Simple beginner questions - Page 1
Forum section: electronics / beginners

author: Mighty Burger
author_rank: Regular Contributor
timestamp: February 21, 2016, 11:20:56 pm
content: As you could probably tell from the title, I'm a beginner. This is my first post. I'm still a teenager. I've taken a semester class of electronics, but it taught me squat. He taught us basically what a couple components look like on a schematic, and some very minor things. We didn't even learn ohms law and some fundamental electronics!Frustrating for someone who wants to be an electrical engineer later in life. At least we were taught how to make good solder joints!So I'm teaching myself through the magical interwebs. I've been doing this for a few days. And gah it's so interesting! So fun too! I haven't played video games for a while because it's not as fun as learning this stuff.Enough of theI have a few questions. It'd be very helpful for me to know these before moving on. All of these should be pretty simple answers but google isn't displaying relevant results to answer my questions, and I don't know how to figure them out myself. It'd be spammy for me to post a thread for every one, but please inform me if I should ask these in another way.1. Silly beginner question. I know volts drive things. Too many volts on an LED, and it burns. Not enough and it's dark. But what about current? I know too much current is also bad, but if there's not enough current will it still function normally, or not? Is this the same for most components or just some?2. Voltage dividers with resistors.So useful, if I could get a better understanding of them! This is apparently the equation to figure out the voltage of one of the outputs.for circuits like thisBut that's only for 2 resistors? What happens when I want to split it 3 ways? Do I just repeat the pattern and apply it again?And the current. Let's say I have 10 volts in. R1 is 75 ohms and R2 is 25 ohms, so 2.5 volts out. So let's use ohm's law to find the current! I=V/R So the voltage along the output is 2.5. Alright. Now the resistance! Um.. err...confusedAnd why does resistor 1 effect the amount of current flowing through resistor 2 vs V out? I'm missing something here.3. Transistors? What do they do and how do I use them? All the information I can find is a simple "they amplify and act as a switch" or info on how they work, not what they do. Let's start with npn. What do I connect to which? I assume connect + to base and collecter and - to emitter. But what does it do? If I have 4 volts 20 milliamps at the collector and 2 volts 10 milliamps at the base, what pops out of the emitter? And then what's the difference with a pnp? Don't think I'm ready yet for the other kinds4. Capacitors. How do I calculate how long it takes to charge one, and how do I calculate it's voltage output and how long it'll output?5. Where do I go to learn about datasheets?I think once I get these questions I am having trouble solving myself, I can really start having some fun with thisThanks for the help! Hopefully this isn't the wrong way to ask questions here

author: DrGeoff
author_rank: Frequent Contributor
timestamp: February 21, 2016, 11:45:50 pm
content: Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmThink of it like water in pipes. Voltage is like the pressure in the pipe behind the tap, and current is like the flow of water when you turn on the tap.The amount of current drawn from a voltage source by a circuit is proportional to the impedance (or resistance for DC circuits). Ohm's Law.Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmOhm's and Kirchoff's Laws. The voltage across the resistor is proportional to the current through it. The proportionality constant is called the resistance (in ohms). How you divide the voltage and current depends on your arrangement of resistors. However if you know the resistance and the current through it then you can calculate the voltage across it. Or if you know the resistance and voltage across it you can calculate the current through it.Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmThere are several types of transistors, but here you are referring to BJT's (Bipolar Junction Transistors). These are current amplifiers. A small current flowing in (or out) of the base terminal results in a larger current flowing in the collector. The amplification is called the Beta of the transistor. A transistor data sheet shows transfer curves which describe this.It's a big topic, so start with a simple arrangement. Single NPN transistor (eg BC547) with a base resistor and collector resistor and emitter as a common ground. Plenty of online resources to teach you this stuff.Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmThere is a formula for the charging curve for ideal capacitors. Vc = Vf(1-e^(-t/RC)). Vc is the voltage across the capacitor, Vf is the final voltage (usually your supply voltage), t is time in seconds, R is the series resistance and C is the capacitance. There are plenty of online resources with detailed pictures showing the charge and discharge curves for capacitors.Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmDatasheets are information about a component. You get them from the manufacturer. Understanding a data sheet requires understanding the electronic theory you are trying to apply and the context of the component's use in your design.

author: tggzzz
author_rank: Super Contributor
timestamp: February 22, 2016, 12:00:23 am
content: You ask good questions. You can take a lifetime to learn all the answers - and that's good, because it means you will never be bored.Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmSuch teachers are indeed irritating. You have a good attitude: "I'm going find a way to learn even if they can't teach me".QuoteMore accurately, volts and amps are tied together; it isn't one causes the other. It is equally valid to sayif there is a voltage across a resistor, it will cause a current to flow through the resistorif there is a current through a resistor, it will cause a voltage across the resistorSometimes it is easier to analyse and understand a circuit in terms of currents, sometimes in terms of voltages. Sometimes voltages cause the working and/or failures, sometimes currents.With time and experience, you will learn which is easier in a given case, and will become adept at flipping between the two. To skip ahead, you will find that "Thevenin source" (i.e. a pure voltage in series with a resistor) isexactlyequivalent to a "Norton source" (i.e a pure current source in parallel with a resistor).QuoteExcellent question. I like the way you are "generalising" from one particular example to a more general case.QuoteThink in terms that the sum of all the currents in each node must be zero (i.e. all currents flowing into a node must also flow out of the node). Think in terms that the sum of all the voltagesarounda loop must be zero (i.e. the sum of the voltages across R1 and R2 must be the same as the voltage between Vin and Gnd - but in the opposite direction)QuoteThink of the current in the node joining R1 and R2: it must be zero, so the same current must be flowing thought R1 and R2. Hence VR1= I1*R1and also VR2= I2*R2, and I1= I2, and VR1+ VR2= Vin. All at the same time, with nonecausingthe others - all you have to do is solve those simultaneous equationsQuoteThat's too difficult to answer fully here; you'll have to do some research. There are shortcuts that can be applied in certain circumstances, and you'll learn which circumstances.QuoteStrictly speaking, it never finishes charging. However if you have a definition of "sufficiently charged", you can calculate a finite numerical answer.QuoteManufacturer's websites, but be prepared to be inundated with information you don't (yet) understand. Also look for manufacturer's "application notes".QuoteTake time, don't get discouraged. Given the nature of your questions, I'm confident you will understand - eventually.

author: Whales
author_rank: Super Contributor
timestamp: February 22, 2016, 12:08:32 am
content: Some answers:1. Yes current is important.To start with: an LED emits light (ie energy). It needs to get this energy from somewhere.If you want to know how much energy per second (power, measured in watts) a component like an LED is consuming electrically (or basically anything else with two leads) then you can use this formula: P = V * I (power equals the voltage across the LED times the current travelling through the LED).Note that if you have a tiny current through the LED, it will get hardly any power. If you have too much, you will be delivering way too much power to the LED and it will (mostly) be converted into heat, blowing the LED up.Most 'common' LEDs are ok with current in the range of 5 to 20mA (0.005 to 0.020 amps). If you have bought them from a source that does not give you a part number/datasheet then this is a good assumption to start with. The less power you give the led, the dimmer it will glow.2. Voltage dividersQuoteYou can make another separate voltage divider with another set of two resistors if you want another voltage.Rather than trusting the equation you have posted, it's a better idea to understand exactly how ohm's law applies to circuits like these. Once you have this understanding you will also be able to easily work this equation out.To apply ohm's law: you need to choose two point (wires) in your circuit. The picture you have has three: the top (vin), the centre and the bottom (gnd):Once you have chosen these two points: V is the voltage between the points, I is the current travelling from one point to the other and R is the resistance between the two points.We can apply ohm's law across just the top resistor: V1= I1* R1We can also apply it only across the bottom resistor: V2= I2* R2Most of all, we can apply it across both of the resistors. Choose one point at the top of the circuit (Vin) and the bottom point (gnd) as the other. Now we have Vacross both resistors= Ithrough both resistors* RtotalImportant notes:There is only one path for current to travel in this circuit, so both resistors have the same amount of current flowing through them. I1= I2= Ithrough both resistorsThe voltage Vin is split across both the resistors. Ie Vin= V1+ V2The resistance between Vin and Gnd is the resistance of both resistors added together. Rtotal= R1+ R2From these equations everything else can be worked out, including the equation you have a picture of above. If you are familiar with algebra you should be able to re-arrange these equations and plug them into each other to get your equation above (and other interesting pieces of understanding).QuoteBetween which two points are you applying ohm's law? You need to clearly define them.Also: 75 and 25 ohms are pretty small, you may find your resistors overheat/burn at 10V across them.QuoteBoth resistors affect the current travelling through them. One way to think about it is to combine both of the resistors into one, by adding their resistances together. Now instead of having one 25 ohm resistor and another 75 ohm resistors, we can pretend we only have one resistor and it is 100 ohm. Now apply ohm's law across the leads/wires attached to this resistor. You should see how increasing either resistance increases the total resistance of the circuit, making the current drop.

author: Mighty Burger
author_rank: Regular Contributor
timestamp: February 22, 2016, 12:34:15 am
content: This is all very helpful! Thanks everyone who replied!

author: IanB
author_rank: Super Contributor
timestamp: February 22, 2016, 12:48:41 am
content: One important thing to grasp before you go too far is that voltage is adifferencebetween two points, not a value at a single point. If you have voltmeter or a multimeter on the volts range, this is why you have two test leads and is why the meter measures the difference in potential between the two points where you touch the leads.For convenience, people often pick a point in a circuit and say "this is zero", and then measure every voltage relative to this. But remember that this is only a choice of convenience and any point in the circuit could be picked as an arbitrary zero if you felt like it.(Pedants may say reality is more complicated than this, but you don't need to worry about it while you are learning.)As to current, current tries to flow between two points where there is a difference in potential (a measured voltage difference on your meter). The amount of current that actually flows depends on the resistance to flow of the path and the size of the voltage difference. In ideal circuits the relationship is linear, so that Ohm's law applies and says that current equals voltage divided by resistance.You mentioned LEDs, and unfortunately LEDs are not linear and do not obey Ohm's law. An LED will conduct little or no current until the voltage reaches a threshold, and then it will conduct a LOT of current. So much so that an LED will burn out if you give it only slightly too much voltage. Because of this, LEDs are always connected with some way of limiting the current through them, of which the simplest way is a resistor in series.

author: broz
author_rank: Regular Contributor
timestamp: February 28, 2016, 02:02:13 am
content: Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmThese are a bit tricky to understand, I'm in my third year of EE and we JUST started learning about them. This is due to the nonlinear nature of them, which makes it difficult to understand exactly HOW they work. But you can think of them as back-to-back diodes. If you're not too sure what a diode does, it (basically) allows current to flow in one direction, but not the other. When the current flows through the diode there is a 0.6-0.7  drop across it.Applying this to the BJT (let's start with the npn): in order for the npn BJT to be in its linear region of operation ("on") each of the back-to-back diodes need to have the 0.6-0.7  drop across it. This means that the voltage at the collector needs to be ~0.7  HIGHER than that of the base, and the voltage at the base needs to be ~0.7  HIGHER than that of the emitter. This means that for the npn BJT to be "on" you need the voltage at the collector to be ~1.4  HIGHER than at the emitter (this also depends on how the npn BJT will be used...but a good general rule of thumb).As to "what pops out of the emitter"... That kinda depends on how you will be using it, if you want to use it as an amplifier, for example since it's probably the most common use, then there are various configurations which accomplish different goals (look up Common Emitter, Common Collector, and Common Base).The difference between npn and pnp is basically reverse polarity...in the pnp: voltage at the collector needs to be ~0.7  LOWER than that of the base, and the voltage at the base needs to be ~0.7  LOWER than that of the emitter. This means that for the pnp BJT to be "on" you need the voltage at the collector to be ~1.4  LOWER than at the emitter.I hope that helps!

author: CatalinaWOW
author_rank: Super Contributor
timestamp: February 28, 2016, 02:26:31 am
content: The good news is you like to learn. Most of us are still learning years or decades after we started.It is hard to tell where to point you without knowing you better. There is so much to learn and the challenge is to take small enough bites to not choke, while getting enough to keep your interest and to accomplish something interesting.I would recommend either a book or internet summaries of beginning electrical circuits, starting with DC circuits (which don't use energy storing components like inductors and capacitors). You will learn more about Ohms law, about loops and nodes and other concepts used to calculate large groups of resistors. You can then extend these ideas to time varying circuits. Finally you can fold in non-linear items. The best way to approach these ideas will depend on you and how far you intend to go. Different approaches work better for different people, and some approaches will have to be unlearned to get the most out of later, more sophisticated understandings.If you plan to go far into this field it also pays to start early on your math studies. You will make sure you have algebra and trigonometry under your belt and then head on to linear algebra, calculus and differential equations. Some stop there but there is much much more that is of use to the engineer. Transform methods, variational methods, calculus of discontinuous functions, vector calculus, numerical methods and on and on.Welcome to the journey and good luck.

author: LvW
author_rank: Frequent Contributor
timestamp: February 28, 2016, 09:46:37 am
content: Quote from: Mighty Burger on February 21, 2016, 11:20:56 pmMighty Burger - just a small hint (and warning) regarding transistors.While reading several documents about the working principle of bipolar junction transistors (BJT) you might get confused a bit because you will be confronted with two fundamentally different explanations:(a)The BJT is a current-controlled device (see reply#1 in this thread) and(b)The BJT is a voltage controlled device (as mentioned in reply#6).It is really funny and surprising that more than 60 years after the BJT was invented we still can find two different descriptions "how the BJT works".The question is simply if the output current (collector current Ic) is controlled and determined by the current into the base (Ib) or the voltage across the B-E diode (Vbe).This subject was intensively discussed in the past (also in this forum) and you should try to develop your own view.In this context, I suggest to select a simple pn diode as an example and ask yourself: Can we expect that a pn junction within the BJT behaves completely different if compared with the pn junction of a diode?