# The waveform after replacing 2 diodes with 2 resistor in bridge rectifier

Thread Starter

#### xso111

Joined Oct 12, 2017
24
in our laboratory class we made a simple bridge rectifier and we replaced 2 diodes with resistors and when we placed the probes of the oscilliscope in the load resistor it produced postive DC voltage but every other cycle is smaller to the previous one, and my professor tasked us to figure out why did this happen? can you guys explain it to me what happens in the positive/negative cycle that would result to this waveform?

this is what we made in the lab

* sorry missclicked the post thread before i was able to finish writing the title*

Moderators note: removed unnessary white space from picture

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#### MrChips

Joined Oct 2, 2009
31,089
What do you want for a title?

Thread Starter

#### xso111

Joined Oct 12, 2017
24
What do you want for a title?
"The waveform after replacing 2 diodes with 2 resistor in bridge rectifier "

ty

#### bertus

Joined Apr 5, 2008
22,305
Hello,

Is the title now OK?

Bertus

PS every member of the staff can change it for you

Thread Starter

#### xso111

Joined Oct 12, 2017
24
Hello,

Is the title now OK?

Bertus

PS every member of the staff can change it for you
yes, thank you vry mch

#### crutschow

Joined Mar 14, 2008
34,845
We don't do your homework for you here (unless you'd like us to get the credit from your professor ), so you need to figure out what is happening.

First figure out how the bridge works with 4 diodes [also plotting the voltage across V1 at the same time, (not V1 to ground) may help].
Determine the current path through the diodes for each half-cycle.
If you understand that then it should be simple to determine what happens when two of the diodes are removed, and then with resistors replacing the diodes.

#### WBahn

Joined Mar 31, 2012
30,303
The idea of he professor asking you the question wasn't for you to go find someone to tell you the answer -- if that had been the case the professor would have just told you the answer. The idea was for you to THINK about what is happing and figure out what the answer is. You will learn a LOT more that way, even though it is more difficult.

So walk through how you THINK the circuit is going to behave in as much detail as you can. If you start going astray we can probably spot it and help get you back on track. There's a good chance you will, in fact, figure it all out by yourself just by trying to clearly explain your thoughts.

#### shteii01

Joined Feb 19, 2010
4,644
normal textbook would have explanation for half wave rectifier, followed by the explanation of full wave rectifier. normal people read the textbook. then you can look at your circuit and have at least a clue of what is going on. then you post here what you think is happening and ask for help.

#### rherber1

Joined Jan 6, 2008
27
normal textbook would have explanation for half wave rectifier, followed by the explanation of full wave rectifier. normal people read the textbook. then you can look at your circuit and have at least a clue of what is going on. then you post here what you think is happening and ask for help.
Hint: consider the half wave where the 2 x 2k resistors are carrying the load current.

Thread Starter

#### xso111

Joined Oct 12, 2017
24
We don't do your homework for you here (unless you'd like us to get the credit from your professor ), so you need to figure out what is happening.

First figure out how the bridge works with 4 diodes [also plotting the voltage across V1 at the same time, (not V1 to ground) may help].
Determine the current path through the diodes for each half-cycle.
If you understand that then it should be simple to determine what happens when two of the diodes are removed, and then with resistors replacing the diodes.
if its alright i'd like to confirm my answer.

in the positive half cycle the current flows in a series of resistor hence the peak voltage in the lower

in the negative half cycle alot more current is flowing in the diode instead of the resistor because the forward resistance of a forward biased rectifier diode is much much lower than the 2.2k resistance in the other path. because the 2.2k resistor and the diode are in parallel they share the same voltage and when we use ohm's law the other is divided by 2.2k while the other is divided by a near 0 resistance the current will be alot higher in the diode and the current that will pass in the load resistor wouldn't have passed from the 2 resistors hence the voltage in the load is about the same.

thanks

#### dl324

Joined Mar 30, 2015
17,145
It would be helpful if you used the component designators so we know which diodes and resistors you're referring to.

For the positive half cycle, why does the current only go through resistors? Which resistors?

For the negative half cycle what path does the current take?

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