# Newbie questions

Discussion in 'General Electronics Discussion' started by WeaselNo7, Jun 5, 2012.

1. ### WeaselNo7

4
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Jun 5, 2012
Hey all,

I have a few grey areas about basic electronics I need clarifying! I'm trying to get a 'feel' for it all (as well as the mathematical knowledge), but there are a few stumbling blocks. I'd be grateful if I could get some feedback!

1) How is everyone thinking about current and voltage? Is this relationship just something you have to get used to? Is there a useful analogy?
I know Voltage is Joules per Coulomb and current is Coulombs per second. Do I get any advantage from considering Joules to be passengers and the Coulombs to be buses? Or is this a misleading/unnecessary analogy

2) Too much voltage. If I have a single device that uses 5v and I have a 6v source... what will happen? does the remaining voltage get lost with no side effect? Is the device forced to 'take' 6v?
What happens in the above case if I have multiple devices in series that add up to less than the source voltage?

3) Too little voltage. I'm aware that the device in the above question would get the entire voltage in the case of a 4v source instead. What happens if there are multiple sources in series that require more than the source voltage? Is the voltage round-robinned across the devices? (so they each get a weighted portion of it, based off what they need?), or will the first device in the series get the voltage it needs and the remainder goes to the second device?

4) LEDs are 'voltage limited' devices (did i get that right?), which means they take X volts and potentially demand unlimited current (which is why you need a resistor in series to 'drive' the 'correct' current).
Why are they like this? are there other devices in this 'class'?
What other 'classes' are there?

5) Some LED datasheets show 'maximum forward voltage' as well as 'reverse voltage' and 'maximum forward current'. What do these values mean? Why are they different? (I'd guess the current one is to do with the maximum circuit current that can pass through it without setting fire to it - or similar!)

Thanks for your help everyone, hope my blatherings make some sense!

Sam.

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,489
2,830
Jan 21, 2010
I think of it as Pressure and Flow rate.

Well, yes, you have to have some intuitive understanding.

Many, see mine above.

Great for physicists, and maybe understanding some equations, but hardly ever something you need to think about.

That analogy doesn't help me.

Maybe nothing, maybe *poof* goes the magic smoke.

Well, it can't get lost...

In that you say 6V is applied, yes, it is forced to accept that 6V. No whether or not it has some internal capability to regulate the input voltage or not, that's a different question and one that differs on a case by case basis.

Possibly the same thing.

Anything from working properly, to partially or incompletely working, to not working at all, or even going *poof* and letting the magic smoke out.

Any of the above.

It depends totally on the devices. All that you can be certain of is that the sum of the voltages will add to the power supply voltage and in the real world none of the voltages will likely be negative.

https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html

They're a case of non-linear device where the current demanded rises faster than the voltage across them.

Almost all semiconductor junctions

It all depends on how you classify them.

3. ### WeaselNo7

4
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Jun 5, 2012
Awesome, thanks Steve, I reckon I'll spend the rest of the Jubilee working through that sticky.

4. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I don't find either of those ideas helpful when understanding the relationship between voltage, current and resistance in circuits.

Here is an analogy that I like. I came up with it myself, though I'm probably not the first to think of it. It is not a traditional analogy, but I find it works pretty well for understanding voltage, current, resistance and capacitance in simple circuits. Semiconductors can be modelled too, with some extra imagination, but I don't think inductance has any equivalent in this analogy. Here are the quantities and basic components. I'll explain the model using examples.

- Voltage is vertical position;
- Current is tension or force;
- Resistors are tension springs (lower resistance means a firmer spring) or pieces of elastic;
- Capacitors are cylinders with plungers in them.

----------------------- +12V
|
\
/ R1 (tension spring)
\
/
|
|------- centre node
|
\
/ R2 (tension spring)
\
/
|
----------------------- 0V

The above ASCII art drawing shows the familiar two-resistor voltage divider. Two resistors, R1 and R2, are tension springs. This is the type with a loop at each end that is stretched between two points. You could also use a rubber band or a piece of elastic to represent a resistor in this analogy.

R1 and R2 are joined ("connected in series") and stretched between two voltage rails, +12V and 0V. The node where they join together is part-way between the two voltage rails; the exact position depends on the values of R1 and R2.

1. You can see that the VERTICAL POSITION on this diagram represents VOLTAGE. The 0V rail at the bottom of the diagram has "no voltage" (it is the reference point from which voltage is measured), and the +12V rail at the top has +12V voltage (relative to the 0V reference). The centre node has a voltage in between 0V and +12V. If R1 and R2 are the same, the centre node will be half way between +12V and 0V, i.e. it will be +6V.

2. You can see that if you REDUCE THE RESISTANCE of R1, by making R1 FIRMER (using thicker wire), the centre node will be pulled upwards towards +12V. This is what happens in practice: if you reduce the resistance of R1, the centre node voltage increases.

3. You can see that the CURRENT (tension) in the two resistors is EQUAL, and will ALWAYS be equal in this circuit, because they are connected in SERIES.

4. You can see that if you reduce the resistances of BOTH R1 and R2 equally, the centre node will still be at 6V but the tension (current) through R1 and R2 will be higher.

5. You can see that if you put another "load" resistor (tension spring) in PARALLEL with (i.e. across) R2, the voltage on the centre node will be pulled down, i.e. it will drop. The lower the resistance of this load resistor, the more the voltage will drop.

6. You can see that if R1 and R2 have low resistance (firm springs) and the load resistance is high (a thin, weak spring), adding the load resistor will have little effect on the centre node voltage.

These behaviours of this analogy all exactly model reality, and are useful in understanding voltage dividers and the behaviour of resistances in general.

Now let's introduce capacitance. A capacitor's behaviour is to keep the voltage across it constant, unless a current (tension, or force,) is applied. This can be modelled as a cylinder with a plunger or piston in it. The plunger is represented by the "=" signs.

An important feature of the capacitor is that the plunger will move at a SPEED that is PROPORTIONAL TO the FORCE applied. The higher the force, the faster the plunger will move. There is a linear relationship between force and speed.

| . | . |
| . | . |
|===.|
| . . . |
| . . . |
--------

Sorry about the dots. The Electronics Point software strips out whitespace in postings so I have to use them.

.. ----- 'X'
.. |
.. /
.. \ R1 (tension spring)
.. /
.. \
.. |
.. |--------- centre node
.. |
| . | . |
| . | . |
| . | . |
| . | . |
|===| C1 (cylinder with plunger)
--------
. |
----------------- 0V

Here is a resistor and capacitor in series. The bottom end of the capacitor is connected to 0V and the plunger is at the bottom of the cylinder. The top end of the resistor is initially not connected to anything, but we then connect it to a positive voltage.

1. Initially, C1 is discharged, i.e. the plunger is at the bottom of the cylinder, and there is no voltage on the centre node.
2. When we connect point X to a positive voltage, the tension through R1 will start to pull the plunger in C1 upwards, i.e. C1 will start to charge up.
3. As C1 charges up, the centre node's voltage increases.
4. As the centre node voltage increases, the extension of R1 (i.e. the remaining voltage across R1) decreases. This reduces the tension in R1, which causes C1 to charge at an ever-decreasing rate.
5. As the voltage on the centre node approaches the voltage at point X, there is almost no voltage across R1 so the tension in R1 is almost zero. This means that C1 charges very slowly, much more slowly than it was charging at the start when voltage was first applied to point X.

This behaviour again models reality, and explains the logarithmic voltage curve that is seen when you charge a capacitor from a fixed voltage through a resistance.

You will have noticed some problems with this model. Vertical wires do not follow the analogy. If the top of C1 and the bottom of R1 are connected together, they are at the same voltage, and they should be at the same vertical position in the drawing, but they are not, because of the vertical wire joining them. You have to pretend that they are at the same vertical position. Also, tension springs and pieces of elastic have a fixed length even when they're not being stretched at all, unlike resistors, which have no voltage across them when there is no current flowing. You have to use a little imagination to deal with these limitations in the analogy.

Now semiconductors. These can be modelled more or less usefully, depending on their complexity. I like to use imaginary little men for modelling semiconductors.

Since you asked about LEDs, let's model an LED. This is hard to do in ASCII art; you'll have to use your imagination. Start by ignoring all the dots in the drawing.

. . | anode
. . |
---------------- 2V (top of box)
| . | . . .-- -- (feet)
| . ==== \/
|. . . . . . |
|. . . . . . o (head)
|
---------------- 0V (bottom of box)
. |
. | cathode

Now this one is pretty weird. It's a C-shaped box with a little upside-down man inside it! I've labelled his feet and his head. The cathode connection is the bottom of the box, and I will connect it to 0V, so voltages are measured relative to the bottom of the box. The anode connection comes through the top of the box and attaches to a piece represented by the "=" signs. The man's foot is stuck between that piece and the top of the box. No, I'm not using recreational drugs; this will make sense soon.

If the anode voltage is zero, the '=' piece will sit at the bottom of the box and nothing will happen. As we start to apply voltage to the anode, the '=' piece will move upwards within the box, but it's not restrained in any way, i.e. no current will flow.

When it reaches the little man's foot, it meets a restraint, and the tension will increase, i.e. current will start to flow. Assuming slight tension, the man will find his foot being squashed a little, and his face will go slightly red. This is equivalent to the LED starting to glow dimly. If we increase the tension (current), his face will glow more brightly.

Also, because the little man's foot is somewhat soft, as the current (tension) increases, the voltage across the LED will increase, but only slightly. For example, a current increase from 1 mA to 50 mA might only correspond to a voltage increase from 1.7V to 1.9V.

If the current is increased too far, the little man expires and the box breaks and loses its magic smoke.

You can see that a simple way to make an LED light is to feed it through a resistor from a voltage rail. The resistor provides a relatively constant force (current) so that the little man's foot is squeezed, but not too hard. Such a resistor is called a "current limiting" or "current setting" resistor.

For simplicity, this model of the LED does not include the LED's behaviour in response to reverse voltage.

Other semiconductors can be modelled using little men who do different things according to how hard their feet are being squeezed!

Yes, the device is "forced" to "take" 6V, provided that the driving resistance is low enough.

A simple voltage source can (in simple terms) be modelled as an ideal voltage source and a series resistance. The series resistance is normally quite small. Let's imagine a small 9V battery supplying voltage to a "load".

----------------------- +9V
|
\
/ SERIES resistance (a few ohms - a fairly hard tension spring)
|
|------ actual battery terminal voltage
|
\
/
\
\
/
|
----------------------- 0V

The series resistance is actually part of the battery itself, i.e. it is INTERNAL to the battery. Your load is connected across the battery terminals.

Assuming the load doesn't draw much current, not much voltage will be dropped across the internal series resistance of the battery, and the terminal voltage will be close to 9V. But if you start to draw a lot of current from the battery, the terminal voltage will drop.

If you're powering a semiconductor device that has some internal behaviour that prevents it from having much more than 5V across it, this is similar to the LED example I gave you before. If you connect a 9V battery to it, a heavy current will flow, and will "squash the man's foot". Depending on how much current the battery can supply, i.e. how hard the internal resistance spring is, the little man may take it (the voltage will drop, because of the voltage lost across the internal resistance), or the little man may expire.

So it depends entirely on what's inside the device. If it's just a resistor or a light bulb, it will just draw more current.

Using the LED example, where a typical red LED needs about 1.7V before it will illuminate, if you connect six LEDs in series across a 9V battery, the 9V will be divided roughly equally by six, so each LED will see about 1.5V across it, and none of them will light.

In other cases, all the devices may work, but not properly, because there isn't enough voltage across each of them.

If the devices are all different, or have irregular current consumption, the splitting of the total applied voltage between the different devices may be uneven or irregular, so some of the devices will malfunction or may even be damaged.

It depends on how they behave. You would NOT deliberately use such an arrangement. If the devices are resistive, they get a "weighted portion of it", as you say. You can imagine how this works with weighted springs joined together. But only simple resistive devices (resistors, and to a limited extent light bulbs) are this simple. Semiconductor devices usually have behaviour that is too complicated to model usefully in this way.

I've explained LEDs fairly clearly (I hope). Semiconductor junctions (such as the base-emitter junction of a bipolar junction transistor) behave similarly (with a lower forward voltage). Complete circuits can have similar behaviour to this, but are far too complicated for a simple analogy. They are designed to be supplied from a simple stable voltage source, and that's what you should do.

Some other "classes" of devices, I suppose, would be capacitive and inductive, which can be modelled as single quantities, and various non-linear devices (semiconductors, light bulbs, transient suppressors, ...).

I hope you can answer those questions yourself now, if you understood my weird analogy.

Sure they do. I hope my ANSWER makes some sense!

Last edited: Jun 8, 2012
5. ### Zapdos

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0
May 24, 2012
Hey mate,

1) I came from a physics background, and also learned Voltage = Joules/Coulomb, Current = Coulombs/Second. I think your analogy is fine, but there are better ways of thinking about it IMO, in particular, the water flow analogy that (*steve*) mentioned.

Another way to think about voltage is using the hill analogy - which is more accurate if you're thinking about potential energy changes (changes in voltage).

So a 12V battery starts you off on a hill of height 12 (you have 12 "units" of potential energy), and say you have two devices using 6V each, as you travel around your circuit, you slide down the hill by 6 units (each device "uses up" 6V), and when you come back to the start, you're at zero. Does that make sense?

It's similar to gravitational potential energy if you're at all familiar with that....

2) Voltage cannot be "lost." If you increased the size of the hill in the analogy above to 20V, each device would have to "use up" 10V each. Look up "Kirchoff's Voltage Law," this may be instructional.

Depends how much excess voltage you have and over what device to determine what will happen. It will probably fail, or at least cause problems in your circuit.

3) Voltage splits in ratio with the resistances of the devices in series. Since the same current travels through each device, and V = IR, a large resistance device will "use up" a large proportion of the source voltage.

4) Not too sure, I'll leave it to the experts.

5) LED = Light Emitting DIODE. Diodes are devices that allow current to only flow in one direction, the "forward direction." But if you apply more than the max forward values to the LED, it may fail.

The reverse voltage, I believe, is how much voltage can be applied in the reverse direction before the diode blows up.

Hope that helped!

EDIT: KrisBlueNZ - That analogy is awesome! I need some time to wrap my head around it though

Last edited: Jun 5, 2012
6. ### WeaselNo7

4
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Jun 5, 2012
Hey, two more things:

1) The sticky didn't mention it explicitly, but I gather reverse current is the maximum current the diode can take if the current is reversed (e.g. plugged in wrong way round) and anything above will damage it

2) An LED is a non-linear device, but that isn't the reason for the need for an LED to be placed in series, right? (although it's certainly an extreme case for why) , any device that will potentially pull more current for a given voltage than the entire circuit can handle needs to have the voltage reduced dynamically (in our case, by the resistor). The exponential nature of the 'thermal-runaway' is extreme, but anything that went past a threshold would need to be guarded against.

7. ### BobK

7,682
1,688
Jan 5, 2010
You questions about what happens to voltage when devices are in series it looking at it the wrong way. The definition of things in series is that the current thorugh them must be the same. How the voltage divides is complex and can change with time, but, at anyt time, the current through each device must be the same.

Consider for example, that one of the devices is a blinking light and the other is a light that is on constantly, and let's say that either of them can handle the entire voltage without failing, so nothing bad happens. The blinking light will require more current when it is on than when it is off. So what happens to the other light? And how is the voltage shared between the two devices when the blinking light is on and when it is off. If you can answer these questions, you understand it pretty well.

Bob

8. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
Re WeaselNo7's two more things.

Reverse current, for a diode, is the maximum current that the diode will pass when reverse biased by a specified voltage. An ideal diode does not allow ANY current to flow when it's reverse biased. Any current that DOES flow is called LEAKAGE current. This does not really relate to the diode being "plugged in the wrong way round" - in most applications, a diode will be reverse biased some of the time.

The non-linear behaviour of an LED is not specifically the reason for using a current-limiting resistor. Any device or circuit that provides a reasonably regulated (or limited) current will work. It is necessary to provide a known CURRENT, rather than a fixed VOLTAGE, because the forward voltage of an LED at a particular current is poorly defined - it varies slightly from one unit to the next, from one batch to the next, depending on temperature, and depending on age of the LED; it varies significantly between type numbers, and between manufacturers, and it varies greatly between colours. Forward current is the important factor, regardless of the forward voltage this produces, so it's the current that must be regulated (albeit fairly roughly).

9. ### WeaselNo7

4
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Jun 5, 2012
You lot are awesome. Except you Kris. You're awesome and mental. Great work everyone, I'm getting a feel for it now, you've all been excellent help.

10. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Cool! Don't get the wrong idea because my analogy uses little men to roughly model semiconductors. After all, semiconductors ARE called "active" devices! But even if you don't like that part of it, it's still useful as a model to help you understand voltage, current, resistance and capacitance.