Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Predicting maximum current flow of diode in half/full-wave rectifier

Status
Not open for further replies.

t0mbst0n3

Newbie level 3
Joined
Aug 12, 2018
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
25
Diode forward voltage drop = 0.6 V
R3 = 18 kΩ
Vs = 10sin(2π*10000t) V

Half wave rectifier circuit diagram:

9DRM5Bt.png


Full wave rectifier circuit diagram:

W1HbYmI.png


How do you predict the maximum current flow for the D1 diode for both scenarios? I'm stuck at this part because the Vs value has an unknown variable.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
21,077
Helped
4,530
Reputation
9,072
Reaction score
4,624
Trophy points
1,393
Activity points
139,098
Hi,

What do you mean with unpredictable variable?
In any case you know it's a sinwave and you know it's amplitude.
Thus the max voltage is clear.

And because there is an ohmic load .... just use Ohm's law.

Klaus
 

t0mbst0n3

Newbie level 3
Joined
Aug 12, 2018
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
25
Hi,

What do you mean with unpredictable variable?
In any case you know it's a sinwave and you know it's amplitude.
Thus the max voltage is clear.

And because there is an ohmic load .... just use Ohm's law.

Klaus

Sorry, I don't get you at the "it's a sinwave and you know its amplitude." Amplitude is 10, so therefore maximum voltage is 10? Then what is the point of the Vs equation?
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
21,077
Helped
4,530
Reputation
9,072
Reaction score
4,624
Trophy points
1,393
Activity points
139,098
Hi,

Amplitude is 10, so therefore maximum voltage is 10?
Correct. This is true for every pure sinewave.

Then what is the point of the Vs equation?
That´s not the question.
But maybe it´s just to check if you recognize that the max voltage is 10V.

Klaus
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
49,394
Helped
14,437
Reputation
29,138
Reaction score
13,227
Trophy points
1,393
Location
Bochum, Germany
Activity points
284,279
How do you predict the maximum current flow for the D1 diode for both scenarios?
Diode current is equal to I(R3) = Vo/R3. You want to calculate Vo, which is Vs - diode voltage drop (0.6 or 1.2 V respectively).

There's no D1 in circuit 2, b.t.w. Apparently you mean a different diode.
 

t0mbst0n3

Newbie level 3
Joined
Aug 12, 2018
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
25
Diode current is equal to I(R3) = Vo/R3. You want to calculate Vo, which is Vs - diode voltage drop (0.6 or 1.2 V respectively).

There's no D1 in circuit 2, b.t.w. Apparently you mean a different diode.

Alright, understood.

For circuit 1, Vo = Vs - 0.6 because there is only one diode, yes? Just want to confirm.

For circuit 2, the diode current must be found for either D3/D4/D5/D6, since all four diodes share the same forward voltage drop = 0.6 V. So based on that logic, would the diode current be equal to I(R3,4,5,6) = Vo/R3, where Vo = Vs - 0.6 - 0.6 ? I assume this to be the case because there out of four diodes, two diodes are in series with each other with the other two diodes being in series with each other.
 
Last edited:

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
49,394
Helped
14,437
Reputation
29,138
Reaction score
13,227
Trophy points
1,393
Location
Bochum, Germany
Activity points
284,279
Voltage drop of the bridge rectifier is always 1.2V.
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
21,077
Helped
4,530
Reputation
9,072
Reaction score
4,624
Trophy points
1,393
Activity points
139,098
Hi,

I assume this to be the case because there out of four diodes, two diodes are in series with each other with the other two diodes being in series with each other.

Just to clarify: In an usual bridge recitifier circuit:
* for the positive halfwave diodes right_top and left_bottom are active and thus these two are connected in series.
* for the negative halfwave diodes left_top and right_bottom are active and thus these two are connected in series.
... although these diodes don´t have direct connection at the bridge rectifier circuit.

Klaus
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top