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Old 27th November 2013, 05:04 AM   #1
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Default What Exactly Is Voltage Drop

The key word that I don't understand is "Drop".

This post is my best effort to seek assistance on a topic which is quite vague to me, so that I am struggling to formulate my questions. I hope that someone will be able to figure out what it is I'm trying to articulate.

If we have a circuit with a resistor, we speak of the voltage drop across the resistor.

I understand all of the calculations involved in voltage drop (ohm's law, parallel and series, etc.). But what I seek is to understand on a conceptual level what voltage drop is. Specifically: what is the NATURE of the change that has taken place between a point just before the resistor and a point just after the resistor, as the electrons travel from a negatively to a positively charged terminal.

Now as I understand it, "voltage" is the force caused by the imbalance of charge which causes a potential difference causing electrons to travel from a negatively charged terminal to a positively charged terminal, and "resistance" is a hindrance caused by a material which, due to its atomic makeup, causes electrons to have decreased current flow, thus opposing that flow of electrons, or "current". So I think I somewhat understand voltage and resistance on a conceptual level.

But what is "voltage drop"? Here's what I have so far:

Voltage drop has nothing to do with number of electrons, meaning that the number of electrons in the atoms just before entering the resistor equals the number of atoms just after

Voltage drop also has nothing to do with the speed of the electrons: that speed is constant throughout the circuit

Voltage drop has to do with what exactly? that's my question

Maybe someone can help me understand what voltage drop is by explaining what measurable difference there is between points before the resistor and points after the resistor.

Does the term "Drop" mean the same as "decrease" such that the potential difference ("Voltage") at the point before the resistor is higher than at the point after the resistor?

If there's a 300Volt DC source for a series circuit with 3 resistors totaling 75-Ohms and R2 has 20-Ohms, then R2 has a voltage drop across it of 80 Volts...I can talk the talk with those numbers that I just made as I typed but I can't understand what that quantity of 80 represents? 80 volt drop means that what happened to those 80 volts?

Last edited by Eddiegnz1; 27th November 2013 at 05:20 AM.
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Old 27th November 2013, 06:47 AM   #2
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[QUOTE=Eddiegnz1;3718402]The key word that I don't understand is "Drop".



You are confusing yourself with the thought that you don't understand. You do! In the third last paragraph, your question "What is the measureable difference between the points before and after the resistor?" you then answered in your last paragraph with the number 80. and in the second last paragraph you describe it correctly - Decrease - or you could say loss. However if the energy isn't lost through something like conversion to heat but instead by performing the work we built the circuit to do, it doesn't make sense to call it loss, so the term drop is used. Cuz that's what it measures like.

You might enjoy going through Basic Electricity and/or Basic Electronics, both by the same authors. I found them very helpful. The NEETS series is also very good because there are questions at the end of every section so you get to see what you got and what you missed. Some might think they're too pedestrian but the military put them together and people's lives surely depended on the technician's being clear about what they knew. I found them to be really helpful in getting a place to start from.
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Old 27th November 2013, 09:06 AM   #3
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Quote:
Originally Posted by Eddiegnz1 View Post
The key word that I don't understand is "Drop".

Does the term "Drop" mean the same as "decrease" such that the potential difference ("Voltage") at the point before the resistor is higher than at the point after the resistor?
exactly

Quote:
Originally Posted by Eddiegnz1 View Post
If there's a 300Volt DC source for a series circuit with 3 resistors totaling 75-Ohms and R2 has 20-Ohms, then R2 has a voltage drop across it of 80 Volts...I can talk the talk with those numbers that I just made as I typed but I can't understand what that quantity of 80 represents? 80 volt drop means that what happened to those 80 volts?
THe power arising from the current associated with them turned into heat.
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Old 27th November 2013, 09:18 AM   #4
DF96 is offline DF96  England
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Quote:
Originally Posted by eddiegnz1
Voltage drop also has nothing to do with the speed of the electrons: that speed is constant throughout the circuit
True: false.

Voltage drop is the difference in electric potential between two points. This tells you the amound of energy which is gained/lost by moving a tiny test charge between these two points. Voltage is always between two points. A single point does not have a voltage; although we may be lazy and assume a reference point without saying so every time.
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Old 27th November 2013, 09:50 AM   #5
Merlinb is offline Merlinb  United Kingdom
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Quote:
Originally Posted by Eddiegnz1 View Post
Voltage drop has nothing to do with number of electrons,
Well, it sort of does, in that there is no current unless the electrons are doing something, but it is not very useful to think about electrons.

Quote:
Voltage drop also has nothing to do with the speed of the electrons: that speed is constant throughout the circuit
No, the speed is different in various parts of the circuit. In metal you have a lot of electrons moving very slowly, in a resistor you have somewhat fewer electrons moving a bit quicker, and in valves you have a handful of electrons moving really fast. What matters is charge, not electrons. The amount of charge being moved per second is the same everywhere in a (series) circuit.

Quote:
Voltage drop has to do with what exactly? Does the term "Drop" mean the same as "decrease"
Bascially yes. It is a measure of how much energy was lost while shoving the charge through the component. (Joules per coulomb = volts)

The voltage across a battery is called EMF, and is a measure of how much energy is gained by the charge by pumping it from one terminal to the other. That energy is lost again as the charge flows around the rest of the circuit (where the drops occur).That's why the EMF voltage arrows point in the opposite direction to the voltage drop arrows on diagrams.

Last edited by Merlinb; 27th November 2013 at 09:59 AM.
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Old 27th November 2013, 12:19 PM   #6
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I have printed NEETS and I'm on the third chapter (reading every day).

In a series circuit, if the total voltage drops equal the voltage supplied Rt = R1 + R2...+ Rn, then it would seem that the series of resistors would reduce the supplied voltage to zero. In other words, you should expect measure a zero voltage if you measured at the point immediately before the positive battery terminal which is also the point after all of the resistors....but since voltage is the measure between two (2) points, what would be the other point that would result in this zero measure
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Old 27th November 2013, 12:37 PM   #7
Merlinb is offline Merlinb  United Kingdom
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Quote:
Originally Posted by Eddiegnz1 View Post
In other words, you should expect measure a zero voltage if you measured at the point immediately before the positive battery terminal which is also the point after all of the resistors....
If you did that then you would be measuring between two points that are in fact the same point. In other words, you have connected the voltemeter probes together, so yes, you would get zero volts!
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Old 27th November 2013, 01:52 PM   #8
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Quote:
Originally Posted by Eddiegnz1 View Post
I have printed NEETS and I'm on the third chapter (reading every day).

In other words, you should expect measure a zero voltage if you measured at the point immediately before the positive battery terminal which is also the point after all of the resistors....but since voltage is the measure between two (2) points, what would be the other point that would result in this zero measure
This is not correct. Considering for a moment that your battery does not get discharged over time, the positive terminal voltage value does not vary. The negative battery terminal is your reference, thus considered your zero test point.
Of course you may reverse your reference and take the positive terminal as your zero point, hence you will measure a negative battery value on the negative terminal.
It has been said above and it's most important to get the concept that any voltage value is measured between any 2 points in a circuit with a resistive path in between them.

Having said this, all voltage drops across the resistors in your in series array will sum up (or subtract, depending on which side of the battery you take as reference) for a total voltage value as that of the battery.
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Old 27th November 2013, 01:55 PM   #9
BIGslim is offline BIGslim  United States
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Sometimes using the water analogy can help with the concepts of electricity. Here is an example from google.

Water circuit analogy to electric circuit

If you open your kitchen faucet halfway, half the water pressure (voltage) is "dropped" across the faucet. You would see this if you put a water pressure gauge (DMM) on each side of the faucet.
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Old 27th November 2013, 05:24 PM   #10
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Quote:
Originally Posted by Eddiegnz1 View Post
I have printed NEETS and I'm on the third chapter (reading every day).

In a series circuit, if the total voltage drops equal the voltage supplied Rt = R1 + R2...+ Rn, then it would seem that the series of resistors would reduce the supplied voltage to zero. In other words, you should expect measure a zero voltage if you measured at the point immediately before the positive battery terminal which is also the point after all of the resistors....but since voltage is the measure between two (2) points, what would be the other point that would result in this zero measure
The battery has two terminals, + and - . The resistors are joined end to end in a string that goes from one to the other. You start measuring the voltage across the entire string from "top" (V+) to "bottom" (V-) with your red plus probe on the top V+ and the black minus probe on the bottom V-. There, you'll measure the total voltage output of the battery. Then, keeping the black probe where it is, if you start moving the red probe downward, one junction at a time, the voltage will decrease each time by the voltage drop across the resistor you have just omitted. Eventually, you end up skipping the last resistor in the chain - Rn - and now both probes are on the same V- rail. So there is no difference between them, so you measure 0 V.

You could also do this in reverse, keeping the red probe at the top and moving the black probe upward, one resistor at a time.

Do you have a volt/ohm meter of any kind? If not , you should get one. It's clear you're working at this so I'd strongly recommend you get a copy of Starting Electronics by Keith Brindley, published by Newnes. It's a well ordered little book that starts with just a very few parts and tools and then gradually builds up. Everything is taught with little experiments. (You could do it with LTspice but where I went to school the teachers didn't let you use CAD until you were really familiar with the real objects the icons represented and working with guys who learned with CAD I see why. It makes a difference.) If you don't have a meter you should have one regardless. Measure everything that doesn't move !
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