What's the difference between using a simple series resistor and a voltage divider to reduce voltage

WBahn

Joined Mar 31, 2012
29,978
If everyone would stop arguing with me, and simply elaborate on the point I was trying to make, the answer would be obvious. Case 2 degenerates into case 1 when you perform the thevinin analysis. From there, the answer to your question is elementary.
And your answer that they are the same is highly misleading. You choose option 1 and I'll choose option 2 and I guarantee mine will perform better than yours (or, in the extreme limit, no worse than yours). If they are the same, then that would not be possible.
 

joeyd999

Joined Jun 6, 2011
5,237
And your answer that they are the same is highly misleading. You choose option 1 and I'll choose option 2 and I guarantee mine will perform better than yours (or, in the extreme limit, no worse than yours). If they are the same, then that would not be possible.
So, you drive an LED with a two resistor network? No? Why not?
 

joeyd999

Joined Jun 6, 2011
5,237
Because you are regulating current, not voltage.
But you can do exactly the same thing with case 2. The load lines are equivalent for the same Thevinen values.

The point is that OP asked a generic question. I gave a generic answer and have been criticized for it. The reality is that both circuits are functionally identical, but there are practical reasons for using one topology over the other, depending on the situation.

In any case, analysis is expedited by reducing to the thevinin equivalent, so it is important to understand that the two networks are functionally identical.
 

WBahn

Joined Mar 31, 2012
29,978
So, you drive an LED with a two resistor network? No? Why not?
What has THAT got to do with anything???

Even though it is entirely unrelated, in the LED case you are not trying to establish a desired voltage across the LED. Instead you are trying to establish a desired current in the LED and are exploiting the fact that the LED voltage drop will be nearly constant regardless of current. The series resistor solution has the benefit that if you can establish the desired current in the resistor, then that IS the current in the LED.
 

WBahn

Joined Mar 31, 2012
29,978
But you can do exactly the same thing with case 2. The load lines are equivalent for the same Thevinen values.

The point is that OP asked a generic question. I gave a generic answer and have been criticized for it. The reality is that both circuits are functionally identical, but there are practical reasons for using one topology over the other, depending on the situation.

In any case, analysis is expedited by reducing to the thevinin equivalent, so it is important to understand that the two networks are functionally identical.
They are "functionally identical" ONLY if you get to pick different and arbitrary values for Vin for both circuits. But if you can do that, then why use EITHER circuit? Why not just pick Vin to be the voltage that you want to have across the load?

If you can do that, then do that. If you can't do that, then they are NOT functionally identical.

So the bottom line is that they are functionally identical is exactly one case and that case is of no value because, in that case, you don't need either circuit.
 

joeyd999

Joined Jun 6, 2011
5,237
What has THAT got to do with anything???

Even though it is entirely unrelated, in the LED case you are not trying to establish a desired voltage across the LED. Instead you are trying to establish a desired current in the LED and are exploiting the fact that the LED voltage drop will be nearly constant regardless of current. The series resistor solution has the benefit that if you can establish the desired current in the resistor, then that IS the current in the LED.
So then you are suggesting one choose case 1 vs. case 2 based on the linearity of the load? Hmmmm. I'll have to think about that.
 

WBahn

Joined Mar 31, 2012
29,978
So then you are suggesting one choose case 1 vs. case 2 based on the linearity of the load? Hmmmm. I'll have to think about that.
Whether to choose case 1 or case 2 depends on a number of factors (as you correctly pointed out).

In the case of an LED, if you are primarily interested in having the current be the desired value, then what you really want is a current source. Unless you have an ideal current source handy, that means you want the equivalent circuit for the source to be modeled as an ideal current source in parallel with as large an equivalent resistance as possible (i.e., a Norton equivalent circuit). When you convert that into a Thevenin equivalent that means that you need the Thevenin voltage to be large so that the equivalent resistance is the same very large value as the Norton equivalent. That means you want Option 1 because it will let you get the highest Thevenin voltage possible (namely Vin) with the largest equivalent resistance possible.
 

joeyd999

Joined Jun 6, 2011
5,237
Whether to choose case 1 or case 2 depends on a number of factors (as you correctly pointed out).

In the case of an LED, if you are primarily interested in having the current be the desired value, then what you really want is a current source. Unless you have an ideal current source handy, that means you want the equivalent circuit for the source to be modeled as an ideal current source in parallel with as large an equivalent resistance as possible (i.e., a Norton equivalent circuit). When you convert that into a Thevenin equivalent that means that you need the Thevenin voltage to be large so that the equivalent resistance is the same very large value as the Norton equivalent. That means you want Option 1 because it will let you get the highest Thevenin voltage possible (namely Vin) with the largest equivalent resistance possible.
[SARC]Ah. So all I've learned about biasing common emitter amplifiers has thus been incorrect. [\SARC]

Look, the point of my original post was not to start this argument, or even to have one. I was hoping that OP could gain an understanding that a complex network could be modeled as a simpler, equivalent one in order to expedite the analysis. I was too brief and should have elaborated further myself. Now our friend is likely hopelessly confused, and for that, I apologize.
 

WBahn

Joined Mar 31, 2012
29,978
[SARC]Ah. So all I've learned about biasing common emitter amplifiers has thus been incorrect. [\SARC]

Look, the point of my original post was not to start this argument, or even to have one. I was hoping that OP could gain an understanding that a complex network could be modeled as a simpler, equivalent one in order to expedite the analysis. I was too brief and should have elaborated further myself. Now our friend is likely hopelessly confused, and for that, I apologize.
When you bias a common-emitter amplifier, aren't you trying to establish a voltage at the base of the transistor? So in that case you want a bias circuit that has a low Thevenin resistance and a Thevenin voltage that is close to the voltage you want on the base, hence you would probably choose Option 2 because it gives you the ability to do that. You are NOT trying to establish a current into the base (or, if you are, then you are relying on the beta of the transistor to be a known, fixed value and that almost always constitutes poor circuit design).
 

hp1729

Joined Nov 23, 2015
2,304
If everyone would stop arguing with me, and simply elaborate on the point I was trying to make, the answer would be obvious. Case 2 degenerates into case 1 when you perform the thevinin analysis. From there, the answer to your question is elementary.
Your math is correct but irrelevant. When would you use one method over the other?
 
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