From II KVL we find Va voltage

Va = V1 - Id * R1

And

Id = (V1 - V2 - Vd)/(R1 + R2) = (12V - 6V - 0.65V)/1K = 5.35V/1K = 5.35mA

So

Va = 12V - 5.35mA*500R = 12V - 2.675V = 9.325V

We know you can do it. The point was to try and have the OP get there on his own. Since he still couldn't see it even after you did it, maybe there will be some remaining struggle.

 

Your "mesh" looks good. But we don't need mash here.
All we need here is II Kirchhoff's law.
http://forum.allaboutcircuits.com/showpost.php?p=463363&postcount=19

And your circuit look like this

So we can write KVL

V1 = VR1 + VD1 + VR2 + V2 and from this we can find Va.

Va = V2 + VR2 + VD1

or

Va = V1 - VR1

Ok, I almost got it.

So, you built that equation having GND reference always in mind, right?

From GND and following by the left side we have V1 and then in opposition we have Vr1, according to the arrows you defined, right?

Or going by the right side, we have all voltage drops in the same direction, according to the arrows until Va... Is this "thinking" correct?


Just one more question:
When you write Va = V2 + Vr2 + Vd1, why it's not -Vd1 if the voltage drop at the diode is at the opposite direction compared to the arrows you have drawed?

We know you can do it. The point was to try and have the OP get there on his own. Since he still couldn't see it even after you did it, maybe there will be some remaining struggle.

What is an OP?

 

hi Psy,
Look at this marked up image.

The arrows are voltage drops NOT current flow.

Sum the voltage drops from Va to 0V


An OP is you the Original Poster
E

 

Ok, I think I got it.

Now, Vb = -V1+VR1+Vd1 = -8.65V. In terms of absolute value, it's ok, but LTSpice plots a positive amplitude.

Note:
The problem is using V1 = 120t, 0ms<t=<100ms ^ V1 = -120t, 100ms=<t<200ms, which is a pulse of 12V of amplitude and a period of 200ms.

So, my Vb will have -8.65V of amplitude by my calcs but LTSpice plots it with positive value for amplitude!

 

hi Psy,
Look at this marked up image.

The arrows are voltage drops NOT current flow.

Sum the voltage drops from Va to 0V


An OP is you the Original Poster
E

Yes, but shouldn't the voltage drop arrow at the diode have the opposite direction? Does the diode allows voltage goes for - to +???

 

Yes, but shouldn't the voltage drop arrow at the diode have the opposite direction? Does the diode allows voltage goes for - to +???

Va is positive with respect to V2, so the diode is forward biased and has voltage drop of ~0.7V

 

So, you built that equation having GND reference always in mind, right?

Yes, because we always measure all the voltage with respect to ground.
http://forum.allaboutcircuits.com/showpost.php?p=485862&postcount=14

From GND and following by the left side we have V1 and then in opposition we have Vr1, according to the arrows you defined, right?

right

Or going by the right side, we have all voltage drops in the same direction, according to the arrows until Va... Is this "thinking" correct?

Just one more question:
When you write Va = V2 + Vr2 + Vd1, why it's not -Vd1 if the voltage drop at the diode is at the opposite direction compared to the arrows you have drawed?

Ok I see your problem. In my diagram the arrows next to the component represent a voltage drop across this component.
And because I assume that current flow from "+" to "-". So the arrow head (tip) points to the most positive end (highest potential) of the component.
And this is why I have Va = V2 + Vr2 + Vd1

What is an OP?

OP = Original Poster

 

.
.
.

Ok I see your problem. In my diagram the arrows next to the component represent a voltage drop across this component.
And because I assume that current flow from "+" to "-". So the arrow head (tip) points to the most positive end (highest potential) of the component.
And this is why I have Va = V2 + Vr2 + Vd1

.
.
.

So, you have not drawed the green arrows randomly, right? You have followed the correct direction of voltage. This means you already knew how each voltage drop would behave according to both Voltage sources present in the circuit, right?

I mean, that the voltage sources will define each component's voltage drop direction, right?

This is confusing to me because I know that in reality, voltage/current goes in one direction but in theory and for academic purposes, the other way around is considered! So, I'm always struggling to know when should I use a minus or a plus sign when analysing circuits...

 

So, you have not drawed the green arrows randomly, right? You have followed the correct direction of voltage. This means you already knew how each voltage drop would behave according to both Voltage sources present in the circuit, right?

Yes, I did exactly that.
And we allowed to do so because V1 > V2, so we already knew the current direction. And the diode will conduct only if V1>V2.

I mean, that the voltage sources will define each component's voltage drop direction, right?

Yes

 

Yes, I did exactly that.
And we allowed to do so because V1 > V2, so we already knew the current direction. And the diode will conduct only if V1>V2.


Yes

I'm sorry but there is a detail that I still don't fully understand.

Why do you have the Vd1 arrow pointing in the direction where the diode doesn't allow voltage to flow??? Diodes current/voltage always flow from + to -, right? So I can't understand why you draw the arrow from - to +.

 

I'm sorry but there is a detail that I still don't fully understand.

Why do you have the Vd1 arrow pointing in the direction where the diode doesn't allow voltage to flow??? Diodes current/voltage always flow from + to -, right? So I can't understand why you draw the arrow from - to +.

A "Voltage Drop" is positive when the voltage on the "Anode (Arrowhead)" side of the diode is greater the voltage on the "Cathode (horizontal line)" side of the diode. In this condition the diode is forward biased and will conduct any amount of current allowed by the external components and power supplies. It is the two 500 ohm resistors that "limit the current" through the diode at the FIXED voltage drop of approximately 0.7V.

 

First of all the voltage don't flow, only current can flow in the circuit. Voltage don't flow. Voltage is present across the component. Diode is forward biased when voltage at anode is more positive than voltage at the cathode.
And also notice that the arrow head (tip) points to the most positive end (highest potential) of the component. For the diode anode is more positive, and this is why tip of the arrow is pointing toward anode.

 

Hi,

It's interesting that if you hold the current constant and start with a temperature of 300K then increase to 301K still holding the current Id constant, the voltage *rises* using that equation. For a real diode, the voltage *decreases* with increasing temperature and constant current.

 

Hi,

It's interesting that if you hold the current constant and start with a temperature of 300K then increase to 301K still holding the current Id constant, the voltage *rises* using that equation. For a real diode, the voltage *decreases* with increasing temperature and constant current.

True - but the simplified equation ignores the temperature dependence of the diode reverse saturation current.

See here ...

http://dspace.mit.edu/bitstream/handle/1721.1/52058/rle_qpr_045_xi.pdf?sequence=1

 

Morning guys...

Appreciate all the help...

New problem... This one is almost understood. Just a small detail that I couldn't find out...

See attached *.asc and *.png.

It is a battery charger (kind of academic version).

We were asked to plot (manually, not with LTSpice), the curves for Vin, diode current and the voltage that the battery that is being charged is "receiving"!

We have this problem solved by the teacher but as usual, our teacher doesn't gives us all the steps but final ones. So there is something I couldn't find out!

When we plot the curves, there are some key time moments that we need to calculate and mark them along with the curves.

These key time moments are marked in the picture attached.

Our teacher calculated them as being θ1=1.1156rad and θ2 = 2.026rad using the expression 12.7=14.14sin(2*pi*50*t) where 14.14 is (220*sqrt2/11)*2. 11 is the transformation ratio and the 2 on the right side is because we're using a transformer with a middle point. I don't know if is said like this!

I don't understand how we calculate those 2 values...

We used them later to calculate the average current at the diodes using the expression (1/T)∫_θ1^θ2 iD1(t)dt. (It's the integral from θ1 up to θ2).

Can you help?

 

morning Psy,
The region between the two 'marks' on the plot is the period in which the Vout exceeds the battery voltage, so the diode battery charge current only flows into the battery during that period.

E

EDIT:
Check the polarity of the V2 voltage source!!!! in your asc sim circuit.

 

morning Psy,
The region between the two 'marks' on the plot is the period in which the Vout exceeds the battery voltage, so the diode battery charge current only flows into the battery during that period.

E

EDIT:
Check the polarity of the V2 voltage source!!!! in your asc sim circuit.

Thanks Sir Eric. I have already noticed that I forgot to add the "-" in V2 amplitude value or to switch V2 polarity!

But what I was asking for help was to find out values of θ1 and θ2.

I could find the value in milliseconds for θ1 which is 3.55ms and it matches LTSpice. But I don't know how to find θ2 neither to convert them into radians!

 

Surely you realise that you place your calculator in radian mode and calculate

\(\theta_1=sin^{-1}(\frac{12.7}{14.14})=1.1156 \ radians\)

and

\(\theta_2=\pi-\theta_1=2.026 \ radians\)

 

I could find the value in milliseconds for θ1 which is 3.55ms and it matches LTSpice. But I don't know how to find θ2 neither to convert them into radians!

hi,
You know the supply frequency is 50Hz , 1/Ft = 20mSec period.

There are 2 *pi radians in One Period of 360 degrees, so One Radian is ~57.3 degrees

Using LTS double cursors to measure the diode conduction period gives approx 2.9mSec.

So 2.9mS/20mS = 0.145 *360 = 52.2degrees , so Rads = degrees/57.3 = .........

E

 

Surely you realise that you place your calculator in radian mode and calculate

\(\theta_1=sin^{-1}(\frac{12.7}{14.14})=1.1156 \ radians\)

and

\(\theta_2=\pi-\theta_1=2.026 \ radians\)

Didn't know the forum supports Latex!!! That's nice!

Well, I found out the 3.55mS by solving the following equation:

\(14.14=\sin(100\cdot \pi \cdot t) = 12.7\)
\(\frac{12.7}{14.14}=\sin(100\cdot \pi \cdot t)\)
\(\sin^{-1}\left ({\frac{12.7}{14.14}}\right )=100\cdot \p\cdot t\)
\(t=\frac{\sin^{-1}\left ({\frac{12.7}{14.14}}\right )}{100\cdot \p}=3.55mS\)

With calculator in radians, t=3.55ms.