# diode confusion

#### Transatlantic

Joined Feb 6, 2014
44
Given the following voltage divider :

I understand that the voltage across each resister is the same due to their values being the same, and so the voltage is evenly weighted.

But if I then add a diode, like this :

I don't really understand why the voltage across each resister is now different?

My guess is that it is because the diode has enough forward voltage such that it's resistance is now very small, so the voltage divider now has more resistance in the top half, hence more voltage.

Is that correct?

My real confusion of all this though is how you would have computed the voltages manually, I find my self getting into a loop. I can't compute the voltage across r3 because I don't now the resistance across d2, and I can't compute the resistance across d2 because i don't know the current through d2 etc etc

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#### AnalogKid

Joined Aug 1, 2013
10,943
When it is forward biases, a diode is a nearly constant voltage drop. If it were an ideal diode, then the voltage across it would be 0.65 V with 1 mA current, 10 mA, 100 mA 1 A, etc. No matter what the current, the voltage drop would be the same. For a real diode the voltage drop does increase with current, but not much over reasonable ranges of values. So in your 2nd drawing, the voltage across R3 is the diode voltage no matter what the value of R3 is. Of course the current changes with the resistance, and this is how you solve the problem. Start with the diode and work backwards toward the battery.

ak

#### shteii01

Joined Feb 19, 2010
4,644
Given the following voltage divider :
View attachment 86274
I understand that the voltage across each resister is the same due to their values being the same, and so the voltage is evenly weighted.

But if I then add a diode, like this :
View attachment 86275
I don't really understand why the voltage across each resister is now different?

My guess is that it is because the diode has enough forward voltage such that it's resistance is now very small, so the voltage divider now has more resistance in the top half, hence more voltage.

Is that correct?

My real confusion of all this though is how you would have computed the voltages manually, I find my self getting into a loop. I can't compute the voltage across r3 because I don't now the resistance across d2, and I can't compute the resistance across d2 because i don't know the current through d2 etc etc
When you added diode, you created current divider. When diode is On, it has very low resistance so the bulk of the current goes through the diode. Apply Ohm's Law to the resistor and you get lower voltage across the resistor.

#### irobot

Joined May 16, 2015
24
Your measurements are what we would expect to see.

By placing the diode in the circuit, you are effectively "shorting" resistor R3. With R3 shorted there is now more of a voltage drop across R2, hence the reading of 5.384 Volts. ( In practice, the diode has a tiny amount of resistance when forward biased, as in your diagram).

The voltage across R2 would be exactly your supply voltage of 6V if you simply shorted R3 with a piece of wire.

So, yes, your "guess" or hypothesis is correct.

Basic Ohm's Law: V = I R

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#### irobot

Joined May 16, 2015
24
Also, to find the exact voltage drop across the diode (forward resistance) you (or any engineer) would consult the data sheet for that particular device.

#### ErnieM

Joined Apr 24, 2011
8,373
You can analyze circuits like this with simulators or with advanced mathematical means but I prefer just using a basic understanding of these things, which I thing is what you are after.

A forward conducting diode will have about the same voltage across it over a large range of current. As a shorthand engineers will just estimate this voltage between 0.6 to 0.7 VDC. What exact value will one day be clearer (it's more intuition then calculation) but for today just guess 0.7V (a very popular estimate).

So look at your circuit again and know R3 has 0.7V across it. That means R2 has 6 - 0.7 = 5.3 volts across it, as the sum of the voltages around the loop is zero.

Knowing the voltages across the resistors you can calculate the currents from Ohm's law:

Ir2 = 5.3 / 1K = 5.3 mA

Ir3 = 0.7 / 1K = 0.7 mA

How much current flows into the diode D2? Well the sum of the currents into/out of a node is also zero so:

Id2 = Ir2 - Ir3 = 5.3 mA - 0.7 mA = 4.6 mA

That's just an estimate, but look how close it matches your simulation. The point here is to see how the diode voltage drives the rest of the voltages and currents.

#### Transatlantic

Joined Feb 6, 2014
44
Your measurements are what we would expect to see.

By placing the diode in the circuit, you are effectively "shorting" resistor R3. With R3 shorted there is now more of a voltage drop across R2, hence the reading of 5.384 Volts. ( In practice, the diode has a tiny amount of resistance when forward biased, as in your diagram).

The voltage across R2 would be exactly your supply voltage of 6V if you simply shorted R3 with a piece of wire.

So, yes, your "guess" or hypothesis is correct.

Basic Ohm's Law: V = I R
I'm trying to explain the values I'm seeing without the ammeters/voltmeters readings in other parts of the circuit to help.

But I don't understand why I'm just able to assume the diode is shorting? Isn't the diode only able to short R3 if it has enough voltage across it (06-0.7v), to be in its conductive state?

Meaning I can only assume it is conducting if I first compute the voltage across the potential divider to see if a suitable voltage falls across it? Where the upper resistance of the divider is R2, and the lower resistance is the parallel components r3|D3. The upper resistance is obviously 10k, but the lower resistance I don't know, because the diodes resistance is dynamic.

It feels like a chicken/egg thing

Is it a 2 part process? Can I do the following?

- I first assume D3 doesn't have enough voltage to conduct, and so it's resistance is infinite, and then let the lower part of the divider be roughly 1k (parallel resistance of R3/D3). I then know that the lower part of the divider has about 3v across it (as the upper part has roughly the same resistance), which is then enough to enable to the diode to conduct.

At this point, D3 shorts R3, and so the lower part of the divider is the parallel resistance of R3 and D3, where the resistance across D3 is now very small, and so the lower part of the divider is now roughly the resistance of D3 in it's conductive state.

I'd really appreciate it if someone could run through exactly how the voltages across R2 and R3 are computed without assuming there is enough voltage across D3 for it to short (without the aid of ammeters/voltmeters)

#### MikeML

Joined Oct 2, 2009
5,444
First, here is the diode voltage V(d), and the three branch currents I(R1), I(R2), and I(D1) as the supply voltage V1 is increased from 0V to 6V in steps of 10mV. Note what is happening for V1<~0.6V, where I(D1) = 0, and the voltage at V(d) is just a simple voltage divider consisting of two 1KΩ resistors in series...

For V1>0.6V, things change. To make the analysis easier, suppose you replace the diode with a voltage sensitive switch that turns on at 0.65V and a 0.65V voltage source that holds the voltage at V(d) constant if it tries to go higher than that. Now look at the voltages and currents and compare them to what happened with the diode in place.

With the switch/source substituted for the real diode, it is easy to algebraically calculate the currents at any given input voltage... Do you understand why the knee is at exactly 2x0.65V?

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#### irobot

Joined May 16, 2015
24
The above examples given above by the others are very good.

Here is yet another one that I hope may help:

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#### Transatlantic

Joined Feb 6, 2014
44
Thanks for your help guys, I really appreciate the answers given. The graphs were especially useful, and I have now discovered how to generate DC sweeps to see how things change over a given range

But there is still one major aspect that I don't understand. In all the answers, you know the voltage across the diode is 0.6-0.7 given by the meters, but how would you know that without the aid of the meters?

To work out the voltage across the diode you need to know whether the lower half of the potential divider has enough voltage to "Switch the diode on", i.e more than 0.6v, but to work out the lower half of the potential divider you need the diodes resistance, and to work out it's resistance, you need the voltage across it ... enter loop

As I was trying to get at in my previous reply, is there some initialization going on, and so things can be broken into time steps?

At (Time zero, circuit initializing)
- R(R2) = 1k, R(R3) = 1k, R(D2 = ∞Ω), V(R2) = 3v, V(R3) = 3v, V(D2) = 3v
- lower half of potential divider forms a parallel resister pair of R(D2 = ∞Ω) and R(R3) = 1k

.. diode now has enough voltage to forward conduct and so it's state changes (resistance drops)

At (Time One, circuit initialized)
- R(R2) = 1k, R(R3) = 1k, R(D2) = 127Ω, V(R2) = 5.3v, V(R3) = 0.6v, V(D2) = 0.6v

Or does it happen another way, where when the circuit is closed, the battery voltage goes from 0-6v over the course of a very small time frame, and so the state changes over time as per the graph posted by MikeML?

.. it's so frustrating when you don't understand simple things

#### MikeML

Joined Oct 2, 2009
5,444
...
But there is still one major aspect that I don't understand. In all the answers, you know the voltage across the diode is 0.6-0.7 given by the meters, but how would you know that without the aid of the meters?

To work out the voltage across the diode you need to know whether the lower half of the potential divider has enough voltage to "Switch the diode on", i.e more than 0.6v, but to work out the lower half of the potential divider you need the diodes resistance, and to work out it's resistance, you need the voltage across it ... enter loop
I showed you that the diode resistance is ignored in the analysis...

#### dl324

Joined Mar 30, 2015
16,642
But there is still one major aspect that I don't understand. In all the answers, you know the voltage across the diode is 0.6-0.7 given by the meters, but how would you know that without the aid of the meters?
Books on semiconductor devices have information on diode characteristics. For a silicon diode biased past the knee of the IV curve, forward voltage is generally considered to be 0.7V. You can see from the IV curve below that at 0.1A, the voltage is 0.8V.

This is from a 1N4001 datasheet:

You can see that at the max average forward current spec (1A), forward voltage is typically 0.92V at 25C; and that it has a negative temperature coefficient (decreases with increasing temperature).

#### shteii01

Joined Feb 19, 2010
4,644
Thanks for your help guys, I really appreciate the answers given. The graphs were especially useful, and I have now discovered how to generate DC sweeps to see how things change over a given range

But there is still one major aspect that I don't understand. In all the answers, you know the voltage across the diode is 0.6-0.7 given by the meters, but how would you know that without the aid of the meters?

To work out the voltage across the diode you need to know whether the lower half of the potential divider has enough voltage to "Switch the diode on", i.e more than 0.6v, but to work out the lower half of the potential divider you need the diodes resistance, and to work out it's resistance, you need the voltage across it ... enter loop

As I was trying to get at in my previous reply, is there some initialization going on, and so things can be broken into time steps?

At (Time zero, circuit initializing)
- R(R2) = 1k, R(R3) = 1k, R(D2 = ∞Ω), V(R2) = 3v, V(R3) = 3v, V(D2) = 3v
- lower half of potential divider forms a parallel resister pair of R(D2 = ∞Ω) and R(R3) = 1k

.. diode now has enough voltage to forward conduct and so it's state changes (resistance drops)

At (Time One, circuit initialized)
- R(R2) = 1k, R(R3) = 1k, R(D2) = 127Ω, V(R2) = 5.3v, V(R3) = 0.6v, V(D2) = 0.6v

Or does it happen another way, where when the circuit is closed, the battery voltage goes from 0-6v over the course of a very small time frame, and so the state changes over time as per the graph posted by MikeML?

.. it's so frustrating when you don't understand simple things
Manufacturer provides the information regarding the voltage (Vf) needed to turn diode On.

#### Transatlantic

Joined Feb 6, 2014
44
Manufacturer provides the information regarding the voltage (Vf) needed to turn diode On.
I know this. As I keep saying (but am failing to get across), my confusion is how you work out if there is enough voltage across the divider in the first place to 'turn the diode on' (I know it needs around 0.6v), when the components across the divider include the diode itself.

But as MikeML pointed out, he has already provided the answer, I'm just too stupid to see it, so I shall keep reading until it all sinks in

#### BillB3857

Joined Feb 28, 2009
2,570
What would the voltage be if the diode were not in the circuit? If it is above the breakdown voltage of the diode, the diode will conduct.

#### shteii01

Joined Feb 19, 2010
4,644
I know this. As I keep saying (but am failing to get across), my confusion is how you work out if there is enough voltage across the divider in the first place to 'turn the diode on' (I know it needs around 0.6v), when the components across the divider include the diode itself.

But as MikeML pointed out, he has already provided the answer, I'm just too stupid to see it, so I shall keep reading until it all sinks in
Ok. You are missing the Current Divider part.
Notice that the voltage across the resistor is 0.616 volts. Applying Ohm's Law, i=0.616/1000=0.616 mA. So what happened? Where did all the current go? The current was increasing until it reached 0.616 mA, at this point the voltage across the resistor became 0.616 volts, the diode is in parallel with the resistor which means that the voltage across resistor and the voltage across diode are the same! Like you already noticed, "typically" diode turns On when there is 0.6 volts across it. So as soon as the voltage across resistor-diode parallel combination became 0.616 volts, the diode turned On and the second (low resistance) branch of your Current Divider turned On.

Voltage Divider works when you have components in Series.

When components are in Parallel, Voltage Divider goes out the window. Parallel Components form Current Divider, that is why your voltage divider approach became useless.

#### WBahn

Joined Mar 31, 2012
29,823
I know this. As I keep saying (but am failing to get across), my confusion is how you work out if there is enough voltage across the divider in the first place to 'turn the diode on' (I know it needs around 0.6v), when the components across the divider include the diode itself.

But as MikeML pointed out, he has already provided the answer, I'm just too stupid to see it, so I shall keep reading until it all sinks in
With just a bit more experience, you will be able to examine most circuits you will encounter and tell whether the diode is conducting or not conducting by inspection. But, of course, you might be wrong. So you analyze the circuit based on your assumption (and, for now, that assumption can be the result of a coin-toss) and then see if the results are consistent with that assumption. So if you think the diode is on, that means that it will have roughly one forward diode voltage drop (~0.7V for silicon) and when you do your analysis you determine what the diode current is. If it is not strictly positive (meaning that at least some current is flowing forward through the diode) then you know your assumption was wrong and you take a step back and make the opposite assumption, namely that it is not conducting and no appreciable current is flowing in it. If you had assumed that the diode was non-conducting, you analyze the circuit and determine the voltage across the diode. If it is not less than the forward diode voltage drop (which includes any negative voltage), then you know that this assumption is wrong.

Another way to approach things, at least for sufficiently simple circuits, is to break the circuit into two pieces in which the diode is one piece and everything else is the other piece. You then determine the voltage-current characteristics of each piece at the terminals where they connect. If you draw these on a graph the solution is where they intersect because there can only be one combination of voltage and current at those terminals and that combination must satisfy both pieces of the circuit. This is called a "load-line analysis".

Yet another way is to use the exponential equation for the voltage-current characteristic of the diode and develop a set of equations using KVL/KCL. The resulting equations will be non-linear and probably cannot be solved algebraically or in closed form. But there are many techniques you can use to iteratively find the solution to whatever degree of accuracy you desire and you can usually get there with just a few iterations making it doable even if you just have a simple scientific calculator handy.

#### t_n_k

Joined Mar 6, 2009
5,455
I "support" the load line approach to determine whether the diode would conduct with a given equivalent source voltage and series resistance.
What do I mean by equivalent source voltage & series resistance?
For the simple case at hand with a voltage divider connected to a diode load this is a simple exercise. One removes the diode and determines the Thevenin equivalent at the output terminals of the voltage divider. With the diode replaced and the Thevenin equivalent circuit supplying the diode one can readily draw the load line on the diode characteristic (presumably available). This should provide an estimate of the likely diode current and voltage drop. If the load line does not intersect the diode characteristic curve at any point then one must conclude that the diode won't conduct.
On a related matter - What does one mean by the diode being in a conducting state? If 1mA of current is a true conducting state what about 1uA? What about 1nA? Does a real diode ever not conduct in the forward bias direction with some non-zero forward bias applied?

#### irobot

Joined May 16, 2015
24
my confusion is how you work out if there is enough voltage across the divider in the first place to 'turn the diode on' (I know it needs around 0.6v), when the components across the divider include the diode itself.
I better understand what your confusion is by this quote . . . . kinda like the answer being part of the question, or which came first-chicken or the egg . . . .

We know from the datasheet that the diode, when conducting in it's "on" state, that the voltage across it and the current will be a certain amount, as in the previous graphs posted by dl324. We plug this parameter into our Ohm's and Kirchoff's equations to determine the value of R2 or whatever other components in the circuit.

Keep in mind, in your circuit, things happen instantaneously. We do these necessary calculations to derive voltages and currents needed to satisfy conditions when our circuit is operating.

Side Note: There are many situations where things don't happen instantaneously in a circuit, such as found in alternating current(s) - there is a whole sub-set of some challenging mathematics for those applications!

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