# Teaching an old dog new tricks.

#### dl324

Joined Mar 30, 2015
17,144
That looks like fun. Is there instructions for creating something that online?
You can make it up as you go along. There's not much to it, but I'll draw up a schematic when I have a chance.

The guy who built it used red paint to mark '+' and made multiple notations on the back to make sure he kept left and right straight.

The solderless breadboards are held in place with the adhesive backing they came with. He used hot glue to hold a lot of the stuff in place. I used my hot air tool to remove most of it.

This is my first power supply and the business end of my first digital voltmeter:

I put the reversed biased diodes on the supply to have a convenient place to attach mini-grabbers.

#### dl324

Joined Mar 30, 2015
17,144
Is this the like the "Bible" of Electronics?
I'm not a fan of AOE. You can probably find free PDFs. I have a couple versions. Don't remember where I got them and they're too large to attach to a post.

Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
Read the thread on LaTeX for formating equiations.

Some will try to write the euation for charging a capacitor as something like:
Vf=Vi(1-e^(-t/RC))

This is much easier to read:
$$\large V_f = V_i(1-e^\frac{-t}{RC})$$

Also be careful with units and the proper names for things.
Volts is abbreviated with a capital V (5V).

A pet peeve of mine is people calling MOSFETs (Metal Oxide Semiconductor Field Effect Transistor) FETs. The first available field effect transistors were junction FETs (JFETs), so when I see/hear FET, I think JFET. There are many kinds of field effect transistors (IGFET, MESFET, etc), so FET isn't descriptive.

In general, acronyms are all caps.
Thank you for the Tips.
I have no problem being corrected, I endeavor on being concise and using the proper naming conventions, to not do so only causes communication issues and problematic misunderstandings.

#### Tonyr1084

Joined Sep 24, 2015
8,002
The water analogy is good but also flawed. The water flow is analogous to current, same as in a stream or river. There is current. In a hose, if you kink the hose you restrict the flow of water. That's analogous to a resistor, since a resistor resists the flow of current in an electric circuit. However, the water PRESSURE is akin to voltage. Higher pressure and higher voltage are similar. But here's the major difference: In a hose there's a source and an outlet. But the outlet doesn't return to the hose. In an electric circuit no current flows despite the amount of pressure (under normal circumstances) if there is no return to the source. In the circuit I drew, if you eliminate the bottom line then no current will flow. Unlike a water hose, the circuit can't work unless it returns to the source.

A battery (or other power supplies) have a potential and a ground. Ground is a term that in itself can be confusing, but let's just leave that alone for now. The battery has a positive end and a negative end. On the negative end there is an abundance of electrons that are looking for a way to get back to the positive end where there is an absence of electrons. This is known as "Electron Flow". While that's a real thing - FORGET ABOUT IT FOR NOW. The "Conventional" take on current is that it flows from positive to negative. Stick with that. You'll probably never need to deal with Electron Flow. Not unless you want to argue with someone. Then that becomes a futile argument.

In the battery circuit current flows with a pressure of 12V. That current encounters resistance that reduces the amount of current to 10mA that the LED can handle. The LED also has that forward voltage that also comes into play, but if the cathode end (the end with the bar) does not return to the battery negative then the electrons go nowhere.

Funny thing about LED's, they almost don't care about the voltage. As long as the current is not too high. There are limits to be found on both extremes so don't take that as an absolute. You can power an LED from a 240VAC source but you need to limit the current to the operating current of the LED. Since AC is a wilder beast, in AC the current wants to flow back and forth. The LED, being a Diode, wants to conduct in one direction but block current in the opposite. LED's have a breakdown reverse voltage rating. Exceed that and you blow up the LED. So in an AC circuit of considerable voltage you need either a blocking diode or a reverse conducting diode. But lets not go there just yet. First learn the DC side of it. The AC side will come later.

Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
Reading schematics is the key. Understanding what the symbols mean will show you how a circuit works. There are simulator programs you can get - but I'd suggest you first get the basics down before you try your hand at sim's.

The most basic circuit is lighting an LED. A simple 5mm LED that operates on 5 to 20 milliamps is an easy enough place to start. You need a voltage source, a resistor and an LED. But you need to understand what the max and recommended amperages are for the LED you're using. Not all LED's operate the same. That's a little deeper.

Suppose you have a 12 volt battery and an LED that you want to power it with 10mA (milliamps). Diodes, LED's and other components have a forward voltage that comes into play. The below circuit shows the LED as having a 2Vf. That Vf must come into the calculations. ( 12V - 2Vf ) ÷ 10mA = 1KΩ. Different LED's will have different Vf's, so you must know what that is. That information usually comes from a Data Sheet. But not all LED's come with a data sheet, which can make it tricky. Let's leave that for another day. For now the circuit below shows 10mA flowing through the circuit. The LED is lit and it is not burning out because you're not exceeding its maximum amperage rating. Again, that comes from a data sheet.

Not to get bogged down, reading a schematic will take some time. But when you learn it will become like reading sheet music to a musician. The afore mentioned circuit looks like this:
View attachment 309628
Anywhere you measure the current it will always be the same throughout the whole circuit. BUT WAIT! There's more to it. The battery has its own internal resistance. In most cases that will not be critical. You can expect the above circuit to behave as is shown.

Look at the schematic. It tells the story. Even without the 10mA designations inside it - the story should be clear. 12V, 1KΩ, 2Vf. You can calculate the amperage from there and come up with 10mA. If you want or need to know the wattage (important for selecting an appropriately rated resistor) multiply the voltage times the amperage. 12V x 0.01A = 0.12W (or 120 milliWatts). A 1/8W resistor can handle the job but it will get hot. A 1/4W resistor will also handle the job and it will not get as hot because it's rated for 250mA, whereas the 1/8W is rated for 125mW.

Don't let this scare you. It may seem like a lot but it's really among the most basic of circuits, and you'll get to a place where you look at this and intuitively know it's running at 10mA and 125mW.

I'll come back to this a little later and pull out my 9V battery and a breadboard and my multimeter.
I don't have any data sheets on my existing LED's. Is there a standard means to use in case you have no idea what the LED is rated at?

#### Tonyr1084

Joined Sep 24, 2015
8,002
If you've done the math challenge I gave you, swapping the 12V battery for a 5V battery, you should have come up with the following:
( 5V - 2Vf ) ÷ 10mA = ? See if you come up with the same answer I do. I came up with 300Ω.

Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
Welcome to AAC.

We all had a beginning. And Dennis beat me to the welcome.

There are things over my head. Some way over and some that just skim across the top - as you can see from the polished scalp.

Basically it comes down to math. Ohms law basically states that all things are related. A given voltage and a given resistance will have that specific amount of current. DC is the easier part but AC is a bit more advanced. Just give it time and don't be afraid to ask questions.

Breaking that down into something simple; suppose you have a resistor in a circuit and a voltage. The resistor is 100Ω (ohms) and the voltage is 10V (volts). That means that 10V ÷ 100Ω = 0.1A (amps) designated as the letter "I" Those numbers can be turned around and still come out the same. If you have a circuit with an unknown voltage but you know the resistance is 100Ω and you know there is 0.1A then you multiply. 100 x 0.1 = 10. You have 10 volts. Since resistance is the thing that is most often a fixed value (doesn't change) you can change the amperage by changing the voltage. More volts and you'll have more amps. 100V ÷ 100Ω = 1A.
E (volts) is equal to I (amps) times R (ohms) ( ( E = IR ) )

That's the most basic part of a beginning. There's an educational section where you can learn from the very most basic to the very advanced. You can go as far as you like.
Thank you for your reply!
Yes, My understanding of Ohm's Law is weak I understand the math of it but one thing still puzzles me about it. I was told to think of voltage as water in a pipe, and a resistor as a smaller pipe to restrict flow but what I don't understand is that regardless if you have one resistor or 10 in a circuit you will still have 0 VOLTs after the last component in the circuit, but even if water flows through a smaller pipe you will still have some water coming out of that pipe. How come in a circuit that only has one resistor there are 0 volts as it returns to ground but in circuit with 10 resistors they all have different voltages after each resistor until the last one? See, my concept is fouled.
Do you have a DVM? Solderless breadboard/jumpers? Soldering iron?

I did electronics as a hobby for more than 30 years before I got a scope (now I have 8 or 10).

Thanks again for helping out.
Also, learn how to draw readable schematics. The preferred flow is left to right and top to bottom. Avoid scenic routing, unnecessary wire jogs/crossings/whitespace. Most old timers like me prefer monochromatic drawings; because that's what we were weaned on.

Here's a recent example from a student:
View attachment 309629
Here's how I would have drawn it:
View attachment 309633
Are those schematics drawn with a program? I've seen stencils with "And and Or Gates" on them. Do you hand draw them?

#### Tonyr1084

Joined Sep 24, 2015
8,002
Is there a standard means to use in case you have no idea what the LED is rated at?
Yes and No. Different manufacturers will come up with different numbers. Here's the numbers I have for my super bright LED's:
White ---- 2.97Vf (average out of 100 pieces)
Blue ----- 2.82Vf (average out of 100 pieces)
Green --- 2.92Vf (average out of 100 pieces)
Yellow --- 2.01Vf (average out of 100 pieces)
Red ----- 1.95Vf (average out of 100 pieces)

MY white, blue and green can be considered to be around 3Vf while the yellow and red can be considered to be around 2Vf.
Again, other manufacturers LED's will have different voltages. Search the internet for LED forward voltages and see what you find.

Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
Here's a quiz for you: What resistor would you need if you had a 5 volt battery instead of a 12V battery?
Wouldn't you need to know the (I) Current because R = V divided by I

#### Ian0

Joined Aug 7, 2020
10,277
Is this the like the "Bible" of Electronics?

I'll put it on my wishlist.

View attachment 309639
It's useful at every level. I bought my copy as an undergraduate in 1985 (same year as I bought my HP 15C calculator). H&H is looking a bit tatty with the spine held together by gaffa tape, but I still refer to it from time to time, even though I have an MA in the subject and work as a senior electronics engineer.
(HP15C calculator is still going strong)

#### Tonyr1084

Joined Sep 24, 2015
8,002
Wouldn't you need to know the (I) Current because R = V divided by I
Yes, you need to know - in this case - the desired current you're building for. Your target current is 10mA.

You already know the LED has a Vf of 2 volts. You have a 5 volt source. You subtract the lower from the higher and come up with 3 volts. Knowing you want the circuit to operate at 10mA you divide 3 by 0.01 (10mA) and you come up with the answer.

I'm going to lunch now.

#### nsaspook

Joined Aug 27, 2009
13,577
Thank you for your reply!
Yes, My understanding of Ohm's Law is weak I understand the math of it but one thing still puzzles me about it. I was told to think of voltage as water in a pipe, and a resistor as a smaller pipe to restrict flow but what I don't understand is that regardless if you have one resistor or 10 in a circuit you will still have 0 VOLTs after the last component in the circuit, but even if water flows through a smaller pipe you will still have some water coming out of that pipe. How come in a circuit that only has one resistor there are 0 volts as it returns to ground but in circuit with 10 resistors they all have different voltages after each resistor until the last one? See, my concept is fouled.
...
I think you're actually too advanced for the brain dead water analogy. Most people also don't understand the closed-loop hydraulic circuits needed to make the analogy useful. IMO don't use it to build foundational information about the electric circuits because it leads to misconceptions about the nature of electrical energy.
https://en.wikipedia.org/wiki/Hydraulic_analogy
The electronic–hydraulic analogy (derisively referred to as the drain-pipe theory by Oliver Lodge) [1] is the most widely used analogy for "electron fluid" in a metal conductor. As with all analogies, it demands an intuitive and competent understanding of the baseline paradigms (electronics and hydraulics), and in the case of the hydraulic analogy for electronics, students often have an inadequate knowledge of hydraulics.[2]
The math is important for answers but IMO the key to understanding electronics and electrical science in general is to see energy and how ohms law calculates the amount and proportion of energy in electrical components using voltage ,current and resistance. Like when you popped the meter fuse. In that circuit configuration, the amount and proportion of energy (that could be calculated using ohms law) moved from the source supply into the fuse, causing it to heat and open the circuit.

We have a cause and effect that can be predicted in advance using Ohms Law.

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Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
Here's a quiz for you: What resistor would you need if you had a 5 volt battery instead of a 12V battery?

OK, Dealing with the current Schmatic you gave in your example ( I got confused )

5V supply
LED has 2Vf
With 10 mA of Current.

(5V - 2Vf ) = 3V

Using Ohm's Law
R = V x I
R = 3 ÷ 0.01
R = 300K Ohms

#### dl324

Joined Mar 30, 2015
17,144
I don't have any data sheets on my existing LED's. Is there a standard means to use in case you have no idea what the LED is rated at?
I'd suggest you start buying ones that can be referenced back to the manufacturer's datasheet.

If you're talking about low power (standard or ultrabright), maximum continuous current is usually around 20mA. Forward voltage depends on color and there's variation within the same lot. More LEDs are binned (sorted) by brightness than by forward voltage. Binning by more than one parameter makes them more expensive.
I was told to think of voltage as water in a pipe
I'd recommend that you forget about the water analogy and just use conventional current. That water analogy is imperfect and can be confusing. Conventional current is the opposite of electron current. You can use either, but stick with one. Most will use conventional current even though, technically, it's wrong. There are only a few cases where using electron current makes something clearer.
Are those schematics drawn with a program? I've seen stencils with "And and Or Gates" on them. Do you hand draw them?
I don't know what the student used. When I was in college, I used stencils. Hand drawn are fine. Many use schematic editors that will also do board layout. I use an old version of Eagle.

Here's a schematic for an LM317 based regulator that can be powered by an adapter or batteries. The power jack does the switching if you study the wiring.

I included the recommended protection diodes. C3 is optional. If you don't use it, you don't need D2.
EDIT: Just noticed that I put D2 in the wrong place. Corrected above, but not below.

Eagle uses nasty, disagreeable colors, so I print to black and white PDF and take clips from that.

Nasty colors from screen capture:

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#### dl324

Joined Mar 30, 2015
17,144
Using Ohm's Law
R = V x I
This is wrong. V = IR. Algebra gives you:
$$R = \frac{V}{I}$$

Use units.
$$R=\frac{V}{I} = \frac{3V}{10mA} = 300\Omega$$
and sanity check your answer. 5V across 300k is going to give you much less than 1mA.
EDIT: 1/60 of a mA to be precise.

3V across .3k will give you 10mA. I used the k suffix for the resistor because 1V across 1k gives 1mA (when you invert take the reciprocal of k, you get milli). That makes it easer to do the calculation mentally.

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Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
I think you're actually too advanced for the brain dead water analogy. Most people also don't understand the closed-loop hydraulic circuits needed to make the analogy useful. IMO don't use it to build foundational information about the electric circuits because it leads to misconceptions about the nature of electrical energy.
https://en.wikipedia.org/wiki/Hydraulic_analogy

The math is important for answers but IMO the key to understanding electronics and electrical science in general is to see energy and how ohms law calculates the amount and proportion of energy in electrical components using voltage ,current and resistance.
I get it now. Explained very well.
The Hydraulic Analogy made it very explainable to me.
Thank You, I watched a video about it that was very visual to me and made a difference.

#### dl324

Joined Mar 30, 2015
17,144
The Hydraulic Analogy made it very explainable to me.
Thank You, I watched a video about it that was very visual to me and made a difference.
If you use it, it's going to cause nothing but grief. When you talk to an electronics type, they'll smile (if you're lucky) and dismiss you.

They teach conventional current and electron current in schools and books, but no serious people use water.

Thread Starter

#### RexR35630

Joined Dec 9, 2023
17
This is wrong. V = IR. Algebra gives you:
$$R = \frac{V}{I}$$

Use units.
$$R=\frac{V}{I} = \frac{3V}{10mA} = 300\Omega$$
and sanity check your answer. 5V across 300k is going to give you much less than 1mA.

3V across .3k will give you 10mA. I used the k suffix for the resistor because 1V across 1k gives 1mA (when you invert k, you get milli). That makes it easer to do the calculation mentally.
OK, I get it.
Thank you.

#### nsaspook

Joined Aug 27, 2009
13,577
I get it now. Explained very well.
The Hydraulic Analogy made it very explainable to me.
Thank You, I watched a video about it that was very visual to me and made a difference.
And that's were the analogy should stop. Once you understand the electrical principle don't use the Hydraulic Analogy in the thought process of problem solving or to explain things.

#### dl324

Joined Mar 30, 2015
17,144
@RexR35630 If you copied the schematic I posted earlier, I made a correction. D2 was in the wrong place.

Here's the wiring from that homemade experimenter. Original wiring, on the left, and with my modifications. I added the labels when I was tracing the wiring:

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