# The true meaning of voltage drop

1. Aug 31, 2016

### Grim Arrow

In CE amplifiers with negative feedback I can't understand what it means that the emitter becomes more positive (when hfe increases and VE increases with it). Does it mean that there is less negative charge after the emitter resistor (going from ground to Vcc) and the emitter feels some positive charge and the potential difference between the emitter and the base is decreased or what? Physicaly I can't explain it. Not theoreticaly, but physicaly. Help me with this one if you can
EDIT#1 I understand that electrons in the - terminal of the battery, in order to flow through the circuit, they must be given the needed potential energy. This potential energy - the energy density of a charge J/C is the thing the electrons need so that they can actualy start a current. By flowing from ground to the high potential, the electrons inevitably encounter resistance and they lose this energy to heat. And when they arrive at Vcc they have already spend all of the potential energy given to them to fulfill this "journey". But how can I imagine the rising emitter voltage VE duo to negative feedback?

Last edited: Aug 31, 2016
2. Aug 31, 2016

### Paul Colby

Not certain I understand the question. I assume an NPN and a bias network. Also a Vcc to collector resistor and an emitter to ground resistor. Where is the negative feedback resistor?

3. Aug 31, 2016

### Grim Arrow

This is it. When Hfe increases, Ic increases and so does Ie and the drop across RE also increases, making the emitter more positive and you know how the feedback works from there on. Similar thing is when in a voltage divider the output falls below a certain level and there is not enough voltage to say turn on a led or a transistor. Im not sure how the voltage manifests. Is it the emitter actualy gaining some positive charge or I dont know.

4. Aug 31, 2016

### Averagesupernova

So are you saying that you don't understand how an increase in emitter current through RE can cause a larger voltage to be developed across RE?

5. Aug 31, 2016

### anorlunda

Trying to think of circuits in terms of electrons is seldom helpful. You might be better off using transistor equivalent circuits. Have a look at the link below, and see if that helpe.

Transistor Equivalent Circuits and Models

6. Aug 31, 2016

### Grim Arrow

Not in that sense. I know that V=IxR. I dont understand the physical process behind it.

7. Aug 31, 2016

### Paul Colby

So increased current through $R_E$ reduces the $VE$ voltage drop

8. Aug 31, 2016

### Grim Arrow

What I understand - Voltage drop is the loss of potential energy in resistors duo to their resistive nature. You increase the resistance - you increase the voltage drop. Not sure though why increasing I will increase the voltage drop. But thats not the question. The question may have routes in quantum physics. What happens in RE, that VE increases. I need to understand this in terms of electrons flowing and such.

9. Aug 31, 2016

### Paul Colby

So the question is why do resistors work the way they do? A hydrostatic analog any help to you. The more flow more pressure drop?

10. Aug 31, 2016

### LvW

For my opinion - two basic questions:
1.) Is there any negative feedback caused by RE?
Answer: Of course - that is the only purpose of RE. Any unwanted increase of IC resp. IE (temperature or beta-variation) increases the voltage drop VE across RE and, thus, enhances the emitter potential. The emitter current IE is determined by VBE (exponential relation) - and the voltage VBE=VB-VE decreases correspondingly because the base potential VB is kept relatively constant (for this purpose the resistors R1 and R2 are chosen such that any change in the base current has only minor influence on the base potential). Hence, we have current-controlled voltage-feedback. This feedback stabilzes the currents IE and IC against unwanted influences.

2.) What is the meaning of "voltage drop"?
Answer: From the physical point of view this cannot be answered because there is not such a thing like "voltage drop". The basic question behind this problem is "can a current cause a voltage if it goes through a resistor"? And the answer is NO! There is no current without voltage! That means: Voltage first and current as a result. This is easy to understand if you remember that current is "movement of charged carriers". Which force causes the carriers to move? Answer: The electrical field inside the conducting body (resistor) which depends on the voltage.
Of course, I am aware that we perform all circuit calculations under the assumption that a current can produce a voltage. And it works fine. And that is the "secret" behind the term "voltage drop". But - because you have asked for the physical background - the answer is as explained above.

Further explanation: Lets analyze a simple voltage divider. Before the resistors are connected, the driving voltage source produces a widespread electric field between both connecting pins. When the resistors are connected, this field is concentrated within the connected parts - in proportion to the conductivity of the parts. This can be easily verified if we think of a potentiometer with a pick-up which also defines two resistors (to the left and right of the pick-up). Finally, the E-field within these two resistors together with the geometrical dimensions define the voltage across each part of the potentiometer (resistor). And this E-field allows a current in accordance with Ohms law.

(PS: I am aware that my explanations are somewhat uncommon and I would not be surprised if I will get some opposition from other forum members)

Last edited: Aug 31, 2016
11. Aug 31, 2016

### Grim Arrow

Okay. I now understand that Voltage drop unlike current, or voltage, or resistance, is not a physical property, but rather a way to say, that if you have a circuit with a resistor and you suddenly increase the resistance of that resistor, the current will decrease and you must apply aditional voltage (the voltage drop across that resistor) to get the same current. Thus voltage drop can be thought as a pure loss of electrical energy.

12. Aug 31, 2016

### LvW

I dont know if there is a misunderstanding between us.
"Voltage drop" is, of course, a "physical quantity". It is simply the voltage (potential difference) which can be calculated and measured between both ends of a resistor.
However, the additional term "drop" comes from the fact that we treat this voltage as if it would be produced by a current (and increases in proportion to the current) - and if this resistor is only one of several other similar elements connected in series.

13. Aug 31, 2016

### Grim Arrow

Oh, so from here comes my second confusion that current can increase the voltage drop via the formula V=IxR when in reality it's not the increased current that causes voltage drop, its the voltage across R that caused such a current to flow. But as you said we like to imagine that current can cause a voltage across a resistor, while in reality its the other way around. Its the potential difference between the ends of that resistor that cause a current. And voltage drop across any resistor is just the same as the potential difference across it.
Thanks! :)

14. Aug 31, 2016

### anorlunda

No. V=IR is Ohms law. You should not read any before-after, chicken-egg, cause-effect relationships in it. By analogy time T, distance D and speed S are related by D=S*T, but you can't say that time causes distance. Speed is defined as S=D/T. Resistance is defined as R=V/I.

You should also stop thinking of electrons and of voltage as energy, not because it is wrong but because it is not helpful at the beginner stage. In circuits, the flow of energy per unit time is power. Power P=V*I. It takes both voltage and current to get power. Focus on that.

15. Aug 31, 2016

### Averagesupernova

Anorlunda is quite right about overthinking ohms law in that which caused what first vs. the other way around. It is helpful in analyzing circuits to just use ohms law for what it is.

16. Aug 31, 2016

### Grim Arrow

Okay, I view it this way: It takes electrical energy to get current through a resistor. The more the resistance the more energy it is required to support electron flow. The more the resistance the more potential difference you need to apply across it. When we aply voltage across a circuit with two or more resistors in series, then depending on V and R, we will get a certain current. The bigger resistance will cause more electrical energy to be lost across it in order to let the electron flow pass which means that the greater will be the voltage drop across it (I found out that voltage drop and potential difference are realy just the same thing). Am I correct?
EDIT: I know you advised me not to think of voltage drop as an energy loss, but it is easy for me to use this way of thinking.

Last edited: Aug 31, 2016
17. Aug 31, 2016

### anorlunda

No, not right, at least not necessarily right. Consider a resistance R connected across a constant voltage source V. The power dissipated is $\frac{V^2}{R}$. If you increase R, then power goes down. If you increase R to infinity (open the circuit) then P=0.

You continue using the wrong units. Power, not energy is what you should focus on.

But in general increasing R is the same as increasing V/I. So if we increase V/I what does that say about V or I. V and I could increase, decrease or remain constant; only the ratio V/I is increasing. Your attempt at cause-effect reasoning is trapped in a circular logic loop.

But I should be angry at you for ignoring what I said in #14. Your reply in #16 repeats what I advised you not to do and @Averagesupernova confirmed in #15.. What's the point of asking questions if you ignore the answers?

18. Aug 31, 2016

### Grim Arrow

Im not ignoring your answer, but I don't quite get why I should consider the power dissipation. Can you give more in depth example or explanation?

19. Aug 31, 2016

### anorlunda

Energy is not created or destroyed. It flows from one place to another, or it is transformed into another form such as heat energy. Such flow is called power, and it includes the all important parameter time.

20. Aug 31, 2016

### davenn

that is incorrect

you have a 12V PSU you can put any resistor value you like across that 12V, the voltage drop will ALWAYS be 12V across said resistor
If there are 2 ( 3, 4 etc) resistors in series, the 12V drop will be shared across the resistors
if the 2 resistors are the same value then the drop across each resistor will be the same .... 6V
if the 3 resistors are the same value then the drop across each resistor will be the same .... 4V
both examples adding up to the 12V total supply value

Dave