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[SOLVED] Diode Bridge Rectifier Circuit

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wfg42438

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Hello ,

Can someone please explain why the current at the output is considered to a pulse??

I don't understand how this the case if out input voltage is sinusoidal, or is the pulse obtained not necessarily the output?

Please refer to the document and the schematic attached


https://user.unob.cz/leuchter/Mohan_usmernovace.pdf

Capture.PNG11Capture.PNG
 

Diagram B appears to have a current source for the load. Therefore it draws constant current from the diode bridge. (Most power supplies have a smoothing capacitor, and draw bursts of current instead of constant current.)

So the diagram is theoretically correct, although it shows a power supply that can provide enormous current near 0V crossings. Real power supplies are not usually capable of that.
 

wfg42438 said:
Can someone please explain why the current at the output is considered to a pulse??
Click to expand...

Anything that is changing, vibrating is considered a pulse. In this context, it means a ripple plus a DC. Pulse is notionally related to impulse: a large force that acts for a short time. In this context, pulses will have no negative component.
 

"Pulse" refers to the square wave current waveform in case b, I presume.

It has been already explained that the waveform is cause by connecting a current source as load, which is more a theoretical model than a real load case. You can however come quite near to it by using a large inductor with series resistance as load.
 

In figure 5-6 a) is a simple resistance as load, then the output wave forms are sine waves rectified.
In figure 5-6 b) as load is a DC current source (I was confused at the beginning because they used notation from AC). If there is a DC current load, then the current drawn from the AC source is square wave just as shown.
 

CataM said:
In figure 5-6 a) is a simple resistance as load, then the output wave forms are sine waves rectified.
In figure 5-6 b) as load is a DC current source (I was confused at the beginning because they used notation from AC). If there is a DC current load, then the current drawn from the AC source is square wave just as shown.
Click to expand...

Hello Everyone, First i would like to thank you for all of your feedback. It has helped me form a hypothesis on how this circuit is behaving as shown below:

CataM,

Let me see if i understand correctly, in circuit b the resistive load has been replaced with a current source. My assumption is that this current source load acts as follows:

As Vs becomes positive ( after wt=0 where Vs=0V ) the current source begins deliver current. Here as Vs increases the current being delivered by the source approaches some intended max current which is reached when Vs reaches some threshold voltage. Shortly after this point it begins to deliver a constant current . However as Vs decreases there will be a point where Vs < V_threshold and some wt after the current source begin to deliver a smaller current and will fall from its constant value it to 0A. This is then repeated for the negative cylcle of VS

Is this the case and if not can you please correct me?

Thank you in advance!!
 

wfg42438 said:
Let me see if i understand correctly, in circuit b the resistive load has been replaced with a current source. My assumption is that this current source load acts as follows:

As Vs becomes positive ( after wt=0 where Vs=0V ) the current source begins deliver current. Here as Vs increases the current being delivered by the source approaches some intended max current which is reached when Vs reaches some threshold voltage. Shortly after this point it begins to deliver a constant current . However as Vs decreases there will be a point where Vs < V_threshold and some wt after the current source begin to deliver a smaller current and will fall from its constant value it to 0A. This is then repeated for the negative cylcle of VS !
Click to expand...

Now you have confused me very successfully. I do not understand what you are saying. Current sources and voltage sources are complementary devices. Both need a load to deliver. Of course you can have lots of fun if you have both a current source (that has infinite impedance; has infinite compliance voltage) and a voltage source (that has zero impedance; can deliver infinite current) in the same circuit.

If the sine wave is having considerable amplitude (>20V or so) we can ignore the diode drop. Otherwise (if you are rectifying a 5V AC) you need to chop off 2 diode drops (for a bridge rectifier).
 

As Vs becomes positive ( after wt=0 where Vs=0V ) the current source begins deliver current. Here as Vs increases the current being delivered by the source approaches some intended max current which is reached when Vs reaches some threshold voltage. Shortly after this point it begins to deliver a constant current . However as Vs decreases there will be a point where Vs < V_threshold and some wt after the current source begin to deliver a smaller current and will fall from its constant value it to 0A. This is then repeated for the negative cylcle of VS

Is this the case and if not can you please correct me?
Click to expand...
Your description is correct for a technical current source which needs a positive voltage to "source" (actually sink) current.

If we perform a simulation we an ideal current source, the current will flow independent of the applied voltage, even if the voltage is negative and the current source delivers power instead of consuming it. Respectively there will be no threshold voltage and the rectifier input current jumps from +id to -id in no time.

Of course, this ideal current source component doesn't exist in a real world. But as mentioned before, an inductive load can achieve a similar behavior during rectifier commutation, although the current isn't exactly constant.
 

wfg42438 said:
Let me see if i understand correctly, in circuit b the resistive load has been replaced with a current source. My assumption is that this current source load acts as follows:

As Vs becomes positive ( after wt=0 where Vs=0V ) the current source begins deliver current. Here as Vs increases the current being delivered by the source approaches some intended max current which is reached when Vs reaches some threshold voltage. Shortly after this point it begins to deliver a constant current . However as Vs decreases there will be a point where Vs < V_threshold and some wt after the current source begin to deliver a smaller current and will fall from its constant value it to 0A. This is then repeated for the negative cylcle of VS
Click to expand...
I will refer to circuit 5-6 b) and hence, graphs b).

"id" is always constant, which means, it is a DC Current source. You are thinking like it is an AC source. It's value can not change, always will be a constant flow of current from it. I also was confused when first saw the circuit because the notation "id" is used for AC sources (both letters in small). You can also see that in the graph. Whereas "is" has a shape of square wave (see graph) because it is not a current source, but a voltage source with the purpose to maintain sine wave voltage. This is what FvM said too.
 
CataM said:
I will refer to circuit 5-6 b) and hence, graphs b).

"id" is always constant, which means, it is a DC Current source. You are thinking like it is an AC source. It's value can not change, always will be a constant flow of current from it. I also was confused when first saw the circuit because the notation "id" is used for AC sources (both letters in small). You can also see that in the graph. Whereas "is" has a shape of square wave (see graph) because it is not a current source, but a voltage source with the purpose to maintain sine wave voltage. This is what FvM said too.
Click to expand...

Thank you for pointing out that i had confused the current source to be an AC source. I know understand the load is a DC current source

Now you mentioned earlier "If there is a DC current load, then the current drawn from the AC source is square wave just as shown."

Can you please explain why the the sinusoidal voltage source delivers current in the form of a square wave?
 

wfg42438 said:
Now you mentioned earlier "If there is a DC current load, then the current drawn from the AC source is square wave just as shown."

Can you please explain why the the sinusoidal voltage source delivers current in the form of a square wave?
Click to expand...

If the electric potential is higher at anode than cathode => DIODE ON
Diodes are ideal ones because the author assumed that.

PART 1:



During this time, voltage of the sine wave is bigger at terminal "A" than terminal "B", meaning:
D1,D2 --> ON (Ideal Diode ON = Short circuit) ; D3,D4 --> OFF (OFF = open circuit).

Current from the AC source goes from positive ("A") to negative ("B") meaning that has the same direction as the reference "is" and is equal to DC Current "load", "id". (How current flows, is represented with red).

PART 2:



This is the same as part 1 with the difference that potential at terminal "B" is higher than potential at terminal "A". So now the current goes from "B" to "A" following the path drawn in red and is equal to the DC current "load", "id".

The reference current is "is" in the direction drawn by author, but now the current in the circuit (red line) has opposite direction resulting in the graph to have negative value.

Of course now in part 2, D3 and D4 ON and rest OFF.

CONCLUSION: part 1 --> current drawn in red has same direction as reference "is". Positive in graph.

part 2--> current drawn in red has opposite direction from "is". Negative in graph.

TOTAL GRAPH --> current is a square wave.
 
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CataM said:
If the electric potential is higher at anode than cathode => DIODE ON
Diodes are ideal ones because the author assumed that.

PART 1:



During this time, voltage of the sine wave is bigger at terminal "A" than terminal "B", meaning:
D1,D2 --> ON (Ideal Diode ON = Short circuit) ; D3,D4 --> OFF (OFF = open circuit).

Current from the AC source goes from positive ("A") to negative ("B") meaning that has the same direction as the reference "is" and is equal to DC Current "load", "id". (How current flows, is represented with red).

PART 2:



This is the same as part 1 with the difference that potential at terminal "B" is higher than potential at terminal "A". So now the current goes from "B" to "A" following the path drawn in red and is equal to the DC current "load", "id".

The reference current is "is" in the direction drawn by author, but now the current in the circuit (red line) has opposite direction resulting in the graph to have negative value.

Of course now in part 2, D3 and D4 ON and rest OFF.

CONCLUSION: part 1 --> current drawn in red has same direction as reference "is". Positive in graph.

part 2--> current drawn in red has opposite direction from "is". Negative in graph.

TOTAL GRAPH --> current is a square wave.
Click to expand...

CataM,

Thank you so much for the detailed description, it has helped clear up quite a few things

Now the only question that comes to mind is if a circuit with an AC Voltage source (such as the one in this discussion) has a DC current source as a load would that then result in a current delivered clipping (turning the sine wave to square wave ) at a value of "Id" ?
 

If it is a different circuit than the full wave rectifier, the current will not result in a square wave in general.. maybe there are circuits that does that..
 

Thank you for all of the help,

Let me ask you one last question,

When looking at a question in the text where they seek the current across the diodes (lets call that i_d) i thought i_d would track i_s however i see the text shows that id is zero when Vs is negative

Do you know why this is the case?

I thought i_d would track i_s since when VS is positive D1 and D2 are on (D3 & D4 open) so i_d flows and is positive (so far this agrees with their plot)

Now when Vs is negative both D3 and D4 conduct (D1-D2 open) now i thought we would have a negative current flowing which is why i thought i_d = -i_s

9330091100_1456805554.png


 
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wfg42438 said:
When looking at a question in the text where they seek the current across the diodes (lets call that i_d) i thought i_d would track i_s however i see the text shows that id is zero when Vs is negative

Do you know why this is the case?
Click to expand...
The book is right. You showed in graph i_d across diode D1 and D1 only (in this case is the same as i_d through D2). As told you before in other post, when VS is negative D1 is OFF wich means no current passes through it.

wfg42438 said:
I thought i_d would track i_s since when VS is positive D1 and D2 are on (D3 & D4 open) so i_d flows and is positive (so far this agrees with their plot)

Now when Vs is negative both D3 and D4 conduct (D1-D2 open) now i thought we would have a negative current flowing which is why i thought i_d = -i_s
Click to expand...

Your first part with VS positive is correct. Second part, VS negative, is also correct for the circuit in general, I mean, if you want to know i_s (square wave) but i_s and ID (DC current) are different from i_d which is current through 1 diode.
So in conclusion: do not confuse yourself with i_s, i_d, ID. All have diferent wave forms.

For diode D1 and D2, wave form for their i_d is like you showed but for D3 and D4 is phase shifted one semi period... However, they still have the same Average and RMS values for current.
 
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Since the book did not specify to focus on on D1 what tells us i_diode is the current flowing through D1 and D2 rather than D3-D4?

Is it simply because i_diode would begin to flow after t=0 when Vs is positive suggesting to focus on D1?
 

Post #14 say this: "... current through each diode...."

In this case, RMS and avarage current are the same for D1,D2,D3 and D4, but their shape is slightly different. I think he wants you to "calculate for all" but you can say that is the same. However, if they want wave form for each diode, then draw it.

They want you to calculate for all of them, not just D1.
 
Thank you for clearing that up!!

I ran into one small issue recently i was hoping i can ask about please let me know if you'd prefer i start a new thread

If for this same circuit Vs is a square wave with a source inductance we have commutation losses which leads to a reduced Vd voltage output as there is volt-radian area (Au) lost between the switching of diodes

For some reason i cant seem to recall how to find the average value of Vd of this rectifier

From what i remember to find the avg value of a voltage you would integrate the voltage over half of its period and multiply it by 1/(T/2) please correct me if that's wrong

under this assumption Vd_0( Ls=0) = 1/(pi) * (integral of (Vspeak) from 0 to u) = Vspeak (Based on the text Vd_0=Vspeak)

where Vd= Vd_0 -Au/pi (why is it that to convert this area to a voltage we divide Au by pi instead of 2pi ?)





If the circuit was a half wave rectifier with the same source how would Vd (avg) change?

 

wfg42438 said:
If for this same circuit Vs is a square wave with a source inductance we have commutation losses which leads to a reduced Vd voltage output as there is volt-radian area (Au) lost between the switching of diodes

From what i remember to find the avg value of a voltage you would integrate the voltage over half of its period and multiply it by 1/(T/2) please correct me if that's wrong
Click to expand...
NO. Average value is as you said but integrating over 1 period and multiply by 1/T.

wfg42438 said:
under this assumption Vd_0( Ls=0) = 1/(pi) * (integral of (Vspeak) from 0 to u) = Vspeak (Based on the text Vd_0=Vspeak)

where Vd= Vd_0 -Au/pi (why is it that to convert this area to a voltage we divide Au by pi instead of 2pi ?)
Click to expand...
They did not give you some explanation on the text book ?


wfg42438 said:
If the circuit was a half wave rectifier with the same source how would Vd (avg) change?
Click to expand...
Half wave rectifier with Voltage source (Vs) square wave and Ls=0, makes Vd(avg) = Vs/2 (if ideal diodes)
 
CataM,

With your help and some reading of the text things are very clear!

Thank you so much for the detailed and prompt responses!

Please let me know if there some way for me to rate your answers to help boost your reputation here
 

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