As im writting my tutorials as i find time, and energy. Im intending to continue where i left off time to time..

So if you dont mind, please post your comments/suggestions etc in either this area, or in a separate thread. This will help me produce a clear line of thought from start to finish in the thread. After its done intend close them, and make changes as needed (from your comments).

This way some future reader may have a good set of tuts to work with. (I dont like how the Main's Iczelion tuts are in total disarray now, so im trying to prevent this.

:alright:
NaN
Posted on 2003-04-10 20:54:27 by NaN
Keep up the good work:alright:
Posted on 2003-04-11 02:18:43 by clippy
Hey NaN, keep it up, I pay more attention to your tuts than I pay to my high school teachers LOL! (I know scary), but anyway can I ask questions in the Theory section or do I make a new thread?
Posted on 2003-04-12 13:43:13 by x86asm
Either or...

I have finished the Ohms Law thread...

I expect alot of questions to follow.. I might add more questions if people are still having a hard time... These questions tho might be better suited for this area, and outa the Theory section.

:NaN:
Posted on 2003-04-12 14:50:44 by NaN
I've also posted the resistance color code table in this thread

:alright:
NaN
Posted on 2003-04-12 23:39:57 by NaN
Just feeling around for peoples take on the tutorials...

I have been going pretty slow and basic. I can step up the pace a if people want me to... or are people waiting for more challenging stuff, since Ohms Law is kinda simple?
Posted on 2003-04-18 23:43:09 by NaN
Hi Nan,
Just finished your Ohm's Law tut. Very nice and easy to follow. Waiting for more tuts.

But just one thing. The voltage is still troubling me.

As you said
Voltage is Potential Energy Available (or Used) on a per Coulomb basis

How does this apply to voltages in parallel or in series.

Does it mean that in parallel the enery available is the same and in series it is not ? But why? Why cant it be the opposite.
I mean, i have a hard time just sort of mugging any law. There must be some reason as to why voltages are equal in parallel. If so why?

In your thread Theory: Ohms Law , in the first circuit diagram you show, the one with a voltage drop of 12v with 6amp current and 2 ohms resistor.

Now what if the resistance were made 3 ohms instead of 2 ohms, then according to the ohms law formula,
``````v = i*r
i=6
r=3
so, v = 6*3 = 18 volts
``````

So now there is a voltage drop of 18 volts. But if battery can supply only 12v then what happens to the extra 6v ?
Posted on 2003-04-19 08:02:09 by clippy
If I can answer the question:

The problem with the situation you outlined, gladiator, is you presumed the current will stay the same. With a higher level of resistance, less current will flow. The energy source only has a potential of 12V. So, i=e/r, 12/3 would equal 4 amps of current.

Energy must be conserved and you can't create energy from nothing. If we were able to leave the current at 6amps then the total power available will also have increased, p=e*i = 12*6 = 72watts compared to 12*4=48watts.

Don't know if he's covered power yet.
Posted on 2003-04-19 09:02:38 by drhowarddrfine
So now there is a voltage drop of 18 volts. But if battery can supply only 12v then what happens to the extra 6v ?

Oops, not quite. The 6 amps was based on 12 volts across 2 ohms. If you change the resistance to 3 ohms, the current drops to 4 amps. The 12 volts stays the same.

WHY? well, first off, schematic symbols can mean many things, as they are a shorthand. Sometimes a wire means a real wire, certain size, made of some metal, or some length. then again, it may be a mathematical wire, which is pure, has no resistance, shows no length effects, and several other non-physical properties.

So the first type of schematic consists of ideal mathematical elements that interact in very simple straightforward ways. The 2nd kind is more of a connection diagram between real physical components. The 2nd kind will require a model of it, or another schematic, before it can be analyzed.

OK, voltage. Voltage is the force that drives the current. The EMF, Electro Motive Force. Keep that definition in mind while you learn this stuff. A voltage can only exist between two points. If you hear someone talk about the voltage at some point, they really mean referenced to some assumed point, such as referenced to ground (which is why ground is so useful as a shorthand).

In NaNs circuit, the voltage is a given, he tells you it is a 12 volt battery. So you can't compute the voltage, it's stated. On the other hand, had he instead told you the resistance was 3 ohms and the current was 6 amps, you could compute the battery was 12 volts.

Put 4 D cells in series, each about 1.7 volts, you get a battery of 6.8 volts. (A battery is defined as 2 or more cells in series. A cell is created by a chemical reaction, and the physics determines the voltage it supplies).

Voltages in parallel are EQUAL.

Put 4 D cells in parallel and they get HOT, and the voltage is still the 1.7 volts of each cell. They get hot because the individual cells have slightly different voltages (they are real world things after all), so the one with the highest voltage tries to charge the lower voltage batteries. So don't try this one at home kids.

One important point for analysis is the voltage around any closed loop is zero. This can be most useful in analyzing a circuit.

I'll have to make some tuts on how to do that. ;-)
Posted on 2003-04-19 09:13:01 by Ernie

drhowarddrfinedrhoward,

No, i havent gotten to power yet.. I dont want to provide more confusion than is already needed ;) . Im sure we both agree its not tough, but still its another variable to add to the confusion.

Everyone else:

So what would people like to see from here, a short list of progressively challenging ideas are:

[*]Power I (DC power)
[*]Mesh / Nodal analysis (would probably get into some Matrix techniques also)
[*]Simple Silicon devices (Diodes / Transistor / Zenors)
[*]Inductors / Capasitors
[*]Transient Analysis of DC circuits
[*]AC circuits and Impedance analysis (gets fun from here ;) )
[*]AC Filters (first order)

Posted on 2003-04-19 10:38:09 by NaN
Hey NaN just checking say I want a voltage drop of 2V using a resistor (say for interfacing ) and I know the current I would rearrange V=IR so that I isolate the R, so I assume it would be V/I=R , if I go through the calculation I will get the resistance needed to drop the voltage correct? Say for example lets say my 16F628 (since I will be designing ciruitry to play around with this chip) I don't have time to look through the but lets assume the current it can take is 250mA.

So then I would do as follows
V/I=R so
2V/0.250=R
what the hell I get 8, 8 ohms resistance is there a resistor that overs that small resistance or is there an error in my calculations? Help before you go on :D
Posted on 2003-04-19 19:10:30 by x86asm
I dont understand what circuit you have in mind, so its hard to give you an answer.

However, my best guess is you have a bright led that runs at 3V max, and you have to drop the outputs from 5 V to 3V to meat its spec.

If so, your saying that the CHIP supplying voltage has 5V with no more than 250mA MAX. Your math is then correct. At 250mA you will get 2 Volts across a 8 Ohm resistor. However, realize this is the MAX current. You should not get to these extents deliberately in design! Its hard on the silicon devices.

In this example your also forcing 250mA through the LED diode. Which is quite high for the average LED. You should also check the spec on the diode and make sure your meeting its spec. Most diodes can operate happily at 10mA. This changes the design of the resistor considerably!

(5V - 2V)/10mA = 300 Ohms.

In practice most people choose to use a 470 Ohm resistor. This means less current in your case. But often the doide doesnt drop 2V. Often its around a Volt.

Hope this helps.
:alright:
NaN
Posted on 2003-04-19 22:19:32 by NaN

I dont understand what circuit you have in mind, so its hard to give you an answer.

However, my best guess is you have a bright led that runs at 3V max, and you have to drop the outputs from 5 V to 3V to meat its spec.

If so, your saying that the CHIP supplying voltage has 5V with no more than 250mA MAX. Your math is then correct. At 250mA you will get 2 Volts across a 8 Ohm resistor. However, realize this is the MAX current. You should not get to these extents deliberately in design! Its hard on the silicon devices.

In this example your also forcing 250mA through the LED diode. Which is quite high for the average LED. You should also check the spec on the diode and make sure your meeting its spec. Most diodes can operate happily at 10mA. This changes the design of the resistor considerably!

(5V - 2V)/10mA = 300 Ohms.

In practice most people choose to use a 470 Ohm resistor. This means less current in your case. But often the doide doesnt drop 2V. Often its around a Volt.

Hope this helps.
:alright:
NaN

Yup thats what I was looking for NaN, thanks :)
Posted on 2003-04-19 22:32:12 by x86asm

Yup thats what I was looking for NaN, thanks :)

Just for your own interest NaN and everyone else, I want to interface my PIC16F628 to 8 LED's to one of its output ports and program a cool light show with it (Remember that guy on the old Star Trek with those lights on his visor type thingy thing going back and forth etc.,I want to imitate that in a sense, man Data was the best robot :) ) , so I would preferably like to drive the LEDS directly from the PIC's output PORTA. But it seems I may need a limiting resistor.
Posted on 2003-04-19 22:47:51 by x86asm
Unless things have changed a led is just a diode and runs at .7v So if the pic has a cmos output close to 5v you need to drop 4.3v across the resistor. If you can find out the maximum safe current of the led then it would be 4.3v divided by that current. NaN is right and I think 470 ohms works in most cases.
Posted on 2003-04-19 23:13:12 by drhowarddrfine
I was puzzled by the 2V requirement myself ;)

However, i have seen bright light LED's that can handle ALOT of current and drop higher voltages... so this was the best assumption i could take. All silicon diodes drop approx 0.7V and Germanium Diodes drom approx .3 V.

LEDS are diodes, but they have other charateristice that gives them the ability to defy this rule of thumb. Its up to the manufacturer really. But ya, most 'standard' red LEDs can be approximated to about 0.7->1.0V drop.

x86asm,

One last thought. If you are using high current LED's beware you can still damage your chip if your not careful. If you design the resistor to pull near the LED's rating, say 200mA, your chip has to also handle 200mA out the port pin.

So what happens when you put eight of them on your chip? You have 8 port pins all allowing 200mA each. While each pin may be ok with 200mA individually, the VDD pin may not allow more than 250mA! This means your pulling 8 * 0.2 A = 1.6 Amps through a power source pin that can handle no more than 1/4 Amps!

In this scenario your turning your nice PIC chip into a toaster... really!

You should be aware of such things! I would try to limit each pin to around 10-15mA to keep things safe! Also read your spec sheets to see what your chip will tolerate!

Just keep in mind: In the electronics world, 250mA is alot of current! Which demands extra care in your design so you dont damage chips.

:NaN:
Posted on 2003-04-19 23:55:20 by NaN
In practice most people choose to use a 470 Ohm resistor.

I must not be most people... I just spec'ed a 332 ohm for that (bi-color LED off 2 PIC pins for 3 colors (4 counting dark))
Posted on 2003-04-20 12:47:37 by Ernie

Why yes, things have changed, LEDs need more then .7 volts. They are not silicon junctions, then use exotic stuff to make the colors... the color being determined by the energy difference (hence emmitted) by an electron filling a hole... or something like that, haven't reviewed the internal physics for 20 years.

Do check the spec sheet for the LED you want to use, you usually have to play with 2 graphs, one for intensity vs current, another for forward drop vs current. Pick an intensity and wade thru for the voltage, pick a resistor to drop the drive voltage (usually Vcc) minus the desired drop.

So in this case, 2V makes sense. Burn 3 volts on the resistor, 2 on the LED. Terrible efficiency, but practable.

PICs get two ratings for output current, one per pin, and per pacjage total. Usually its 25mA per pin, 250mA per package, but do check your specific part.

Pins may be paralled for extra current, just make sure they are on the same port and all changed in sync. Also avoid read/modify/write instructions for setting them, if the load is pulling them one way or the other (logic high or low), an XOR isn't going to change the state.
Posted on 2003-04-20 12:52:23 by Ernie

I was puzzled by the 2V requirement myself ;)

However, i have seen bright light LED's that can handle ALOT of current and drop higher voltages... so this was the best assumption i could take. All silicon diodes drop approx 0.7V and Germanium Diodes drom approx .3 V.

LEDS are diodes, but they have other charateristice that gives them the ability to defy this rule of thumb. Its up to the manufacturer really. But ya, most 'standard' red LEDs can be approximated to about 0.7->1.0V drop.

x86asm,

One last thought. If you are using high current LED's beware you can still damage your chip if your not careful. If you design the resistor to pull near the LED's rating, say 200mA, your chip has to also handle 200mA out the port pin.

So what happens when you put eight of them on your chip? You have 8 port pins all allowing 200mA each. While each pin may be ok with 200mA individually, the VDD pin may not allow more than 250mA! This means your pulling 8 * 0.2 A = 1.6 Amps through a power source pin that can handle no more than 1/4 Amps!

In this scenario your turning your nice PIC chip into a toaster... really!

You should be aware of such things! I would try to limit each pin to around 10-15mA to keep things safe! Also read your spec sheets to see what your chip will tolerate!

Just keep in mind: In the electronics world, 250mA is alot of current! Which demands extra care in your design so you dont damage chips.

:NaN:

Ohhh, I see, I will remember that (write it in my notes :) ). Thanks NaN
Posted on 2003-04-20 14:18:59 by x86asm

Why yes, things have changed, LEDs need more then .7 volts. They are not silicon junctions, then use exotic stuff to make the colors... the color being determined by the energy difference (hence emmitted) by an electron filling a hole... or something like that, haven't reviewed the internal physics for 20 years.

Do check the spec sheet for the LED you want to use, you usually have to play with 2 graphs, one for intensity vs current, another for forward drop vs current. Pick an intensity and wade thru for the voltage, pick a resistor to drop the drive voltage (usually Vcc) minus the desired drop.

So in this case, 2V makes sense. Burn 3 volts on the resistor, 2 on the LED. Terrible efficiency, but practable.

PICs get two ratings for output current, one per pin, and per pacjage total. Usually its 25mA per pin, 250mA per package, but do check your specific part.

Pins may be paralled for extra current, just make sure they are on the same port and all changed in sync. Also avoid read/modify/write instructions for setting them, if the load is pulling them one way or the other (logic high or low), an XOR isn't going to change the state.
Thanks I'll go order some bulk LEDs now I'm running out or I could use the ones in my electronics kit :)

Thanks Ernie I'm learnding :tongue:
Posted on 2003-04-20 14:22:25 by x86asm