# CVD model and Diode Voltage

Thread Starter

#### Hadams0002

Joined Sep 14, 2020
2
Hello,

This is my first post, but I have an issue understanding a topic. In my class, we discuss the constant voltage drop model which assumes a 0.7 voltage drop across silicon diodes (which is what we supposedly use in the class). However, in my labs, we are asked to calculate the diode voltage. Would this not be the 0.7 volts? If not, how would I go about calculating this value?

#### ericgibbs

Joined Jan 29, 2010
17,440
hi H2
Welcome to AAC.

Do you have more details of your question.? However, in my labs, we are asked to calculate the diode voltage.

E

Thread Starter

#### Hadams0002

Joined Sep 14, 2020
2
hi H2
Welcome to AAC.

Do you have more details of your question.? However, in my labs, we are asked to calculate the diode voltage.

E
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We are evaluating this circuit using the CVD model at different supply voltages, for example Vs1 =0.5 V and Vs2 = 1V. We are asked to find the diode voltage for each supply voltage, but to my understanding, the diode voltage should always be 0.7 V (assuming it is a silicon diode). We are asked to calculate other values as well, but I am comfortable with those.

#### ericgibbs

Joined Jan 29, 2010
17,440
hi,
When doing rough calculations a 0.7V diode forward voltage is used.

The Vfwd [ Voltage forward] is current dependent, thats why you need to measure from 0.5v thru 1v.

The plot is always shown in the datasheet for a particular diode type.

Post your test results and we can check for you.
E

#### crutschow

Joined Mar 14, 2008
32,067
The diode forward voltage is actually has logarithmic relation to the diode current, it's just that at typical small diode currents, the forward drop is about 0.7V.
It's not a constant.

#### WBahn

Joined Mar 31, 2012
28,494
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View attachment 218891
We are evaluating this circuit using the CVD model at different supply voltages, for example Vs1 =0.5 V and Vs2 = 1V. We are asked to find the diode voltage for each supply voltage, but to my understanding, the diode voltage should always be 0.7 V (assuming it is a silicon diode). We are asked to calculate other values as well, but I am comfortable with those.
If the supply voltage is 0.5 V and the voltage across the diode is 0.7 V, what is the voltage across the resistor? What direction is current flowing in the resistor? Does this make sense?

#### WBahn

Joined Mar 31, 2012
28,494
hi,
When doing rough calculations a 0.7V diode forward voltage is used.

The Vfwd [ Voltage forward] is current dependent, thats why you need to measure from 0.5v thru 1v.

The plot is always shown in the datasheet for a particular diode type.

Post your test results and we can check for you.
E
The diode forward voltage is actually has logarithmic relation to the diode current, it's just that at typical small diode currents, the forward drop is about 0.7V.
It's not a constant.
The lab he is doing is specifically using a constant-voltage model for the diode. Look at the voltage supply values. There is no need to go to better non-constant voltage drop models to see the point of the lab.

#### crutschow

Joined Mar 14, 2008
32,067
Look at the voltage supply values. There is no need to go to better non-constant voltage drop models to see the point of the lab.
He mentions 0.5V and 1V, but also other unstated voltages.
Certainly the diode voltage will change with those voltages.

#### WBahn

Joined Mar 31, 2012
28,494
He mentions 0.5V and 1V, but also other unstated voltages.
Certainly the diode voltage will change with those voltages.
Doesn't matter. The lab that he is doing specifies the use of the constant-voltage-drop model for the diode with a forward drop of 0.7 V.

The whole point of the lab is to hit home the point that even with that model, you can't just blindly assume that the voltage drop across the diode is always a constant 0.7 V. There is a HUGE caveat in that model that people new to the topic frequently miss completely -- like the TS did here.

#### MrAl

Joined Jun 17, 2014
10,091
The way i read this problem, and i could be wrong, is that the 0.7v constant voltage model assumes 0.7v but ONLY if the circuit allows. If the circuit does not allow, then the diode is OPEN.
So the two choices are either 0.7v or OPEN.

Now if the diode is an open circuit, it could have LESS than 0.7v across it but it is not conducting.
So part of the problem is to determine if the diode is conducting or not. Once it conducts, the voltage must be a constant 0.7v, but before that it could be something else including a negative voltage like -5vc but of course ONLY if the circuit dictates that.

I wont give the two answers just yet, but it should be obvious now.

@WBahn
I agree that we do not want to get into the log or exponential models, in fact we cant really do that because the model has already been specified.

#### crutschow

Joined Mar 14, 2008
32,067
I see the purpose of the experiment being to show that the constant voltage model is not accurate/sufficient.

#### WBahn

Joined Mar 31, 2012
28,494
I see the purpose of the experiment being to show that the constant voltage model is not accurate/sufficient.
How would it show that?

Not accurate/sufficient for what?

The constant voltage model is used for the overwhelming majority of design and analysis problems involving large signals, even in the real world. Even most small-signal design and analysis work uses the constant voltage model as the large-signal (typically the bias or quiescent point) component. There are certainly situations in which it is not sufficiently accurate for the problem at hand, but that is true of pretty much any model.

The first time I taught Circuits I, I through together a largely random circuit with six diodes and a bunch of resistors in it and asked the students to, as a group, draw the Vout vs Vin characteristic on a piece of graph paper, over a range of Vin from -30 V to +30 V, using the constant voltage model using a particular scaling for each axis. After they were done I put their chart on an X-Y recorder and then pulled out that circuit, made with 5% resistors and several different small-signal silicon diodes and connected an adjustable voltage supply to the input of the circuit and the X-input of the recorder and the output port of the circuit to the Y-input and then walked the voltage up and down from over the specified range. The actual characteristic followed quite close to the one they had prepared, even at the transition points as diodes transitioned (there was certainly a rounded transition at each one, as you would expect) between 'off' and 'on'.

#### bertus

Joined Apr 5, 2008
22,143
Hello,

Read the PDF about the CVD diode.

Bertus

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#### ericgibbs

Joined Jan 29, 2010
17,440
We are evaluating this circuit using the CVD model at different supply voltages, for example Vs1 =0.5 V and Vs2 = 1V. We are asked to find the diode voltage for each supply voltage, but to my understanding, the diode voltage should always be 0.7 V (assuming it is a silicon diode). We are asked to calculate other values as well, but I am comfortable with those.
Hi Bill,
The way I read the question, is that the TS has to use 0.5 and 1v , but also 'other values as well.

I appreciate that the question says CVD model, but IMO doing the test with fixed voltages of 0.5v and 1v is pointless, it will not give the TS an insight into the actual operational plot of a diode. ie: Vfwd wrt to Ifwd.

E

BTW: Do we know the Author of the CV Diode PDF.?

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#### Wolframore

Joined Jan 21, 2019
2,590
since the voltage drop depends on the supply voltage and current after this lab it would be interesting to change the resistor value.

#### crutschow

Joined Mar 14, 2008
32,067
We are asked to calculate other values as well
What are those values?

#### MrAl

Joined Jun 17, 2014
10,091
I see the purpose of the experiment being to show that the constant voltage model is not accurate/sufficient.
Sorry but i believe you are over thinking this.
Two very simple problems, two very simple answers.

#### WBahn

Joined Mar 31, 2012
28,494
Hi Bill,
The way I read the question, is that the TS has to use 0.5 and 1v , but also 'other values as well.

I appreciate that the question says CVD model, but IMO doing the test with fixed voltages of 0.5v and 1v is pointless, it will not give the TS an insight into the actual operational plot of a diode. ie: Vfwd wrt to Ifwd.

E

BTW: Do we know the Author of the CV Diode PDF.?
I don't think that this lab is intended to give any insight into the actual operational plot of a diode -- usually there are other labs that are intended to get at the exponential nature of the diode characteristic.

I think the intend of this lab is to drive home the point that the CVD model of the diode does NOT say that the forward voltage drop across a diode is ALWAYS the same constant value in ALL circumstances. This is a very common misconception that students new to working with diodes make. Like the TS did, they blindly replace the diode with a voltage source without considering whether the diode in question is actually conducting. The voltage drop across the diode is Vd only if Id > 0. If Id = 0, then the voltage drop can be any voltage up to Vd.

#### crutschow

Joined Mar 14, 2008
32,067
I think the intend of this lab is to drive home the point that the CVD model of the diode does NOT say that the forward voltage drop across a diode is ALWAYS the same constant value in ALL circumstances.
So how is that different from showing that the CVD model is not sufficient to understand the operation of a diode?
Seems like semantic hair splitting.

#### WBahn

Joined Mar 31, 2012
28,494
So how is that different from showing that the CVD model is not sufficient to understand the operation of a diode?
Seems like semantic hair splitting.
Two completely different things. A lab that illustrates correct versus incorrect application of a device model has nothing to do with showing the validity of the model itself.

You seem to be under the same miscomprehension as the TS regarding just what the CVD model for a diode is.

Again, the CVD model does NOT claim that the voltage across the diode is always a constant under all circumstances. If you apply a forward voltage of 0.5 V across a diode, the CVD model for a diode that has a forward voltage of 0.7 V says that the voltage across that diode is 0.5 V; it does NOT claim that it somehow has a voltage of 0.7 V across it. Anyone that says that it does is merely demonstrating their lack of understanding of what the CVD model is.

To show that the CVD model is insufficient, you would need to show that the voltage and current predicted by it are too far in error to be useful under the circumstances related to that problem. Well, is 0.5 V across the diode so far in error as to not be useful?

Let's use the data for a 1N4148 diode and see. Here's the applicable graph from the ON Semi datasheet:

At a voltage of 500 mV, the typical current at room temperature is 100 μA. If that current is lowing through a 330 Ω resistor, the voltage drop across the resistor is 33 mV.

Clearly this voltage is too high, since that would leave no voltage across the diode to drive that 100 μA.

At a voltage of 475 mV, the current is about 60 μA yielding a drop of about 20 mV.

This voltage is too low, since the 25 mV across the resistor would result in 76 μA of current, which would result in a higher diode voltage.

If we interpolate on the graph a bit it appears that 70 μA is at about 482 mV and this voltage is too high because that would only yield a resistor current of about 55 μA.

So the typical room-temperature diode voltage for a 1N4148 in this circuit would be somewhere between 475 mV and 482 mV.

It sure seems like the CVD model's prediction of 0.5 V isn't that bad.

But it would seem that you are claiming that measurements taken in this lab are sufficient to show that this model just isn't good enough, even for this lab.

So what are the parameters for the exponential model that you seem to be claiming the TS needs to be using in order to come up with a decent answer to this lab question when Vs = 0.5 V?

We can see that V(100 μA) = 500 mV and V(1 μA) = 275 mV (as good as we can claim from the graphic, anyway).

The diode equation is

$$i_d \; = \; I_S \left( e^{\frac{v_d}{nV_T}} \; - \; 1\right)$$

Hopefully you will agree that we can ignore the -1 at the levels we will be operating at.

With those two data points, we can see that the voltage needed to increase the current by a factor of 10 is 112.5 mV.

With this we can deduce that the change in voltage, ΔV, away from Vo (the voltage that produces a current of Io) needed to produce a current of α·Io is

$$\Delta V \; = \; 112.5 \, \text{mV} \; \frac{ln(\alpha)}{ln(10)}$$

If we use Vo(Io = 100 μA) = 100 mA, then the voltage across the resistor will be -ΔV and the current will be -ΔV/R = α·Io, so the load line requires that

$$-\Delta V \; = \; 112.5 \, \text{mV} \; \frac{ln \left( \frac{-\Delta V}{I_0 R} \right) }{ln(10)}$$

Solving this yields V = 478.7 mV

However, before we start claiming that this proves that the exponential model must be used in order to get useful results and that the CVD model just isn't sufficient, how good is the exponential model at accurately representing reality? What about the impact of temperature. If the diode is cold, say -40 °C, then the diode voltage would be around 497 mV, putting it much closer to the CVD's prediction than the basic exponential model ones, even if you took the thermal voltage into account since the impact of temperature on the saturation current is the dominant effect.