Electronic circuit help (about diodes)!!!

[Sukhdeep]

I am new here so just trying the site if it could help me with Electronic Circuits problems for me.

I made an Circuit with some given parameters in LT Spice. In the file Attached below i have tried to give the overall view of the Circuit that Works on changing capacitance Attached to the bias voltage (voltage Source). I am having difficulty in understanding that when will the diodes D! and D@ will conduct. As far far i understand the D! will start conducting when voltage at Vp1 will be large than the voltage at Storage capacitor Cs (Vb) i.e. bias voltage in turn. I really dont have any Clear idea that when will D2 conduct.
Can anyone help me with it??

Thanks a lot in advance.

 

[Harald Kapp]

Welcome to our fourm.

You have to take into account the inductor. As long as D1 is conducting, current will flow from Vp1 through D1 and L1 into the output (Cs).
If D1 stops conducting (e.g. by Vp1 becoming lower than the voltage on Cs), The magentic field in the inductor wil induce a negative voltage at the junction of D1 and D2. This negative voltage will make D2 conducting.

You can look at this phenomenon in another way, too: An inductor resists change of current due to the energy stored within the inductor (this is why V=L*dI/dt). If the current cannot flow through D1 (due to low voltage on Vp1), it will be supplied by C2 via D2.

As you have modeled the circuit in LTSPICE, you can easily see this effect by substituting a square wave function for V1 and monitoring the currents through D1 and D2 as well as the voltage at the junction of D1, D2 and L1.

Both are accpeted views.

 

[Sukhdeep]

Thanks a lot Harald...

The phenomena sounds correct but i just wonder that id D2 really required here as the flow of current is contineous. I feel that i this particular case d2 will never conduct at any point...

 

[Harald Kapp]

Without D2 the curent can't flow when D1 is off. You need a closed circuit for current to flow. This closed circuit is created by D2, L1 C2 (and in parallel to C2 R1 and C3). Run the simulation and monitor votages and currents in the different circuit components.

 

[Sukhdeep]

Hi,

Thanks a lot for Your suggestions as i am New to Electronics so Your replies are really hellping to understand the Things more clearly HARALD.
With that note here comes one more questions related to Attached model. In following model i Wonder about D2 again. do it help in same way as it do in the previous model Attached With old question? Do the following model called Half wave Bridge rectifier?

 

[Harald Kapp]

Since there is no inductor, you don't need D2 in this circuit.

 

[Sukhdeep]

yes, but is it called half wave rectifier?

 

[Harald Kapp]

D1 makes a half wave rectifier. D2 is useless here.

 

[Sukhdeep]

OK..Agree and thanks .I studied some halfwave reactifier on Google. I am really confused that if D2 is useless then why do i get different outputs With and without D2. And also without D2 the output dont look like a halfwave reactifier as i get some negative peaks too.

What if i add inductor to it keeping the position of D1 and D2 as it is?Will it then be halfwave rectifier With D2?

 

[Harald Kapp]

I am really confused that if D2 is useless then why do i get different outputs With and without D2

When I say D2 is useless for a rectifier, this doesn't mean that it has no influence on the circuit. A negative input voltage will be shorted out by D2.

What if i add inductor to it keeping the position of D1 and D2 as it is?Will it then be halfwave rectifier With D2?

No. An inductor is not part of a rectifier. An inductor could be used to filter the rectified signal, but then the inductor comes after the diode (to the right in your schematic).

In any way your schematic looks curious to me. Why do you use a DC voltage source and a variable capacitor? If this is done to simulate an AC signal as input to the diodes, there is a much simpler way in (LT-)SPICE: use an AC voltage source
- place a voltage source on the schematic (or use the one that is already there)
- right click into the voltage source. A window wil pop up where you can enter "DC value" and "series resistance". This window has an "Advanced" button, click on it.
- a new window will open. Select "sine" from the list at the top left.
- set the parameters amplitude and Freq as desired (you can leave the other entries blank).
You now have a sine voltage source with configurable amplitude and frequency. For more details see the LTSPICE help.

 

[Sukhdeep]

Thanks a lot for taking interest in the circuit. Actually i know the way to put in ac source in spice but this circuit is not just what you are looking upto but far more complex. So i just took a part of it and modified a little bit to make things bit clear and asked my confusions/questions related to the part appearing difficult to me to understand.
Reality being someone very experienced said that in the circuit attached orginally if i move D2 before D1 like i did in second circuit, then it will turn out to be halfwave rectifier but im not getting how? I simulated but the output doesnt look at all like halfwave rectifier. So im confused if im wrong somewhere or the person is wrong!!!

 

[Harald Kapp]

A half-wave rectifier uses only one diode, see the lionk in my reply #8.

 

[Sukhdeep]

Yes...i have already seen that..thanks

Another question related to SPice comes out that if i use switch in the Circuit before D1 that i would like to Close when Cvariable is at its minimum otherwise switch always be at off state. I have seen People using Voltage pulse Source to Close the switch but my question is that if i use the behvioual Source instead of voltage sourec like IF(Cvariable>-minimum, 10) can that work? i found that Three agreements are required in behavioural souce but can you help me With other two arguments along With the one i mentioned in IF statement?

I hope i made it Clear!! Or may be you can suggest me some other good way of doing that!!
(Assume one dont know when the Cminimum is reached , so switch should take care of it itself)

Thanks

 

[Harald Kapp]

A behavioral source should work fine for controlling the switch. Give it a try.

The general syntax of the if statement is:
if(x,y,z) If x > .5, then y else z

BV n0001 n0001 V=(if((Cvariable/Cminimum)-0.5, 1, 0))

where
(Cvariable/Cminimum)-0.5 is an expression that evaluates to >=0.5 for Cvariable >=Cminimum and <0.5 for Cvariable < Cminimum
1 is avoltage of 1V for y if the condition is true
0 is a voltage of 0V for z if the condition is false.
You may need to swap y and z to match the switch model in your circuit.

If you are stuck, post the .asc file so someone (not necessarily me) can inspect it.

 

[Sukhdeep]

I made the switch work following Your suggestion..thanks

Can you please explain the Circuit Attached which is half wave rectifier but With Inductor in. The plot shows greenutput across loda and red: Input voltage.
Can you please explain the ripple kind of formation in the green plot? I hope its visible.

 

[Harald Kapp]

The inductor here is of no use.
The "ripple" is called ringing. It is a damped oscillation due to the inductance and parasitic capacitances of the circuit (e.g. in the diode) which form a resonant circuit. This resonant circuit is "triggered" by the turn-off of the diode. If you add a capacitor from the connection of D1 and l1 to GND, e.g. 100nF, you will see a much stronger version of this oscillation.

 

[Sukhdeep]

Ok...so that means LC forms a ringing type behavior but RC dont?? Am i correct? So i Guess that the ringing mainly comes because of Inductor in the Circuit.

Thanks a lot whosoever you are. You are helping a lot to make me feel comfortable about Electronics which i always wanted to learn.

 

[Sukhdeep]

Can you please have a look on the Attached file and explain that why diode is conducting here during the negative half of the voltage cycle?

 

[Harald Kapp]

Ringing does not come from an inductor alone nor from a capacitor alone. You need an LC-circuit for oscillation to occur with passive components.
A resistor will not add to oscillation, on the contrary, it will dissipate energy and therby will damp the oscillation of an LC circuit.
In your circuit there are "hidden" capacitances e.g. in the diode, in the windings of the inductor (yes, there is capacitance and resistance in an inductor, too) and even in the resistor (although these may not be modeled for this simulation).

 

[Harald Kapp]

Look at the volate at the connection of D1 and L1. Watch it go negative when the diode turns off. That's because the inductor tries to force the current from the on phase of the diode to flow, see expalantion at the beginning of this thread.

You'll learn more if you do not just simulate circuit by circuit and ask why it behaves as it does. Think first what a circuit is expected to do,design the circuit accordingly, then simulate.

Also be aware that a simulation is only a reflection of the real world. If you build the same circuit on a breadboard you may find that it exhibits evn other, unexpected behavior. A simulation can only model the real world to a certain degree (with sometimes astonishing results), but a simulation is not a tool to design a circuit. You design the circuit using your skills. A simulation is a good tool to verify your design and to look for possible errors.