when I power an LED (a little one used as one indicator) through a 5VDC, the blows up, but when I use a resistor in series with the LED, that lights.

You are watching: Does current change through a resistor

My instructor calls this resistor a present limiting resistor.

It is recognized that current in a collection circuit stays the same.

How walk the existing limiting resistor job-related in this case, and also why walk the LED blow without the resistor, also though the current remains the exact same in both cases?

Below is the schematic of the circuit

simulate this circuit – Schematic created using CircuitLab

present resistors current-limiting
re-superstructure
mention
monitor
edited january 26 at 21:58

MarkU
inquiry Oct 28 "20 at 7:18

SaqlainSaqlain
7755 bronze title
\$endgroup\$
6
| present 1 more comment

13
\$egingroup\$
It is well-known that current in a collection circuit continues to be the same.

The existing is the very same at any part of a simple collection circuit. However that doesn"t median the present stays the same if you readjust the circuit.

An LED drops approximately 1.5 come 3V once it"s working normally. If you connect it straight across a 5V supply, one excessively huge current flows, and also the LED blows.

Put a 100 ohm resistor in, and also the extra voltage is dropped throughout the resistor. This will be somewhere between 2 and 3.5V. Given I = V/R, a 100 ohm resistor will pass in between 0.02 and also 0.035A (or 20 come 35mA). That"s much better for one LED.

share
mention
monitor
answered Oct 28 "20 in ~ 8:42

Simon BSimon B
\$endgroup\$
2
include a comment |
11
+25
\$egingroup\$
Another question about LEDs and resistors...Ok, I"ll take it it.

I believe you might benefit from the visual aid associated through the usage of load lines. Shot to check out your circuit together the combination of two one-port elements: the generator and also the load. Each of these one-ports will have its own Voltage-Current characteristic and also when you incorporate them together,by connecting the respective ports, you end up in share both voltage and current.Note that it is customary to take on two antithetical conventions because that the authorize of the current: in the situation of the source, the present is exiting the one-port native the point at greater potential, whereas for the fill the existing is entering the one-port from the allude at greater potential (where the + is or the voltage arrowhead has its allude in the non-German convention).

Now, due to the fact that both one-ports re-superstructure the very same voltage V and the same present I (here the choice of present convention mirrors its usefulness), the operating point of the circuit drops on the intersection that the two features curves.

You have the right to ascribe the summary each harbor V-I characteristic come the applications of a generalized type of Ohm"s regulation (allowing because that nonlinear and active elements), when the relationship between the electric variables that the ports developed by their interconnection deserve to be viewed as an application of Kirchhoff voltage and also current laws.

Let"s check out what this entails by assessing some basic circuits wherein we affix a diode to various kinds that sources. In all the following circuits, the load will be a diode (an LED, if friend will) i m sorry is usually a nonlinear resistor through an exponential V-I characteristic.

Let"s start with an ideal voltage source. The V-I characteristics is a vertical heat at \$V_batt\$, meaning the generator is capable of offering the same exact voltage \$V_batt\$ no issue the current drawn through the load. If the value of \$V_batt\$ is sufficiently higher than the "nominal" threshold voltage the the diode, the intersection the the generator and also load qualities will be in ~ the suggest (Vq, Iq) wherein the existing Iq is high enough to destroy your diode.

Ideal voltage sources execute not exist in real life, and any source you will discover will exhibit an interior resistance \$R_s\$ that will reason the V-I characteristic to bend towards the i axis. The intersection that the resource characteristic with the i axis is the brief circuit current \$I_sc\$ (not displayed in the following snapshot because it"s method up ~ above the i axis). In some instances the combination of the open up circuit voltage \$V_batt\$ of the source and its interior resistance \$R_s\$ might reason the intersection of source and load curve to take place at a point (Vq, Iq) through low enough existing not to destroy the LED.

This is the case for example of those silverish switch cells: you have the right to safely attach an caused them there is no any additional limiting resistor. But with most other voltage resource - uneven they space regulated through other means around the nominal threshold voltage that the LED, you will usually finish up death the diode.

So, why no take manage of the situation and use a collection resistor \$R_lim\$ in stimulate to limit the value of the present to a for sure value? This is the easiest method to power a LED (a far better way would be to use a current source). Usually this collection limiting resistance \$R_lim\$ is much greater than the interior resistance \$R_s\$ the the source so the the last parameter have the right to be for sure ignored. The "limited" resource characteristic would bend even further towards the ns axis and the intersection of resource and fill lines have the right to be made to happen at a predetermined suggest (Vq, Iq).

See more: Him And His Or He And His Or He And His, Use Of The Words He/She, Him/Her And His/Hers Etc

Given the form of the exponential diode characteristics (here not drawn to scale), the value Vq won"t stray too much from the in the name threshold voltage \$V_gamma\$ (or \$V_threshold\$) the the LED, however the necessary thing is that by including a series resistance \$R_lim\$ us gained control on the otherwise insanity variable worth of the current.The mathematics to identify the exact value of the intersection of the direct load line through the exponential diode curve, while tho being an easy calculus, is not precisely elementary (you might be forced to include the Lambert W function to her vocabulary) and it is customary to resort to numerical or graphical approximations as displayed in some of the other answers.(These approximations amount come swap the nonlinear exponential curve through a piece-wise linear curve either rising vertically in ~ \$V_gamma\$ or slanted v a slope established by the diode"s differential resistance \$r_d\$)

From a qualitative point of view, though, you can appreciate the role played by the limiting resistor by drawing the V-I characteristics of the source for different worths of \$R_lim\$:

As you deserve to see, by raising the worth of \$R_lim\$ you bring down the operating point(Vq, Iq) at lower and also lower worths of the current.Note that for a given, set, addressed value of \$R_lim\$ the present is the same for all series elements (source, limiting resistor and also LED) however if you adjust \$R_lim\$ (by one of two people swapping the resistor, or acting on the knob of a potentiometer, or also by transforming its temperature through blowing hot air over it) friend will watch the current adjust in all elements of the series.

Appendix:This is what the application of the load lines technique looks like once the exponential diode characteristics is approximated through a piecewise direct characteristic with solved \$V_thereshold\$ and zero differential resistence: