Welcome back to electronics, this is Doctor Ferri.
In this lesson we're going to look at a,
a model called the ideal diode plus voltage source model.
In our previous lesson we looked at ideal diode model and
we came up with a systematic way of analyzing circuits that way and
that was called the Assumed States Method.
In this lesson we will introduce a better model which is
called the ideal diode plus voltage source model.
In particular, let's look at a diode.
And say this is the actual diode right here and this is its iV characteristics,
so this is the actual diode iV characteristics and
then this is going to be our new model.
If you remember back
with what the model looked like with the ideal diode it came straight up here.
So that was with the ideal diode.
And all we're doing is shifting this to the right.
To get our new model here.
So, in this case, we're going to replace the,
the actual diode, this is the actual diode, right here,
in our circuit with a voltage source, or battery, essentially.
And our ideal diode.
So this voltage source essentially takes the ideal diode and
shifts it to the right by V sub f.
Let's look at a same example that we've done before.
So in our last lesson we did this example with the assumed states method.
And what we found here is at
the consistent state with both diodes conducting.
So the ideal diode model had both states conducting, on on.
And in that case we were interested in solving for V sub 1.
And what we found out was V sub 1 is equal to 5 volts, with that ideal diode model.
So now we're going to redo this model with the better approximation.
So let's do the analysis with the ideal diode plus voltage source.
In that case, we just, first step is to just replace the diode,
the actual diode, with the voltage source plus an ideal diode.
There's my voltage source, there's my ideal diode, and I've got my resistors.
V sub f, we'll assume that these are the same types of diodes.
So they have the same value of V sub f.
Once I make that first step then this is analyzed using the assumed states model,
treating these as ideal diodes.
While in the other case, in the last lesson, we found that the On-On state was
consistent so we'll try that one here and see if we get a consistency.
If so then I don't need to look at any of the other states.
With the diode on then I'm going to replace the ideal diode
with a wire but I keep that voltage source in there.
And I can go ahead and analyze the circuit.
I can do KVLs around this similar to what I did in my last lesson.
When I analyze this circuit I can do kv all around the outer loop and
I can do a kvl here and what I'll find is that if I solve for
i sub d2is equal to 5 minus V sub f over 2,000.
Well that's greater than zero if V sub f is less than 5.
And if I do the same thing solving for i d1, I get 5 minus V sub f over 1,000.
And that's greater than 0 if again V sub f is less than 5.
So let's try silicon.
With silicon V sub f is equal 0.7 volts.
So, if that's the case, then certainly it's conducting, this is
this is consistent, so this case is a valid case.
If V sub f is small enough, and
in that case I'm ultimately interested in solving for V sub 1.
If I solve for V sub 1, I can look around this outer loop.
I now know what, well, let me analyze it this way, in this case.
This being V sub 1 right here, if I do a KVL around the outer loop,
I will solve for V sub 1 is equal to 4.3 volts.
With the ideal case, we solve for V sub 1 is equal to 5 volts.
This model's just a better model,
because it's a better approximation for the i V curve.
Now let's look at a different model.
This is, or, a different circuit.
This is a simpler circuit, but a more complicated voltage sour, supply.
The voltage supply is a ramp.
But I've only got one diode.
So first step is again to redraw it.
With the voltage source plus the ideal diode.
And then this is V sub R.
So this node if I create,
if I make this my ground node then this node voltage is V sub S minus V sub f.
V sub S minus V sub f is right there.
If that's positive, if that's positive,
then current's going to want to flow in this direction.
So it conducts, the diode conducts,
V sub s minus V f is greater than 0.
So in that case, in that case, I can look at the circuit in this way.
Where I just replaced it with a wire.
And I can solve for VR.
VR is equal to V sub S minus V sub f.
And when this becomes positive, the diodes starts conducting, and
when it goes negative, it stops conducting.
So let me go ahead and plot this V sub, V sub f.
At some point in time, the.
At some point in time, this circuit, the, the voltage goes past V sub f and
that's the point.
That's the point that it starts conducting.
So this is V sub R.
Is when this value starts, becomes positive, this value here is V sub f.
Now, suppose V sub s starts going negative, and goes like this.
And then maybe it goes positive again.
Well, if I follow along with V sub R,
it always retains this difference between them which is approximately V sub f.
As soon as this value goes negative.
As soon as this goes negative it stops conducting, and
it's going to be flat along here.
Because that is.
That value's negative.
And at some point in time when V sub R goes positive again,
enough positive so it crosses the threshold.
And V sub R starts, starts conducting again, or
the diode starts conducting again, and you get this type of signal.
So let's go ahead and build a real circuit.
This is an LED, which is light emitting diode, so we're using that as our diode.
It's in series with this resistor.
Right here is the ground.
All of this, this along here is the ground.
I'm measuring the voltage across the resistor with this line.
And over here I've got the power supply and I've got a a line to the scope.
So I'm also measuring the input voltage.
Now let's take a look at our screen.
We've got a function generator and an oscilloscope here.
My function generator is set to a triangular wave.
So let me go ahead and run that.
And we've got a very small voltage there.
Let me run the oscilloscope.
And I'm seeing that voltage.
This is the voltage of the supply, so V sub s.
The blue, this, that's in green.
The blue will be V, V sub r, the voltage across the resistor.
I'm not seeing the V sub r right now because it's actually too small.
So let me increase the amplitude over here on my input.
I have to get an amplitude high enough that the diode starts conducting.
I have to get past that turn on voltage, and you can see it right there,
I've passed it.
So this distance right here is V sub f.
Once V of S gets large enough, great than V of f, then it starts conducing and
V of, and this blue is see voltage across the, the resistor.
As I keep increasing.
We see that we maintain this distance between these two curves and
that's approximately V sub f.
And as we predicted, it, the diode conducts through this region and
it cuts off when it wants, when V sub R wants to go negative.
So the diode turns off and do not conduct until we get over to here, again where
the voltage source is greater than V sub f, and then it starts conducting.
So, in summary, this particular model, the, the ideal diode plus voltage source
model has a threshold voltage that must be surpassed before the diode is turned on.
And that threshold is different depending on the particular material
the diode is made of.
So silicon it's 0.7 volts, Germanium it's, it's smaller.
And LED is a little bit larger, It's one to four volts.
And what we did in this particular model is that we
replace an actual diode with an ideal diode plus a voltage source.
And then we look at that circuit and analyze the ideal voltage, or ideal diode
using the assumed states method that we used in the previous lesson.
Thank you.