Voltage Dividers - Electronics Basics 12 

I agree w/ you 100%. I watched three other videos before this one, and seeing this explanation it all made sense-it was intuitive.

From the same generation, and totally totally right. Started electronics in 6th grade with Radio Shack Fairchild Heathkits. The only fast way to understand it was by figuring it out yourself, and reading books from city library. We definitely didn't have RU-vid!

@@thurmanwatson9693 From the same generation, and what really kills me is that so many things are availabe for todays Generaion as learning material that is incredible yet they have great difficulty learning,,, I wish I had all those possibilities,,, I guess our generation wanted to make it easier to the next one, but with huge amount of information I guess the students are getting lost.

Yep kids get it easy these days

@@W3TFART Kids nowadays always complain,,, there don't understand how difficult live was back then. I had a house strait outof colege and a stable job and worked hard. If I had this information back then, I wouldve been much better of now,,, Cant change it I guess.

everycircuit. it is $15

@@cachepage6261 no it's free to use. You can pay for more features if you want

@@j5892000 no. it's 24 hour TRIAL. Or you make new accounts every day.

@@GameBacardi no that's false

@@j5892000 everycircuit.com/ has 24 hour TRIAL, after trial you can't do nothing. Period.

Curious to know if you have found a purpose for resistor dividers?

Electron flow is from negative to positive. Conventional is from + to -.

Current flows opposite to electron flow.

I feel "flow" is all about "charge". Conventional flow is the flow of positive charge. Electron flow, is clearly the flow of negative charge. So, conventional flow is the opposite to electron flow. As such, electrons are physically moving. They increase negative charge at their destination, and their absence increases positive charge at their source. So negative charge flows with electrons, and positive charge "flows" into the absence of negative charge.

This app he's using?

Consider a voltage divider as "pass by value" to a function, as opposed to "pass by reference". You're supplying a voltage level to your circuit without changing the current flowing through the R1+R2 divider. Say if you use V2 as the input to an OpAmp. If you connect a circuit or component that DOES draw current, parallel to R2, then that WILL change current through the R1+R2, through R1, and the circuit changes. This is like a "pass by reference", which causes a change in the scope of the "calling" portion.

Use Ohms law. You know the V and the I, so calculate the R. Once you know the R needed at the point of measurement, you can then also calculate the 2nd R value to make up the divider circuit you need.

all parallel things have the same voltage

@@iwantitpaintedblack That's a risky generalisation. Risky in that someone may read too much into your statement.

Aya, I've left a few replies to other comments in this video. If you still need an answer, maybe one of my other replies can help?

@@CollinBaillie *Have the same voltage applied to them but may have a different voltage drop across them corrected? :D

A VU meter is made with a resistor in parallel to the wire wound "motor" of the meter. It basically measures the voltage across the resistor inside the meter. That resistor has a capacity, and running a larger supply through it would break it. So you want to put an extra resistor in parallel with the meter, which will take 1/2 of the Power away. Then you're left with a VU meter which will read 1/2 of the true Power. Your scale would be wrong. The VU meter would work safely, but the reading would be inaccurate because the scale is 1/2 what it should be. Check out some videos on shunt resistors to see how voltage is calculated according to current flow through a resistor, and you will understand how the VU meter works. Apply that knowledge to be able to calculate the necessary resistor to use in parallel with the VU meter, and then try it out. If you get it to work, don't forget to somehow change the scale printed on the meter. Maybe you can find a data sheet for the particular meter which will tell you what the internal resistor value is, and just use the same value resistor in parallel with the meter.

there are many 100W or even 1000W resistors, extremely huge

The 2 ohm resistor will always have 2v and the 10 ohm resistor will always have 10v, using the example circuit, regardless of the order of the resistors.

To reduce the voltage output of the circuit, incase the voltage output previously was too high for the load

I'm not quite sure I understand the question correctly. But, if you want to know the resistance of both resistors to get a desired voltage, there isn't an equation that will tell you what both resistors should be. However, you can work out the ratio between the two. First you need to know, supply voltage (Vin), and the desired voltage across a resistor(Vout). Ratio = Desired Vout/(Vin - Vout). If my supply voltage is 12V and desired voltage is 5V. Ratio = 5 / (12 - 5) Ratio = 1.4 I now have the ratio, but I still need to chose a random value resistor. Let's say I choose 10ohms for R2. Because the ratio should be 1.4, R1 should be 1.4 times the resistance of R2. In this case, R1 = 10 x 1.4. R2 = 14ohms. If we now work out the voltage across R2... V = (12v/24ohms) x R2 R2 Voltage = 5v If you need rounded numbers and you get decimal answers, just round the result up or down to the nearest whole number. Your voltage won't be truly accurate, but good enough. There are different voltage dividers such as accurate potentiometers that can provide much better accuracy and variable resistance. Sorry if i didn't adequately answer your question.

play.google.com/store/apps/details?id=it.android.demi.elettronica&hl=en this tool has a calculator that will do just that

Thx fot the responses. When I wrote the question I was wondering if somebody would understand it, but it happens to be I got my answers. Thx!

no problem M8

The circuit changes (obviously) so it will no longer be 0v. The 0v will be on the negative terminal of the LED.

A voltage divider is all about getting a specific voltage level to part of your circuit. A Linear Voltage Regulator supplies both Voltage and Current to the circuit. You use a voltage divider to set a voltage level to supply a component which "draws no current" like the input of an OpAmp.

i think the best way to think about it is that the current thru each component is constant. V_total/R_total= V/R = I (current) :)

@@andrescoca983 why?

@@andrescoca983 nice, that's elegant!

@@userou-ig1ze because that is the assumption this circuit and formula makes. In the example circuits, the current IS the same through R1 and R2. It cannot be different, in these example circuits. If you use V2 (the portion of Vtotal, measured across R2) as the input to a Hi-Z (or high impedence) component (like an OpAmp), there will be "no" current flow, and so The total current through the divider network (Itotal) will not change, which is where this formula and such circuits are useful. If you parallel circuitry with R2, and it draws current, then Itotal changes, and this change occurs through R1, so I1 will change, and the whole calculation changes. Then you have to mess around calculating the effective resistance of R2 and the circuit part which is drawing current, using the total current through that part of the circuit (R2+parallel circuit) and pump those numbers back into the equation with R1 to get the actual voltage drop across R2. Or just use a Hi-Z component, and be happy.

You can use a buck converter ( step down voltage converter) to reduce voltage from 3.7 to 3V.

is there any other method ? like adding resistors ?

You want to use the formula: V(out) = V(in) x (R1)/(R1+R2). If you plug in the numbers you get 5 = 12 x (R1)/(R1+R2). If you rearrange this you get that R2 = 1.4 x R1. This means you can have any value for the resistors as long as one of them is 1.4 times the other.

You can't, using ONLY 2 resistors. If both resistors must be 5v, the total voltage drop is only 10v. To drop from 12v, you need to arrive at a circuit which has 10v across the divider network, and then the 2 resistor values will be the same value. Are you sure BOTH resistors MUST be 5v?

Simulation software require ground for simulation including in LTSpice ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-tx9vUj2sY_g.html

this is true because the resistors act as loads. You can calculate the power they drain using the equation: power = (Vin*Vin)/R1+R2

well, his total resistance is 12Ohms, and total voltage is 12V, Mr Ohm told us that Total Current (Juice) flowing should bi V / R = 1Amps. so this sample circuit is burning 1 Amps of juice through the resistors and with Power + Volt * Amps you get 12 Watts of hot resistors :D

@@codewithlenny Are voltage bucks more efficient than this method?

@@davey3765 voltage bucks are far much more efficient. Get them

The idea of this circuit is to learn the theory of the circuit. Generally you wouldn't use such low values. Imagine the supply, while being 12v could supply infinite amps. Also imagine you're using 1 trillion watt resistors. Even if it's drawing 10 amps, that's only 120w. To an infinite amp supply and a trillion watt resistor, 120w is nothing. Theory is just theory. It's used to understand HOW is WORKS. It's NOT how you should IMPLEMENT it in your circuit. It applies to ALL circuits, from high voltage high current to extremely low voltage and low current circuits. How you implement the theory in a real circuit is up to you. Let's make a real example: Power supply 12v at 5A (reasonable) Resistor divider using 1k ohm resistors. Fairly typical values, relatively small when you consider that megaohm resistors are common enough. Each resistor gets 6v (using the theory taught in this video) and using Ohms law, we know current would be 6mA through the circuit. That's 36mW through the resistors. That's not going to burn a 1/4w resistor (250mW), and not going to tax the 12v supply. This is an IMPLEMENTATION of the theory. Learn the theory, then apply it to YOUR implementation.

Davey buck converter has higher efficiency

A buck converter is a POWER (voltage AND current) source. A voltage divider is ONLY a voltage source, no current. It's more of a voltage reference, for measuring or to feed high impedance components, like the input of an OpAmp, which draw no current.

Its everycircuit.

Everycircuit.com

Thanks

I works for me on Windows 10 Firefox

@Bloodred Pyrate Ok I'll only play with low voltages like 9v and an arduino

1 resistor would REDUCE the voltage, not DIVIDE it. The voltage drop provided by the single resistor is calculated using the current THROUGH that resistor. A voltage DIVIDER is used to "supply" a voltage WITHOUT current flow, to another component or circuit part. (eg that component or circuit part draws "no" current) Like the input of an OpAmp. A voltage divider doesn't provide POWER (volts AND amps), just Volts.

Kirchoff's Law dictates that the sum of all voltage drops in a series circuit must equal the supply voltage. Therefore, if there was only one resistor, there'd be only one voltage drop and this voltage drop would equal 100% of the voltage in the circuit. The voltage would literally be undivided.

@@CollinBaillie this is fascinating. I haven't heard this, yet. I'm relatively new to matters of circuit theory. One question, if I may... If Vout carries no current, does this also mean that Vout is only used as signal voltage? Data, perhaps?

The sign convention for current flow in a circuit is arbitrary and was defined long before the electron charge polarity could be emprically determined.

The "flow" simulated in the software is NOT electron flow. It is conventional flow, which is the flow of "positive charge" (which is how electricity was thought to work historically) We now know that electricity works by the flow of electrons, which have a negative charge. So electron flow is the flow of negative charge. When an electron moves, it leaves behind an absence of negative charge, which we consider to result in a positive charge, so the flow of positive charge is in the opposite direction of the electrons and their negative charge.

Morshed, check out some of my other replies to comments in this video, maybe the purpose of voltage dividers will become more clear.

The GND reference is a clarification of that node (the one being between the voltage source and the 10 ohm resistance) to be the reference node for the whole circuit, being the reference node a 0v one because GND is generally considered 0v, so all the voltages mentioned in the video (12v and 10v) are measured to the ground reference.

Voltage is also referred to as potential difference. Try searching for some explanation videos about potential difference, it may help. On the bottom of the circuit, there is 0v potential difference to the negative pole of the power supply. At the top, there is a 12v potential difference to the negative pile of the power supply. If you put a meter with both res and black leads on the bottom wire, you'd get a reading of 0v, if you have the black lead on the bottom wire, and the res lead on the top wire, you would get a reading of 12v. The bottom wire has no potential difference to the negative pole of the supply (what you call GND) so it is 0v. You might also call the poles of the supply 12v and 0v. GND is 0v, conventionally, in DC circuits.

Ummm... yeah you can. That's the whole purpose of these videos. That's the whole purpose of electrical theory. Theory is ALWAYD used to design circuits. You don't put parts together and measure them to design a circuit. Do you hook up a power supply to a resistor and measure it EVERY TIME you want to calculate the current drawn through it, or do you use Ohm's law? I know which one I'd prefer. So much cheaper to use the law to calculate the values, and build the circuit AFTER. You KNOW the desired voltage, or the desired ratio you want, and given the supply voltage, you can the work out which resistor values you need to achieve that ratio. Once you have the resistor values, you can calculate the current flow, and then you can calculate the power rating those resistors need to be. Once you get that sorted out and wired up, THEN you can test it with a meter, and see that all is correct. The simulator software does just that... it simulates a real circuit being tested. The style of the video here seems to be more theoretical, but does use a simulation to give an example of a real tested circuit. I'm sure you can find a video of a real person building a physical circuit and testing with actual meters. I don't think you have any right to demand someone make a video the way YOU want it to be made however.

Or, you don't understand the real life application of voltage dividers. Maybe that's it?

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