Diode and resistor series

Question

Can someone show me how to calculate voltage drops and current on this circuit?

^{simulate this circuit – Schematic created using CircuitLab}

I have 9 V source and 100 Ohm resistor.

So for the resistor on its own, the current would be 9 / 100 = 0.09 A across the whole circuit.

But I don't know what to look up to determine the voltage on the diode, or what the current will be on the circuit, now that it is added.

Answer 1

Try to find a datasheet on the internet. Find about which voltage would be across the LED when the current you calculated without the LED (0.09A) runs through the LED now tou added the LED. Next apply Kirchoffs Voltage Law and use this LED voltage, recalculate the current through R1 and repeat the steps above.

You see that after 1 or 2 iterations the LED voltage hardly changed; you're done

Answer 2

The so-called forward voltage can be found in the datasheet. If you don't have a datasheet, than find a few different ones, and check the average for the color of your LEDs (most LEDs with the same color have a (small) range of forward voltage.

Once you know that, use the following formula:

V - Vfw = I * R  where

V : Voltage
fw: Forward voltage of the LED
I : Current
R : Resistance

Assuming you have a LED with forward voltage 1.7 V, than your formula will result in

9 - 1.7 = I * 100 <=> I = (9 - 1.7) / 100 = 0.073 A = 73 mA

Btw, this is probably way too much, since most LEDs (3mm) have a limit of around 20 mA or 40 mA. Assuming you want 20 mA, you have to increase your resistor value (using the same formula above, I will leave this up to you).

Update

I now see the exact type of your LED, using a datasheet of it (e.g. LTL-307EE Datasheet, it shows a forward voltage of typically 2.0 V.

This means in your case the current will be

(9 - 2) / 100 = 70 mA

Although it can stand short peaks of 120 mA (see also datasheet), the continuous forward current is 30 mA. So if you want to have it continously lit at max current, than the resistor value should be:

(9 - 2) / R = 0.03 <=> R = (9 - 2) / 0.03 = 233 ohm

The next common higher resistor value is 270 ohm which results in a current of

(9 - 2) / 270 = 26 mA

Update 2

When looking at figure 2, at 30 mA, the forward voltage is 2.1 V, so the current is actually

(9 - 2.1) / 270 = 25.6 mA. 

Answer 3

This answer is just a (hopefully useful) complement to the other ones, also since the answer by Huisman has already been accepted: however, I think it is worth to detail all possible ways to solve this frequently encountered problem, including the analytic method not described in earlier answers, which I describe below.
When you need to calculate the current in a mesh where one of the elements is nonlinear, you have three ways:

1. The graphical method, exemplified by the answer of @Transistor and by @Michel Keijzers in implicit form. If the producers gives you the \$I-V\$ characteristics of the device in the form of a diagram (see for example fig.2 in the LTL-307EE datasheet), you can draw on the very same diagram the load line, derived as a consequence of the KVL (Kirchhoff Voltage Law) for the mesh, i.e. $$ I=\frac{V_1-V_{{D_1}}}{R_1}\tag{LL}\label{ll} $$ for values of the anode voltage of diode \$D_1\$ (\$=V_{D_1}\$) ranging from \$0\mathrm{V}\$ to \$V_1=9\mathrm{V}\$. Now find out where this load line intersects the characteristics in the diagram: the coordinates \$(V_{QD_1},I_{AQ})\$ of the intersection are the sought for quiescent values of the anode voltage and of the anode current (and thus the mesh current) of \$D_1\$.

2. The numerical approximation method, exemplified by @Huisman. Suppose you have an expression of the \$I-V\$ characteristics of the device in the form of an equation or again of a diagram $$ I_A=f(V_{D_1}),\tag{I-V}\label{iv} $$ and calculate "first guess" approximation \$I_{AQ}^{[0]}\$ of \$I_{AQ}\$ by using the load line equation \eqref{ll} with \$V_{D_1}=0\mathrm{V}\$: $$ I_{AQ}^{[0]}=\frac{V_1}{R_1}. $$ Then, by using this first guess you can calculate a first approximation \$V_{D_1}^{[1]}\$ of the anode voltage of \$D_1\$ by using equation \eqref{iv}: $$ I_{AQ}^{[0]}=f\big(V_{D_1}^{[1]}\big)\iff V_{D_1}^{[1]}= f^{-1}\big(I_{AQ}^{[0]}\big) $$ and use this for the calculation of a \$I_{AQ}^{[1]}\$ $$ I_{AQ}^{[1]}=\frac{V_1-V_{D_1}^{[1]}}{R_1}. $$ Now, if the nonlinear characteristic \eqref{ll} is in some sense well behaved, by iterating the process you get a sequence of values which converges to the desired quiescent point $$ \big(0,I_{AQ}^{[0]}\big),\big(V_{D_1}^{[1]},I_{AQ}^{[1]}\big),\ldots,\big(V_{D_1}^{[n]},I_{AQ}^{[n]}\big)\underset{n\to\infty}{\longrightarrow}(V_{QD_1},I_{AQ}). $$ In practice, after \$n \simeq (3\div4)\$ iterations you have a satisfactory value for most applications.

3. The analytical method. For certain \$I-V\$ characteristics, you can calculate directly the quiescent point \$(V_{QD_1},I_{AQ})\$. For example, the \$I-V\$ characteristics of the diode \$D_1\$ is $$ I_A=I_s\left(e^{\frac{V_{D_1}}{V_T}}-1\right) \tag{D I-V}\label{div} $$ where

   • \$V_T=\frac{k_BT}{q}\simeq 25\mathrm{mV}\$ at \$T=25^\circ\mathrm{C}\$ is the thermal voltage and
   • \$I_s\$ is the saturation current of the diode.

   Applying the KVL to the above mesh, we get $$ V_1-V_{QD_1}=R_1 I_{AQ}, $$ and using equation \eqref{div} to express \$V_{QD_1}\$ we have $$ (I_{AQ}+I_s)\frac{R_1}{V_T}e^{(I_{AQ}+I_s)\dfrac{R_1}{V_T}}-I_s\frac{R_1}{V_T}e^{\dfrac{V_1+R_1I_s}{V_T}}=0\tag{1}\label{1} $$ Equation \eqref{1} has the form \$we^w-x=0\$, which can be solved analytically by using Lambert's \$W\$ function, $$ w=W(x) \tag{L's W}\label{lw} $$ Therefore, this special function can be used to solve equation \eqref{1} above: $$ \begin{split} (I_{AQ}+I_s)\frac{R_1}{V_T}&=W\left(I_s\frac{R_1}{V_T}e^{\dfrac{V_1+R_1I_s}{V_T}}\right)\\ &\Updownarrow \\I_{AQ}&=\frac{V_T}{R_1}W\left(I_s\frac{R_1}{V_T}e^{\dfrac{V_1+R_1I_s}{V_T}}\right)-I_s \end{split}\tag{2}\label{2} $$ To apply this formula, we need to know \$I_s\$: having again a look at the LTL-307EE datasheet, we see that typically \$V_{QD_1}=V_\mathrm{F}=2.0\mathrm{V}\$ when \$I_{AQ}=I_F=20\mathrm{mA}\$: from \eqref{div} we have that $$ I_s\simeq I_{AQ} e^{-\frac{V_{QD_1}}{V_T}}\simeq 3,6097\cdot 10^{-37}\mathrm{A}. $$ Finally, by using a Lambert's \$W\$ calculator (or the nice asymptotic expansion 4.13.10 from the NIST) and the given values \$R_1=100\Omega\$ and \$V_1=9\mathrm{V}\$ you get $$ I_{AQ}\simeq 71\mathrm{mA} $$ which is nearly the value found by @Michel Keijzers in his answer.

Final notes

• Historically, the first method used by engineers in order tho solve this problem was the graphical method, since it requires only a ruler and a nice diagram in the data sheet. With the appearance of small calculators and PCs, the numerical approximation method gained popularity and finally, in the second half of the nineties of the XX century, consequentially to the emergence of new studies on the Lambert's \$W\$ also the analytic method gained popularity (see for example the nice paper by Banwell [1]).

• Each of the three methods has its own merits, the graphical one being the most simple but less accurate, with accuracy rising at the expense of simplicity when passing from it to the numerical approximation and finally to the analytic evaluation. Therefore, given the accuracy of the numerical method, It would seem that the analytical approach is worthless: but it is not so, basically for two reasons

  1. For large values of the parameter \$x\$ (as in the case under study), \$W(x)\simeq \ln x\$ (see again formula 4.13.10 from the NIST) so there is almost always no difficult in evaluating \eqref{2} or similar expression involving \$W\$.
  2. Lambert's \$W\$ function can be used to deduce exact formulas where the effects of the variation of each of the involved parameters is determined exactly. This is very useful in determining how the parameter tolerance, temperatur/age drifts, etc. impacts on the output of the circuit: for example, \$W\$ can be used to design BJT circuits with outstanding performances, especially from the point of view of parametric variation with temperature, as Banwell [1] shows.

• Finally, as an Electronics Engineer I must point out that, however precise, each of these methods gives in practice approximate (thought more or less) results: the spread of devices characteristics and the non constancy of junction temperatures put a lower limit on the precision attainable by calculations.

[1] Thomas C. Banwell (2000), "Bipolar Transistor Circuit Analysis Using the Lambert W-Function", IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS—I: FUNDAMENTAL THEORY AND APPLICATIONS, VOL. 47, n° 11, pp. 1621-1633, DOI: 10.1109/81.895330.

Answer 4

The LED calculations can seem a bit tricky because the forward voltage, V_f, changes with current. A loadline graph and a little maths may help your understanding.

Figure 1. A graph of various coloured LED current versus Loadline resistance graph.

1. Each loadline is drawn from the supply voltage point (5 V in Figure 1) to its short-circuit current point on the I_f axis. So for your 100 Ω on a 5 V supply this would be \$ I_f = \frac {V}{R} = \frac {5}{100} = 50 \ \text{mA} \$.
2. Next pick your LED colour. You didn't specify so we'll go with orange.
3. Then find where the 100 Ω loadline crosses the orange LED curve and read the current. This is 30 mA in our example.

For your 9 V supply you can redraw the 100 Ω loadline out to 9 V and up to 90 mA and do the calculation.

In practice most of us would make an initial calculation, say that it can't be more than 90 mA with 100 Ω on a 9 V supply and that judging by the graph the forward voltage will be about 2.3 V. Therefore you have 6.7 V across the resistor and can work out that the current would be about 67 mA.

Answer 5

The LTL-307EE is the default part number that the schematic capture has for an LED part. Make your circuit, use a 1K resistor, and measure the voltage across the LED to find its Vf. Then stick into the equation to determine current flow.

(Vsource - Vf)/known resistor = currrent

Or flip it around:

(Vsource - Vf)/current desired = resistor to use.

Many LEDs have a 20mA max continuous current, and are quite bright at just 10mA. The old red indicator LEDs needed 20mA, new LEDs are a lot more efficient.