What happens if I feed a diode (in the forward direction) from a DC current source?

Can I forward bias the diode using a a DC current source? Is there a minimum forward bias current that switches on the diode? How would the voltage on the branch be impacted by the diode?

Suppose the same DC source is feeding another branch. How can I visualise which branch would have less resistance for the current to flow through? How would the current decide how to split?

Edit: I am looking for a physical explanation to why and how things happen. I could simulate the circuit but this doesn´t provide the explanation I´m looking for.

The results I get from the simulation with/out the second branch are below the circuit.

• Why not use a simulator and test your ideas out? Commented May 4 at 8:33
• I did, but I am rather looking for a physical explanation , perhaps a good point to add to the post. Thanks ! Commented May 4 at 8:39
• I get your point, I´ll work on it and update the post Commented May 4 at 9:20
• Init, I added some graphs of what you should be focusing on in your simulator. Commented May 4 at 9:27
• A diode with constant forward current makes a decent temperature sensor. Commented May 4 at 19:42

Can I forward bias the diode using a a DC current source?

Yes. It's not uncommon to find diodes biased in such a way. Sometimes the diode may be the B-E diode of a transistor, but either will act just like a diode does.

Is there a minimum forward bias current that switches on the diode?

Not really, since you're the one who chooses what "switching the diode on" means.

How would the voltage on the branch be impacted by the diode?

It will settle to whatever follows Schockley's model. Let's say that we want to know the ratio of the two branch currents.

We can manually solve the dual-branch circuit:

simulate this circuit – Schematic created using CircuitLab

The two branches are constrained by the following equations

\begin{aligned} V(I) &= V_D(I_1) + R \, I_1\\ V(I) &= 2 R \, I_2\\ I &= I_1 + I_2 \end{aligned}

The Shockley formula can be solved for diode voltage $$\V_D\$$: $$V_D(I_D) = n \, V_T \log \frac{I_D+I_S}{I_S}.$$

For 1N4148, $$\I_S\approx 2.52 \cdot 10^{-9} \rm{A}\$$. For $$\I_D\gg I_S\$$, $$V_D(I_D) = n \, V_T \log \frac{I_D}{I_S} .$$

Substituting this into the constraint equations, we get \begin{aligned} n\, V_T \log \frac{I_1}{I_S} + R \, I_1 &= 2 R (I-I_1) \\ \log \frac{I_1}{I_S} &= \frac{R}{n\, V_T} (2I_2-I_1). \\ \end{aligned}

Now let $$\I_1=i_1 I_S,\,\, I_2=i_2 I_S\$$, \begin{aligned} \log i_1 &= R \frac{I_S}{n\, V_T} (2i_2-i_1), \\ \end{aligned} and factor out the diode conductance parameter $$\G_D=\frac{I_S}{n\, V_T}\$$, and we're left with $$\log i_1 = G_D\, R\, (2i_2-i_1),$$ or equivalently $$i_1 = e^{G_D\,R\,(2i_2-i_1)}.$$

For 1N4148, $$\n=1.752\$$, and assuming room temperature $$\V_T=0.0258\,\rm{V}\$$, \begin{aligned} G_D(\rm{1N4148}) &= 55.6\rm{\,nS} \\ R_D(\rm{1N4148}) &= \frac{1}{G_D}=18.0{\rm\,M\Omega}\\ \end{aligned}

For a given circuit, let $$\r=G_D R=\frac{R}{R_D}\$$, and note that it is a constant. For 1N4148, $$r=\frac{R}{18.0{\rm\,M\Omega}}$$

Solving for $$\i_1\$$ and $$\i_2\$$: \begin{aligned} \DeclareMathOperator{\prlog}{prlog} i_1 &= \frac {\prlog \left( r\, e^{2\,i_2\,r} \right)} {r}\\ i_2 &= \frac{r\,i_1 + \log i_1}{2\,r} = \frac{1}{2}i_1 + \frac{1}{2\,r}\log i_1 , \\ \end{aligned} where the product log function $$\\prlog(z)=w\$$ is the principal solution of $$\z=w\,e^{w}\$$.

From the solution for $$\i_2\$$, we get the ratio of the two currents in the two branches as a function of diode branch current $$\i_1\$$: $$\frac{i_2}{i_1}=\frac{1}{2} + \frac{1}{2\,G_D\,R} \frac{\log i_1}{i_1}.$$

For 1N4148, $$\frac{i_2}{i_1}=\frac{1}{2} + \frac{9.0{\rm\,M\Omega}}{R} \frac{\log i_1}{i_1}.$$

We note that $$\I_D\gg I_S\$$ implies that $$\i_1\gg1\$$ and thus $$\\log i_1>0\$$.

We can now look at the last paragraph of Fabio's answer:

For the 1N4148, and with Ibias = 1mA, I would expect that setting R= 1kΩ would go very close to having almost equal current in both branches, some slight tweaking would find the correct result.

First, $$\I=1{\rm\,mA}\$$ means that $$\i_1+i_2\approx 400{\rm\,k}\$$, so we expect $$\i_1\$$ to be around 200k.

The ratio of the currents $$\frac{i_2}{i_1}=\frac{1}{2} + \frac {9.0{\rm\,M\Omega}} {1.0{\rm\,k\Omega}} \frac{\log i_1}{i_1} = 0.5 + 9000 \frac{\log{i_1}}{i_1}.$$

We can now plot the current ratio as the function of $$\i_1\$$:

The currents will be equal when $$\i_1 \approx 220{\rm\,k}\$$, that is when $$\I=2I_1=2\,i_1\,I_S=1.1{\rm\,mA}\$$. Fabio was pretty close!

• @ Kuba... - I consider the current i2 (and the corresponding resistor 2R) as a "loss" if compared with an ideal current source. So we have a non-ideal current source supplying the diode path. And you have shown that (and how) this works. Question: What happens when - according to the TO´s question - an IDEAL current source (which does not exist in reality) is used instead of a non-ideal one?
– LvW
Commented May 5 at 8:06
• Hi @Kuba, I wish I had enough reputation o upvote your response. This is a pretty good explanation that I can understand on the electrical / mathematical side ! Commented May 7 at 17:39
• @LvW The whole answer assumes an ideal current source. If you remove the i2 branch, that will be a whole different question. If you're interested, you'd want to ask separately. Commented May 7 at 17:43

Let's assume you want the current source to split equally between the two branches, ie: 0.5 * Ibias in each branch. The value of R would start at zero, and would need to increase to the point where the diode forward voltage at I = 0.5 * Ibias is the same as that of the resistor:
Vf (0.5 Ibias) = 0.5 * Ibias * R.

The forward current in a diode is a function of the applied forward voltage, temperature, and some diode-specific parameters, as given by the Shockley diode equation which you can look up on Wikipedia:
https://en.wikipedia.org/wiki/Shockley_diode_equation

Let's assume that Ibias is about 1mA. From the diode equation, and by looking up some datasheets of diodes, you will get a good idea of what sort of voltage is needed to increase the current from the nano-amp range to the milli-amp range. Here is the datasheet for a popular small-signal diode, the 1N4148:

For the 1N4148, and with Ibias = 1mA, I would expect that setting R= 1kΩ would go very close to having almost equal current in both branches, some slight tweaking would find the correct result.

• Thank you Fabio for the good explanation (Y) Commented May 4 at 10:45

How to show “why and how things happen”

I am looking for a physical explanation to why and how things happen.

I am not sure exactly what you mean by "physical", but I will offer you a functional, qualitative, intuitive, heuristic explanation “to why and how things happen”. This means that in order to understand the main idea (what you need first), we will not be interested in what is inside the diode, but what is its behavior. The benefit of this is that you will be able to understand what and how it does not only the specific diode but also all elements of this type (voltage-stabilizing non-linear resistors).

The role of simulators

I could simulate the circuit but this doesn´t provide the explanation I´m looking for.

Simulators are just tools; it is up to us how we use them. And they can be successfully used for purposes of intuitive understanding. Here is how we can do it here.

Implementation

First, we will take an IV diode characteristic (curve). On it we will choose three typical points with coordinates (V, I). Next, we will replace the diode with an equivalent variable resistor and set its resistance do that to obtain the same three current and voltage values. In this way, we will replace the study of the obscure diode with the study of something simple and understandable... and we will understand how it does this magic.

Voltage divider viewpoint

Real diode

Let's select three typical points (V, I) of the diode IV curve obtained by the arrangement below: (726.5 mV, 20 mA), (798.7 mV, 60 mA) and (844.4 mV, 100 mA).

726.5 mV, 20 mA: Let's start with a 20 mA current value.

simulate this circuit – Schematic created using CircuitLab

798.7 mV, 60 mA: Then we increase the current to 60 mA...

simulate this circuit

844.4 mV, 100 mA: ... and finally, to 100 mA.

simulate this circuit

Dynamic diode resistor

Now let's see how the diode does this magic of keeping the voltage across itself almost constant. The explanation is that when the current increases, the diode decreases its static resistance RF = VF/IF so that the point with coordinates (VF, IF) is located on the IV curve.

This is the clever diode trick - it keeps the product of the two quantities (current and resistance) relatively constant, changing the resistance in the opposite direction of the current change.

Now we have to repeat the three experiments above with a real diode replacing it with an equivalent variable resistor with resistances R1 = 36.33 Ω, R2 = 13.31 Ω and R3 = 8.44 Ω.

726.5 mV, 20 mA: To set 726.5 mV voltage at this current, the diode in Schematic 1.1 behaved as a 36.33 Ω resistor; so we can replace it with such a resistor. We can either calculate or determine this value experimentally by adjusting the resistance so that the voltage is 726.5 mV.

simulate this circuit

798.7 mV, 60 mA: When the input current increases, RF decreases to 13.31 Ω in order to set 798.7 mV (we adjust it so that the voltage is like so)...

simulate this circuit

844 mV, 100 mA: ... and finally, RF decreases to 8.44 Ω in order to set 844,4 mV.

simulate this circuit

Dynamic source resistor

So far we have assumed that the input current is constant, and the diode is dynamic and changes its static resistance to maintain a nearly constant voltage. In reality, things are more complicated because at the same time the current source changes its internal resistance to maintain a constant current. Thus, two non-linear resistors - one voltage-stabilizing and the other current-stabilizing - change their resistances in the same direction until equilibrium is reached. So a dynamic voltage divider Rin-RF is obtained. Let’s see it for the same input currents of 20, 60 and 100 mA.

726.6 mV, 463.7 Ω:

simulate this circuit

798.6 mV, 153.3 Ω:

simulate this circuit

844.3 mV, 91.53 Ω:

simulate this circuit

Current divider viewpoint

Real current divider

How can I visualise which branch would have less resistance for the current to flow through?

When we connect a resistor (e.g., R = 100 Ω) in parallel to the diode, we obtain a dynamic current divider configuration. To simplify the schematic, instead adding a resistor, we can simply set the voltmeter's internal resistance to 100 Ω. Indeed, this will be a very bad voltmeter, but in this case it is useful for us... we want it.

How would the current decide how to split?

The current is divided into two parts which are inversely proportional to the resistances. To observe the partial currents, we can insert (perfect) ammeters in the branches.

Exploring

726.5 mV, 20 mA: As above, let's start with a 20 mA current value.

simulate this circuit

798.7 mV, 60 mA: Then we increase the current to 60 mA...

simulate this circuit

844.4 mV, 100 mA: ... and finally, to 100 mA.

simulate this circuit

Dynamic current divider

Now we have to repeat the three experiments above with a real diode replacing it with an equivalent variable resistor with resistance RF = VF/IF:

R1 = 703/12.97 = 54.2 Ω

R2 = 787.8/52.12 = 15.12 Ω

R3 = 835.7/91.64 = 9.119 Ω

726.5 mV, 20 mA, RF = 54.2 Ω: As usual, let's start with a 20 mA current value.

simulate this circuit

798.7 mV, 60 mA, RF = 15.12 Ω: Then set IF = 60 mA...

simulate this circuit

844.4 mV, 100 mA, RF = 9.119 Ω: ... and finally, IF = 100 mA.

simulate this circuit

Conclusions

• In this arrangement (constant current source supplying a diode), two non-linear resistors (current-stabilizing and voltage-stabilizing) are connected in series.

• To set the desired current, the current source begins decreasing its internal resistance. It stops when Vin/(RIN + RF) = IF.

• To set the desired voltage, the diode also begins decreasing its internal resistance. Thus it helps the current source in its desire to set the current. It stops when Vin.RF/(RIN + RF) = VF.

• Equilibrium is reached when both the current and voltage stops because they have reached the desired values.

• Cascode circuits are based on this "voltage source helps current source" phenomenon.

• That is a very nice way of using the simulator, kudos for showing how to put the simulator to actual use. Would you mind elaborating a bit more on your last conclusion "cascode circuits are based on this voltage src. helps current src. phenomenon ? I assume you refer to cascoding being a current buffer on a transconductance amplifier ? Commented May 7 at 17:42
• @Init_Eng, You can see more tricks in my CircuitLab bag of tricks. I will write about cascode circuits later. Commented May 7 at 17:47
• Thanks ! I'll be looking forward Commented May 7 at 17:53
• @Init_Eng, Here is a nice CircuitLab story about these weird never understood cascode circuits. I created it a year ago with a lot of desire and enthusiasm and since then I have not visited it. But now I liked it so much that I dare not write anything more :-) Commented May 7 at 18:21

Quote InitEng: "Is there a minimum forward bias current that switches on the diode? ..................... Edit: I am looking for a physical explanation to why and how things happen. I could simulate the circuit but this doesn´t provide the explanation I´m looking for.

I agree, a circuit simulation will only show the correlation between two variables, but it will not reveal - from the physical point of view - the difference between cause and effect.

The problem is the artificial element - the current source.

I think, in reality, there is not such a thing called "current source". Speaking of electronic ciruits and the currents within these circuits, it is always a voltage which will be the cause of the observed/measured currents (not the other way round). Otherwise, there would be no E-field (caused by a voltage) which provides the force to let the charged particles move within the wire as well as parts (resistors, diodes, transistors,...).

Simple example: In a simple voltage divider consisting of two resistors, we say - according to Ohm`s law - that the voltage across one of the resistors is "created" by the current through the resistor chain. And - yes - we are allowed to make our calculations accordingly because of the valid correlation.

However, physically spoken, this is not correct. In fact, it is the voltage source which develops corresponding E-fields within the resistors which allows the current. And the strength of these E-fields depends on the voltage source and the conductivity (resistance) of the part.

Simplified example: Voltage source Vs connected to a potentiometer: Depending on the position of the center tap, the entire E-field is divided into two partial ranges (depending on the partial resistances) and two corresponding partial voltages.

Coming back to your question regarding the diode connected to a current source - In reality, this will be a non-ideal current source - realized with a voltage source and a corresponding large source resistor Ri. And as a result- we have a voltage division between Ri and pn junction of the diode.

Remember: The pn junction needs - as a precondition - an external voltage across the depletion zone before a current can be observed through the device.

• You might not be aware that I was in discussion with the op about using a simulator to help them clarify things before they edited part of the question that you have quoted. I was also in the process of developing my answer when they made that edit. Commented May 4 at 15:25
• @ Andy aka - My contribution was not directed to your answer. It was an answer to the TO. Anything wrong with my answer? I am willing to learn - but I want to know why my answer was downquoted. Just to say "wrong" is not enough - I like to improve my understanding.
– LvW
Commented May 4 at 15:35
• @LvW Thanks for providing an extra point of view. Commented May 4 at 20:28

My answer began in comments underneath the question. Once I'd made my answer I saw that the OP had added to the question saying this: I could simulate the circuit but this doesn´t provide the explanation I´m looking for.

Nevertheless, at the time of creating this answer, It appeared that the OP wasn't familiar with a proper simulation because, directly below the newly added words, the OP said this: The results I get from the simulation with/out the second branch are below the circuit. This indicates to me that the OP thought that a diode was more like a switch and could not be shown to have an analogue characteristic. Hence, my reasons for adding so much detail about making a proper simulation.

A simulation can help you establish the type of relationship between the diode current and the voltage across it. Here is a simulation of the 1st quadrant characteristic for the 1N4148: -

I've varied the voltage from 0 volts to 1.5 volts over a time period of 1.5 seconds. That forms the basis of the x-axis i.e. time becomes voltage. You can even use simulators to plot the more useful graph of current against "voltage" where the current is on a log scale: -

You can see that the diode has a quite significant non-linearity. The log graph demonstrates that. Going back to the previous graph format; if I put a 2 Ω resistor in parallel with the diode and repeated the same experiment we would see this: -

You can now see the point where the current through the 2 Ω resistor equals the current through the diode. This tells you that if you applied a current above twice 700 mA (as per your circuit), the diode would take more than 50% of the current.

Here's my version of your circuit. Current ramps up to 1.5 amps in 1.5 seconds: -

And clearly you see the point at which the diode starts to hog the current. These are the types of experiment you can do in a simulator to help you get a feel for what is happening. You can also raise the ambient temperature to get a broader idea about things. Previously I used 27° C but now I use 127° C: -

So, get a feel for what is happening by using a simulator then when you ask this: -

I am looking for a physical explanation to why and how things happen

The answer is the Shockley diode equation: -

Image taken from Voltage across diode, Shockley equation.

Can I forward bias the diode using a a DC current source?

Hopefully you can see that virtually any current in the forward direction will forward bias a diode. You can make an argument that below around 1 nA it's just leakage current sneaking through of course.

Is there a minimum forward bias current that switches on the diode?

Hopefully you can now see that diodes aren't switches but, we do tend to regard a current of 1 mA as a significant point. For the 1N4148 this occurs when the forward bias voltage is 0.6 volts.

• Hi Andy,I am thankful for you taking the time to generate the plots. I understand the schockley equation, and how the voltage over the diode starts to forward bias it, I'll ask a question to the schockley equation and the the graphs you presented from the circuit with 2 branches. 1- The schockley equation suggests the current is the effect of the diode voltage, but the diode is being biased via a DC current source, can the voltage also be treated as an effect of the effect in this case ? in other words, does forcing the current in the diode reduce the fermi level in the p-n junction ? Commented May 4 at 10:38
• @Init_Eng if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. Commented May 4 at 10:39
• Right, I'm working on it Andy, I just edited my comment for a further question. Commented May 4 at 10:41
• @Init_Eng cause and effect are interchangeable here. You can apply a voltage and get a current or, you can apply a current and get the same corresponding terminal voltage. Questions on fermi level are probably best handled on physics stack exchange. Commented May 4 at 10:47
• Alright, thanks Andy for helping me understand this. Commented May 4 at 10:57

Here's another simulation method that can actually give you I-V curves:

I've used a long time of 1 second for the current ramp (so that dynamic effects are insignificant) and changed the X-axis to be logarithmic I(I1) so that time disappears from the plots.

Here's another way- each diode gets 1mA constant current but I've simulated the diode voltage at temperatures from -20°C to +80°C. If you have a .op simulation at more than one temperature then the x axis defaults to showing temperature.

• Thank you for providing an alternative simulation. Commented May 7 at 17:52