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Can someone please explain why the current at the output is considered to a pulse??
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.
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 !
Your description is correct for a technical current source which needs a positive voltage to "source" (actually sink) current.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?
I will refer to circuit 5-6 b) and hence, graphs b).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
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.
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?
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.
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.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
NO. Average value is as you said but integrating over 1 period and multiply by 1/T.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
They did not give you some explanation on the text book ?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 ?)
Half wave rectifier with Voltage source (Vs) square wave and Ls=0, makes Vd(avg) = Vs/2 (if ideal diodes)If the circuit was a half wave rectifier with the same source how would Vd (avg) change?
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