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

No LTSpice allowed to find those values!!!

t_n_k already told my how to do!

 

No LTSpice allowed to find those values!!!

t_n_k already told my how to do!

hi,
You did post a LTS plot image asking the question, so I replied in the same format.

I thought you wanted to use LTS to double check your calculated answer.

E

 

hi,
You did post a LTS plot image asking the question, so I replied in the same format.

I thought you wanted to use LTS to double check your calculated answer.

E

Ah ok... I misunderstood you...

Sure... I want to double check with LTSpice...

Sorry...

Thanks for replying!

 

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

Hi,

What i was really trying to say was that it makes no sense to me why we would calculate any response to temperature change with the simplified equation when it represents only part of the total equation normally used for a diode, and that the results are so skewed from what we really see it's hard to understand why anyone would even ask this question. This is especially true since in this example (as many others) we are asked to calculate the current, and so the error in the voltage is not noticed. This gives the impression that we have the correct equation already and dont need to modify it, so this is very misleading.

Now i am not saying that including the adjustment for Is over temperature is perfect, but it fits much better with real life observation.

Note that it seems like i might be being a little pedantic here, but because the results are so very different i dont feel i really am. If the results were a little off, or even a little bit more off than we really like to see usually, i might not have brought this up. But because the results are off by a lot and in a totally different *direction* too (positive instead of negative or vice versa), i just had to say something. Maybe just dont calculate the voltage then

 

Ok, now I need help with the attached one.

Is this a full wave or half-wave rectifier? I would say full wave rectifier, but I'm not sure!

I also need help to plot Vout curve... To do this, I need to build Vout equation.

Can you help me identify the current directions???

I cannot understand the circuit because both diodes are only allowing current to flow from the transformer mid point to the circuit and in the previous problem, the current was flowing from the other 2 points to the circuit.

I'll try an approach to find Vout but I'm almost sure that it's not correct!

\(V_{in} = R_{L}*I_{1} + 0.7\)
\(I_{1}=\frac{v_{in}-0.7}{R_{L}}\)

I get an amplitude for current of 0.0142 which I think it matches LTSpice!

 

hi,
Dont forget the Vin is a 50Hz sinewave that swings positive and negative with respect to 0V.[ref]

So the diodes will conduct when the AC swing is on the negative half cycle.

the current was flowing from the other 2 points to the circuit.

It still is, but on the neg half cycle.
E

 

hi,
Dont forget the Vin is a 50Hz sinewave that swings positive and negative with respect to 0V.[ref]

So the diodes will conduct when the AC swing is on the negative half cycle.




It still is, but on the neg half cycle.
E

Frequency is 100Hz. At least is what is said in the problem!

So, I think this is a full wave rectifier. Am I correct?

About the equation, is it correct or not?

 

For the positive half of a cycle D1 is reverse biased and D2 is forward biased

For the negative half of a cycle we have this situation

 

Frequency is 100Hz. At least is what is said in the problem!

You are correct 100Hz, I was still thinking you were working on 50Hz mains.

E

 

Morning guys...

Can I write the following equation for the attached circuit?

Vin - 2V - Vout + 0.7 = 0

 

Morning guys...

Can I write the following equation for the attached circuit?

Vin - 2V - Vout + 0.7 = 0

Morning Psy,

Go have a nice cuppa coffee and re think your equation.

Why are you using a DC sweep for Vin.

Eric

 

Morning Psy,

Go have a nice cuppa coffee and re think your equation.

Why are you using a DC seep for Vin.

Eric

I just can't find a way to analyse circuits and write down their equations.

I've used a DC Sweep to try to see the moment where the diode stops/starts conducting current!

I don't understand why my equation is wrong.

I tried to use Jony130 method using his image

and tried to do the same with my circuit:

 

Ok... I got it now!!!

I was considering Vout as being only \(R_{1}*I_{1}\) but \(V_{out}\) also includes \(V_{2}\)

 

Hi.
Consider the End Limits of the Vin sweep voltage.

When Vin= 0v, the diode is forward biased due to V1 and conducting.
Current is 'flowing' thru the R1 resistor and the diode.

You assume that the Vfwd of the diode is 0.7V, so V1- 0.7 is acting around the circuit.
Thats 1.3V across R1 and the diode, from that you can work out the R1 current.

Now consider when Vin = 2v, no current flows on R1 or the diode.

At some point along the rising Vin sweep voltage, V2 + Diode Vfwd will equal V1.

Does this help.?

E

 

Hi.
Consider the End Limits of the Vin sweep voltage.

When Vin= 0v, the diode is forward biased due to V1 and conducting.
Current is 'flowing' thru the R1 resistor and the diode.

You assume that the Vfwd of the diode is 0.7V, so V1- 0.7 is acting around the circuit.
Thats 1.3V across R1, from that you can work out the R1 current.

Now consider when Vin = 2v, no current flows on R1 or the diode.

At some point along the rising Vin sweep voltage, V2 + Diode Vfwd will equal V1.

Does this help.?

E

Sure it does...

I've already did my calcs:

\(V_{in}+V_{\gamma}+R_{1}\cdot i_{1}-V_{1}=0\)
\(\Leftrightarrow i_{1} = \frac{-V_{in}+1.3}{1000}\)

and:

\(V_{out} = -R_{1}\cdot i_{1}+V_{1}\)

If I set \(V_{in}=1V\)

\(i_{1}= 300\mu A\)

and

\(V_{out} = -0.3V + 2V = 1.7V\)

which matches LTSpice if I replace the Diode for a Votlage Source of 0.7V...

 

hi,
Looks ok to me.

If Vin = 1V and V2 = 2V the voltage across R1 and diode = 1v

As The diode forward bias is assumed as 0.7V, this means 1v-0.7v = 0.3V must be dropped across R1.

Which makes Vout = 2v - 0.3v = 1.7v

E

 

Well, my exam was yesterday but it didn't go very well...

I think I'll try again next year because I need 1st to understand circuits analysis as it's there my main struggle.

Thanks for all the help you guys gave!
Psy