I'm stuck in a digital age!! :( - Page 1
Forum section: electronics / beginners

author: tornup
author_rank: Newbie
timestamp: January 28, 2014, 07:17:49 pm
content: Hi All,First of all, Hi! I’m Tom, 26 years old from the UK, Working in IT Management, Software Dev, and System Administration within our wonderful National Health Service!Been on and off with electronics for 5-6 years, but finally built my lab and am fully concentrating on it for the last 5-6 months!I am here because I am in a bit of a rut(http://idioms.thefreedictionary.com/in+a+rut) with my electronics! I seem to only get my head around the digital side of electronics!(maybe because of my IT background?)For example, recently, I watched a project that came out of the 555 timer contest, the“Le Dominoux“ -http://youtu.be/PQOjkuJtBfMPretty much using a photodiode/LDR to detect the light output of an LED and reproduce it again, put them in a row and you have a kind of light chain/pass it on. Type of thing! Perfect, quite a simple 555 project! Take a LDR “state” if it reaches the fresh hold it fires the LED. Nice!Couldn't get my head around 555 timers!!!Ended up producing it with an attiny85! (What isn’t too bad because I’m going to do as Mike (mikeselectricalstuff) did and create a kind of IR comm’s link to control RGB legs!Now let me explain, I REALLLLLLLY want to understand analogue electronics more! I want to watch Dave’s videos and understand his tear-downs of analogue equipment, or even the analogue side of a digital system..Don’t get me wrong, I have a “mild” grasp of some analogue things, smoothing CAPS on IC’s, so a rough (not as advanced as Dave) understanding of a capacitor.Resistors I have a basic grasp of also (although why I use a 1k pull up resistor instead of a 10k I don’t really know…)However, op-amps, impedance, etc… Confuse me!!….And it’s not from lack of trying to learn either, I've watched 4-5 videos of 555/op-amps timers without any better understanding; the video on impedance blew my mind!Has anyone else encountered this problem?? Does anyone have any advice on where I should look at learning this stuff from? Do I need a book(be warned, I’m personally more of a video person, books I find hard to remain focused on!) surly YouTube with its millions of videos has refined the art of producing instructional videos enough to teach me 555 timers?!HELP!I really don’t want to have to keep through a MCU into everything (there are even areas of electronics where an MCU wouldn't work)TOM!

author: Icarus
author_rank: Regular Contributor
timestamp: January 28, 2014, 07:37:21 pm
content: Microelectronic Circuits - Adel S. SedraAnalog Circuits - Robert PeaseAnalysis and Design of Analog Integrated Circuits - Paul R. Gray

author: Rerouter
author_rank: Super Contributor
timestamp: January 28, 2014, 07:45:06 pm
content: Its the small things that add up, but i'll give you a few search terms to let you work over at your own pace,- Resistance voltage divider,- RC lowpass filter- Transistor as switchThough it does vary when you add capacitors and inductors, impedance refers to the impediment of current, as capacitors and inductors have reactance (they pass a current depending on there value and the frequency) this term reluctance makes up the third side of a triangular diagram, resistance reactance and impedance, with impedance being the squared root sum of the 2, so for a capacitor that has ESR (resistance) and and a value (reactance) you can work out the impedance for a given frequency, (all this ties in with what are called phasor diagrams but that may be going a tad too far at once)now i am going to guess you have been wondering more in the way of controlled impedance traces, if such is the case, this only matters when the length of the wire / trace is more than 10% of the wavelength of the frequency your using (so its either very long wires or very high frequency)as for op amps, they are not that big of a beast, the basic principle is they amplify the difference between the 2 pins, and the resistors around it in the example images you find are to limit and adjust the gain as with no feedback (some voltage from the output summed with the input) it would act like a comparator and slam rail to rail,there are a few parameters in there datasheets that can byte, the main ones being input range, output range, and the current output when approaching those rails, as even a rail to rail type opamp may only have its input OR its output capable of doing so, and for the output may only be able to deliver a tiny fraction of the current it can midrange when a few mV from GND for instance,

author: tszaboo
author_rank: Super Contributor
timestamp: January 28, 2014, 07:55:49 pm
content: my suggestion is to take it easy. First, buy a breadboard and a bunch of sn74, 555, and opamps, understand them. dont just use the computer, build it. if you want the full analog experience, put your PC away, print the page you need, and use that.The problem is that people play safe and take the easy way, which means they use microcontroller for everything. Analog is harder. But more fun.

author: Icarus
author_rank: Regular Contributor
timestamp: January 28, 2014, 08:40:21 pm
content: Here is some videos from the legend

author: tornup
author_rank: Newbie
timestamp: January 29, 2014, 11:09:54 am
content: Hi All, Thanks for all your brilliant comments, its good people understand! and are happy to help (I can't imagine how many people come on here asking the same questions!!)Quote from: Icarus on January 28, 2014, 07:37:21 pmThanks Icarus, I will look over them book, however on first impression... It seems a lot of crazy complex mathematical calculations!?Quote from: Rerouter on January 28, 2014, 07:45:06 pmIv come across these terms a little before! Voltage dividers are something I've got a mildly firm grasp on. i know if the values are the same on both resistors them it half's the voltage?Low pass filters i understand at least what it achieves, but couldn't calculate the values myself! Although I don't really know what the resistor does... i presume the Cap is acting like a Smoothing cap same as used on a MCU VCC?And transistors as switches iv looked at also, however i got lost on understanding the values, for example, i have 5v and with that 5v i want a transistor to "open" and allow 12v through the C to E. but i need a transistor that is "fully open" at 5v, 20mA, I don't know how to calculate that. I could prob guess what value is the Max Voltage/Current you can source across the transistor...Quote from: Rerouter on January 28, 2014, 07:45:06 pmOK, that makes sense (confirming what i gathered from the videos I've watched) however, why would you want to amplify something? Could someone give me some real life examples?I am presuming Dave's uCurrent uses an op-amp to amplify the small voltage levels/currents to a level a DMM is able to pick up? and as long as you finely control the "gain"(Multiplication of amplification?) in a precise way(hence his expensive resistors very close to their value) it is possible to measure this input value on a DMM unable to go that low?Quote from: Rerouter on January 28, 2014, 07:45:06 pmI am more wondering about when people talk about a pin on a MCU being high impedance(primarily when you are Charliplexing and you can sink an LED through a pin! still puzzles me how a pin can provide a voltage, but also sink to ground!?!), or the input on a scope being high impedance. what does that mean!? how does it benefit me/the scope?Quote from: NANDBlog on January 28, 2014, 07:55:49 pmI have a Breadboard, a few 555 timers, a few darlinton arrays(ULN2803A) plus one of them "party packs" of resistors(5%) and Caps(cheap Chinese crap ones prob!!) any recommendation on a good all round Op-amp for prototyping, learning? what is a sn74?So, iv got my breadboard, with all my parts...........now what? lol..i agree 100% that learning through doing is the best way to learn, however ideally, i need a project... any recommended projects?Quote from: Icarus on January 28, 2014, 08:40:21 pmThanks for the video... looks informative. I don't know how long i can manage the vintage feel and crackly audio though! lol.Thanks for all your help!!!! keep it coming!TOM

author: Hideki
author_rank: Frequent Contributor
timestamp: January 29, 2014, 03:30:54 pm
content: Start with voltage dividers and go from a "mildly firm grasp"to really getting it. It's just two resistors in series and you look at the output across one of them.And then.. overcome the mistake that 100% of beginners seem to make: say you have 10V in and made a divider that divides that in two. Yay, it's a 5V "regulator"!Then they connect some device to that 5V and get confused because the output is no longer 5V. Why?To move onto RC lowpass (or highpass) filters you need to understand the divider first. The only difference compared to the divider is that one of the resistors gets replaced by a capacitor.

author: tornup
author_rank: Newbie
timestamp: January 29, 2014, 07:36:02 pm
content: Quote from: Hideki on January 29, 2014, 03:30:54 pmBecause a voltage divider outputs a division of the input voltage depending on the resistor values(equally matched resistors will half it, so 10V down to 5V, but should that input voltage change(as a battery looses its charge) the output voltage will change, however the division value will not(to compensate) resulting in a 9v battery now outputting 4.5 volts from a voltage divider..i understand that bit... but I don't understand the difference between using 2 100 ohm resistors causing a voltage drop of 50% or 1k resistors causing a voltage drop of 50% or is it just "one of them things?"Equally, ask me to calculate the values to make a 9v input voltage divider output 5v (not a direct 50% drop) and i'd fall on the floor and cry!maybe I am just poor at math!!! for example, I don't actually understand this formula:what does the line through the middle dividing them mean???Thanks for everyone being really patient and understanding. I will get there eventually!Tom.

author: c4757p
author_rank: Super Contributor
timestamp: January 29, 2014, 07:43:07 pm
content: Quote from: tornup on January 29, 2014, 07:36:02 pmThe 100 ohm resistors draw more current. It'll just take experience to figure out how much current you need - as a good starting point, try for around 1mA flowing. Just use Ohm's law to compute.QuoteThat's a pain. I sometimes usethis toolmade by jahonen (forum member).Quote"divide"But if you try to memorize formulas all the time you'll never understand what's really happening. Remember Ohm's law (if you don't know it, look it up - I'd define it here, but you'll find some good reading in the process) and you can easily compute it. All you have to know is that two resistors in series are equivalent to a single resistor with the resistances of the two added together (1k in series with 10k is 11k).

author: Rerouter
author_rank: Super Contributor
timestamp: January 29, 2014, 07:51:03 pm
content: Quote from: tornup on January 29, 2014, 07:36:02 pmThe difference in value comes down to the question, what happens if you load your 5V division with a 100 ohm load, or a 1000 or a 1M ogm load to ground, the following formula will let you calculate it,just to lay it out step by step, this is the slow and easy way,- first work out the equivilent resistance of the whole thing (series and parallel resistors should be something i would hope you to know) so in the case of a 1K resistance on the top and a 5K resistance on the bottom its 6K,- Divide the supply voltage by this resistance to get the current, (833uA)- multiply the value your interested in by this current to work out the current flowing through it, (4.17V)now if you added a 1M resistor say a scope probe in parrellel with the 5K, the two values become 1/(1/5000+1/1000000) = 4975 ohm,so your output voltage becomes (5/(4975+1000))*4975 = 4.16Vnow this isnt much, but then again 1M ohm is almost an order of magnitude greater than the value it was in parallel with, try it with some smaller values and some varying loads,

author: Hideki
author_rank: Frequent Contributor
timestamp: January 29, 2014, 11:22:17 pm
content: Quote from: tornup on January 29, 2014, 07:36:02 pmUmm.. no. That wasn't what I meant. Say the input voltage is always 10V and you have two 100 ohm resistors in series. You will see 5V over the bottom resistor.How do you calculate this? Well, there is ohms law, but let's skip that for now. First you need to find the ratio between the bottom resistor and the total resistance.The total is easy: 100 + 100 = 200 ohm. The ratio is the bottom resistor divided by the total: 100 / 200 = 0.5 So whatever you put in, you get 0.5 times that out.Then you connect something like a tiny light bulb (to make it easy to calculate) across the bottom resistor. Let's say the resistance of the bulb is also 100 ohms.The light turns on and you measure the voltage again. Now it's 3.33V.. what happened to the 5V!?You changed the divider! The lower part is now 100 ohm from the resistor and 100 ohm from the bulb in parallel. That ends up being 50 ohms instead of 100.The total is now 100 + 50 = 150, and the ratio is 50 / 150 = 0.333... With 10V in, you get 10V * 0.333 which is 3.33V.Quote from: tornup on January 29, 2014, 07:36:02 pmLet's see what happens if you use two 1k with the bulb connected. Now there is 1000 and 100 in parallel at the bottom. That's 90.909... ohms.The total is now 1000 + 90.909 = 1090.909, and the ratio is 90.909 / 1090.090 = 0.083... With 10V in, you get 10V * 0.083 which is 0.83V. Not much left of that 5V eh?So the lesson is: The voltage divider still works exactly as it should, but you can't connect anything else to it that draws any current out of it (or into it) and assume that it will still divide the same way as before. Even just connecting your multimeter across the resistor to measure the voltage will change the total resistance, and hence the voltage a tiny bit.Quote from: tornup on January 29, 2014, 07:36:02 pmCalculate the ratio! 5 volts out of 9 total? 5 / 9 = 0.555... so you need to make sure the lower resistor is 0.555 times the total resistance. To make it really easy, make the lower resistor 5k and the upper one 4k (that is 9 - 5).Quote from: tornup on January 29, 2014, 07:36:02 pmIt means you divide whatever is above the line by whatever is below the line.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 30, 2014, 03:10:17 am
content: "OK, that makes sense (confirming what i gathered from the videos I've watched) however, why would you want to amplify something? Could someone give me some real life examples?Take the case where you want to connect the output of a microphone to an ADC.I'm not an ADC guru,so I will say for the sake of argument,that the ADC input voltage range for an AC signal is from zero volts p-p to 500mV p-p.A signal of 500 mV p-p will deliver the best resolution that the ADC is capable of.The microphone,however,has an output of 50mV p-p.The ADC resolution is obviously reduced.If we now amplify the mic output voltage by a factor of 10,we can again use the full resolution of the ADC."but I don't understand the difference between using 2 100 ohm resistors causing a voltage drop of 50% or 1k resistors causing a voltage drop of 50% or is it just "one of them things?"Now that previous posters have pointed out the effect of connecting a resistor in parallel with one of the voltage divider resistors,do you think you would get less change by connecting a 1k resistor in parallel with a 1k resistor ,or with a 100 Ohm resistor?Hint:- 100 Ohm in parallel with 1k gives us 90.9 Ohms.

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 30, 2014, 05:52:59 am
content: Quote from: tornup on January 29, 2014, 11:09:54 amHave you taken a calculus course? Do you understand the concept of derivative and integral?If not, you'd be well served to learn those, either in a math class or on your own. Different people have different ways of acquiring their understanding, and I don't know you or your background well enough to say what you should do, but I'll point out that there's a good reason analog electronics books tend to have equations, often involving derivatives and integrals. You can try to avoid them, but if you really want to understand AC circuits, you've got to get a grasp on the basics of calculus, one way or another. You don't need to know a whole lot of techniques, and you don't necessarily need to dive in too deep, but if you have a good understanding of what derivatives and integrals are, you'll be ahead of the game. Some people learn the practical applications first, and then figure out enough math to make sense of it. I think it may be easier to learn the math first, and then the practical electronics aspect will come more naturally. Whatever works for you.AC circuits involve capacitors and inductors, and capacitors and inductors involve derivatives and integrals.The voltage across a resistor is proportional to the current.The voltage across an inductor is proportional to the derivative of current. (or the current is proportional to the integral of voltage.)The voltage across a capacitor is proportional to the integral of current. (or the current is proportional to the derivative of voltage.)When you look at an oscilloscope hooked up to an interesting circuit, you'll often see sine waves, exponentials, or some combination of them. When the derivative of x is proportional to x, you'll get an exponential. When the second derivative of x is proportional to x, you'll get a sinusoid. Don't let me scare you too much when I call the first case a "first order differential equation" and the second case a "second order differential equation".You can try to learn analog electronics without calculus, but I think it will always be frustrating.

author: AG6QR
author_rank: Frequent Contributor
timestamp: January 30, 2014, 06:04:53 am
content: Quote from: tornup on January 29, 2014, 11:09:54 amHere's one example (that doesn't involve calculus, derivatives or integrals, as I wrote about above).People have already showed you how a voltage divider composed of resistors stops working as an ideal voltage divider as soon as you draw current out of it. In order for the voltage divider to work as advertised, the current through both resistors must be equal, and in order for both currents to be equal, no current can enter or leave the junction between the two resistors.What if you want to make a voltage divider that can carry a load? To make it harder, what if you want to carry a varying load?An op amp, set up for unity gain, can take its input from resistor voltage divider. It won't source or sink any current from that divider (in practice, it may source or sink a tiny bit, but ideally it'll draw precisely zero current). Its output can source or sink huge currents. The amplifier makes sure the output voltage is equal to the input voltage.There are many other uses of amplifiers. Most commonly, the output is at a higher voltage than the input (but not in the example I just gave). A common everyday use is to take the output of a little microphone and use it to drive big powerful speakers, so that a voice can fill a stadium.

author: nuhamind2
author_rank: Regular Contributor
timestamp: January 30, 2014, 06:06:56 am
content: look at thevenin equivalen, this will give you perpective with the "5V voltage divider regulator"

author: vk6zgo
author_rank: Super Contributor
timestamp: January 30, 2014, 09:13:42 am
content: Guys,he's having problems with Simple Fractions & you want him to do Calculus & Thevenin?Calculus is essential if he is going to study at EE,or even Sub-Professional level,but you can still have a useful knowledge of Electronics using basic Algebra & Trigonometry.Some people cannot visualise abstract concepts without having it proved to them using quite sophisticated levels of Mathematics.Others can,learning these concepts quite naturally.This initial advantagedoesbecome a disadvantage if higher learning is attempted & they are required to give Mathematical proofs of what they understood through "visualisation".There is always the temptation to say "I know this stuff,why should I have to prove it ?"Many people find the "visualisation "approach quite adequate for their purposes..

author: nuhamind2
author_rank: Regular Contributor
timestamp: January 30, 2014, 05:48:33 pm
content: Actually the thevenin equivalen of the resistor divider make it easier to understand the effect of loading. The circuit became a voltage source followed by series resistor which is similar to real battery with internal resistance which is already taught in highschool.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 31, 2014, 12:14:26 am
content: True enough,but as I said,he is still having a few problems with his low level Maths.To the OP--I would avoid Youtube videos.Some are very good,many are rubbish!Would you suggest I learn your field of expertise in this way?My suggestion is to borrow a copy of the ARRL Handbook from your local Library,(or buy one),& work through the early chapters that deal with basic theory.

author: scientist
author_rank: Frequent Contributor
timestamp: January 31, 2014, 12:50:38 am
content: Quote from: nuhamind2 on January 30, 2014, 05:48:33 pmNot anymore.