinking more current)
When we connect the short from the source to the tap we can also say two things:
1. The source is sourcing more current to the tap.
2. The tap is sinking more current from the source.

So it is a two way street: the node with higher energy potential is said to 'source', while the node with lower energy potential is said to 'sink'.
Note that it's not really based on the polarity of the source, it depends on how we view the source current flow. So for negative sources we might look at a negative source as 'sourcing' even though it is the most negative node in the system. This might happen in a circuit with a battery where the battery positive terminal is the common ground, or in a dual polarity system with two batteries where we might actually have to think of it both ways at the same time (both positive and negative sources as "sourcing" current even though the current flows in different directions for the two sources).

What a dual polarity system means then for the voltage divider is that we might look at it either way. For example, say we have the same voltage divider of two 1k resistors connected across two batteries both are 10 volts each and they are connected in series and the center tap connected to ground. This gives us plus and minus 10 volts, so we have a +10v source and a -10v source. At the resistors center tap we have exactly 0v because they divide 20v exactly in half and ground is 0v too.
Now if we connect a wire across the top resistor, we can say:
1. The +10v source is sourcing more current to the tap.
2. The tap is sinking more current from the +10v source.
But if we connect the wire across the bottom resistor we can say:
1. The -10v source is sourcing more current to the tap.
2. The tap is sinking more current from the -10v source

Conversely, if we call the -10v node ground that makes the top most node +20v and the center tap at +10v, so now when we apply the short across the bottom resistor we might say:
1. The tap is sourcing more current to the -10v supply, and
2. The -10v source is sinking more current from the tap.

So depending how we want to view the system we might actually have to swap what we call sinking and what we call sourcing.

 

In order to go lower than -5V, it would need to go to something like -6V. With the voltage divider made up of the emitter resistor and the load, the only way to have the node point go more negative would be to increase the ohmic value of the load resistor, thus re-defining the ratio of the voltage divider. The transistor will supply a positive pull, but cannot force a negative.

Why not? Sorry if i'm dense but I still don't see how it makes logical sense. In other words I would've never figured it out myself by logical reasoning, I would've needed someone to tell me, "that's how it works".

 

The issue is with that Rload resistor in the circuit.

Remove the Rload and yes, the emitter can go to -10V.

 

Why can't the emitter go below -5V? What's the difference between, say -3V, and -7V? Why is -3V possible but not -7V? Why is the voltage divider an issue in one case, but not in the other?

If the emitter is at -3V, then the GND would "source" current through the load, 3V would be dropped across the load (from 0 to -3V), then from the emitter to the -10V rail, the potential would drop across the emitter resistor from -3V to -10V.

The same reasoning is possible with -7V....Why not?

 

For this circuit

When BJT is off the voltage at transistor emitter is equal to Ve = 10V * RE/(RL + RE) = 10V * 1/2 = 5V.
So what voltage is needed at transistor base to open the BJT?

 

Why not? Sorry if i'm dense but I still don't see how it makes logical sense. In other words I would've never figured it out myself by logical reasoning, I would've needed someone to tell me, "that's how it works".

Okay, I'll take a stab at it.
The NPN transistor connected as an emitter-follower can only supply current (source) from the plus supply to the emitter output (i.e. it acts sort of as a variable resistor between the plus supply and the emitter resistor).
Thus any negative current must be supplied by the emitter resistor.
This means the most negative voltage it can deliver to the load is when the transistor is off.
For the power supply voltages, and emitter and load resistor values shown, that would be -5V (due to the voltage divider action between the load resistor and the emitter resistor values).

If there's anything about that you don't understand I'll try to further clarify.

 

For this circuit
View attachment 74286
When BJT is off the voltage at transistor emitter is equal to Ve = 10V * RE/(RL + RE) = 10V * 1/2 = 5V.
So what voltage is needed at transistor base to open the BJT?

about 5.6V
I know where you're going
when the transistor off, the voltage at the emitter is 5V and the base needs 5.6V to open the BJT. anything lower than 5.6V will not open it up. I agree with this, however...

that doesn't preclude the possibility that we could have 5V at the emitter with the BJT open...

in other words,

transistor off ==> we need a base voltage of more or equal than 5.6V to get it going
doesn't imply "base voltage is less than 5.6V ==> transistor is off" ... maybe the transistor was already running prior to reaching a base voltage of 5.6V (say an AC signal that goes from +10V to -10V)....why should we automatically assume and deduce that once the AC signal reaches less than 5.6V, the transistor will shut down??

 

This means the most negative voltage it can deliver to the load is when the transistor is off.
For the power supply voltages, and emitter and load resistor values shown, that would be -5V (due to the voltage divider action between the load resistor and the emitter resistor values).

Change the configuration of the voltage divider and yes you can go below -5V, like I pointed out in post #11.

 

about 5.6V
I know where you're going
when the transistor off, the voltage at the emitter is 5V and the base needs 5.6V to open the BJT. anything lower than 5.6V will not open it up. I agree with this, however...

that doesn't preclude the possibility that we could have 5V at the emitter with the BJT open...

in other words,

transistor off ==> we need a base voltage of more or equal than 5.6V to get it going
doesn't imply "base voltage is less than 5.6V ==> transistor is off" ... maybe the transistor was already running prior to reaching a base voltage of 5.6V (say an AC signal that goes from +10V to -10V)....why should we automatically assume and deduce that once the AC signal reaches less than 5.6V, the transistor will shut down??

Well the Kirchoff's law force the BJT to cut-off.
Let as assume for the moment that the BJT is ON and the voltage at emitter is equal to 2V. If so, this means that I_RE = 2V/1K = 2mA and RL current is equal to I_RL = (10V - 2V)/1K = 8mA. Now from KCL we have IE = IL + I_bjt
2mA = 8mA + I_bjt ----> I_bjt = 2mA - 8mA = -6mA negative current in NPN ( flow from emitter to collector) ?? How can NPN transistor sink -6mA? Transistor can not do this and this is why the BJT is OFF.

 

Okay, I'll take a stab at it.
The NPN transistor connected as an emitter-follower can only supply current (source) from the plus supply to the emitter output (i.e. it acts sort of as a variable resistor between the plus supply and the emitter resistor).
Thus any negative current must be supplied by the emitter resistor.
This means the most negative voltage it can deliver to the load is when the transistor is off.
For the power supply voltages, and emitter and load resistor values shown, that would be -5V (due to the voltage divider action between the load resistor and the emitter resistor values).

If there's anything about that you don't understand I'll try to further clarify.

Well the Kirchoff's law force the BJT to cut-off.
Let as assume for the moment that the BJT is ON and the voltage at emitter is equal to 2V. If so, this means that I_RE = 2V/1K = 2mA and RL current is equal to I_RL = (10V - 2V)/1K = 8mA. Now from KCL we have IE = IL + I_bjt
2mA = 8mA + I_bjt ----> I_bjt = 2mA - 8mA = -6mA negative current in NPN ( flow from emitter to collector) ?? How can NPN transistor sink -6mA? Transistor can not do this and this is why the BJT is OFF.

Thank you. I think I am starting to get it.
I_emitter + I_emitter resistor + I_load = 0
Assume positive current for anything going into the node, negative current for anything going out
Since the npn in emitter follower can only source current, we can only have I_emitter > 0
If I_emitter < 0, then the npn is sinking current which it cannot do
So the breaking point is at I_emitter = 0, in other words, when the npn is off, at which point the emitter is at -5V (because of the voltage divider)
For anything more negative than -5V, load is sourcing more current than what the emitter resistor is sinking, which can only be possible if the npn sinks current, which it cannot do
For anything more positive than -5V, load is sourcing less current than what the emitter resistor is sinking, the extra current is provided by the npn which is sourcing current.

Right?

 

It seems that you finally got it.

 

Thank you. I think I am starting to get it.
I_emitter + I_emitter resistor + I_load = 0
Assume positive current for anything going into the node, negative current for anything going out
Since the npn in emitter follower can only source current, we can only have I_emitter > 0
If I_emitter < 0, then the npn is sinking current which it cannot do
So the breaking point is at I_emitter = 0, in other words, when the npn is off, at which point the emitter is at -5V (because of the voltage divider)
For anything more negative than -5V, load is sourcing more current than what the emitter resistor is sinking, which can only be possible if the npn sinks current, which it cannot do
For anything more positive than -5V, load is sourcing less current than what the emitter resistor is sinking, the extra current is provided by the npn which is sourcing current.

Right?

Hello,

I am new to electronics, trying to self-study, I amworking through the Arts of Electronics textbook. On page 67 it says:

"In an emitter follower the npn resistor can only "source" current".

What does "to source current" mean?

Here is the circuit the textbook refers to:

Also, it says that this circuit cannot go more negative than -5V output (-4.4V input), and that further negative swing at the input results in reverse biasing the base-emitter junction. How come? Why? It seems to me that there is no problem with a, say -6V input. There is still plenty of leeway to keep the base-emitter junction forward-biased...

Thank you.

Hello,

I am new to electronics, trying to self-study, I amworking through the Arts of Electronics textbook. On page 67 it says:

"In an emitter follower the npn resistor can only "source" current".

What does "to source current" mean?

Here is the circuit the textbook refers to:

Also, it says that this circuit cannot go more negative than -5V output (-4.4V input), and that further negative swing at the input results in reverse biasing the base-emitter junction. How come? Why? It seems to me that there is no problem with a, say -6V input. There is still plenty of leeway to keep the base-emitter junction forward-biased...

Thank you.

The common confusion for first reader is not seeing R_{emitter} is on -10V while R_{load} is on ground (or 0V). If you remove everything else except the two resistors and their wiring, then conventional current flow in the direction from R_{load} to R_{emitter} in series with total voltage drop of 0-10V = -10V. Hence measuring voltage between the resistors in series form a voltage divider, -5V each summing to -10V. Then suddenly everything else in that TAOE page make sense.