Half wave rectifier calculation question

Thread Starter

avjg123

Joined Apr 3, 2018
4
Hey, guys. I've done an experiment on that the subject above last weak and I'm having trouble in the calculation. So, I've made a circuit. And the inputs were 10V sine wave with 60Hz frequency. Used a silicon diode and a 1k Ohm resistor. I've got a voltage output as 2.6. Now, I've got to calculate the error percentage between the theoretical value and 2.6 but I don't know what the theoretical value is. I searched up on the Internet and found it is either (10*rt2-0.7)/pi or (10*rt2-0.7)/2pi. Which one's right? Or if these are both wrong could you tell me what the correct equation is?
 

wayneh

Joined Sep 9, 2010
17,166
...the inputs were 10V sine wave...
Be careful here. What does the 10V refer to? With no further information, we would normally assume that means the RMS voltage. Our wall socket is 120V AC. But in the lab where you can see the peaks at ±10V, it would be easy to call that a 10V sine wave. Obviously the results are much different in these two cases.

As for your formula, you'd be better off to calculate it yourself than to rely on guessing which internet version is appropriate.
 

dl324

Joined Mar 30, 2015
14,313
Yup. It' homework.
For homework, you need to show your work so we can try to guide you to a solution.

I assume the output voltage is across the 1K resistor.

Is the sine wave source ideal (i.e. no source resistance)? Does 10V sine wave mean peak to peak?
 

Thread Starter

avjg123

Joined Apr 3, 2018
4
Be careful here. What does the 10V refer to? With no further information, we would normally assume that means the RMS voltage. Our wall socket is 120V AC. But in the lab where you can see the peaks at ±10V, it would be easy to call that a 10V sine wave. Obviously the results are much different in these two cases.

As for your formula, you'd be better off to calculate it yourself than to rely on guessing which internet version is appropriate.
So is it like you're trying to say both of the equations are correct but in different situations? If this is the case could you tell me what the differences, of meaning, are between the two equations? If I've misunderstood..sorry I haven't used English for about 5 years..
 

Thread Starter

avjg123

Joined Apr 3, 2018
4
For homework, you need to show your work so we can try to guide you to a solution.

I assume the output voltage is across the 1K resistor.

Is the sine wave source ideal (i.e. no source resistance)? Does 10V sine wave mean peak to peak?
Yes the output voltage is across the 1k resistor.
The sine wave source is ideal and when calculating everything's going to be put as ideal except the diode. Whatever the input, the diode is going to have a 0.7V threshold voltage.
The 10V means the amplitude. And umm..about my work..Like what kind of things am I supposed to show?
 

crutschow

Joined Mar 14, 2008
29,489
Used a silicon diode and a 1k Ohm resistor. I've got a voltage output as 2.6.
How did you measure that voltage?
The 10V means the amplitude.
Peak or RMS?
And umm..about my work..Like what kind of things am I supposed to show?
How you might calculate the voltage, besides a random search on the internet. ;)
It other words you need to think about the problem, not just look for an answer somewhere.
 
Last edited:

MrAl

Joined Jun 17, 2014
8,866
Hey, guys. I've done an experiment on that the subject above last weak and I'm having trouble in the calculation. So, I've made a circuit. And the inputs were 10V sine wave with 60Hz frequency. Used a silicon diode and a 1k Ohm resistor. I've got a voltage output as 2.6. Now, I've got to calculate the error percentage between the theoretical value and 2.6 but I don't know what the theoretical value is. I searched up on the Internet and found it is either (10*rt2-0.7)/pi or (10*rt2-0.7)/2pi. Which one's right? Or if these are both wrong could you tell me what the correct equation is?
Hello,

What do you mean by:
"I've got a voltage output as 2.6"

Where did you get that?
 

wayneh

Joined Sep 9, 2010
17,166
So is it like you're trying to say both of the equations are correct but in different situations? If this is the case could you tell me what the differences, of meaning, are between the two equations? If I've misunderstood..sorry I haven't used English for about 5 years..
I'm not saying either of those equations is correct (or incorrect when used properly). I'm suggesting you do the integration of half a sine wave yourself and think carefully about the limits of the integration.

Note that the time-average voltage you calculate from the integral is not necessarily what you would predict your meter to report. The details of how the meter smooths out the waveform is a mystery unless it's documented. I guess that's part of the point of your experiment.
 

MrChips

Joined Oct 2, 2009
25,920
Be very careful with the wording and definition.

When one says, "10V sine wave", that ought to mean 10V rms. In other words, the amplitude is 14.14V and peak-to-peak voltage is 28.28V.

When one says, "sine wave of 10V amplitude", that ought to mean 20V peak-to-peak, or 7.07V rms.
 

MrAl

Joined Jun 17, 2014
8,866
Hi,

Yes as you guys are indicating too, we need more information to solve this effectively .

I took the 10v to mean 10vrms because he multiplied by sqrt(2), but i'd like verification of that also.
As well as what that "2.6" was, where that came from. We dont know that yet either.
 

MisterBill2

Joined Jan 23, 2018
11,582
OK, now I see that it is indeed a very old thread. I would suggest that first, the dates of posts be posted at least as clearly as the date that the poster joined. And second, how do these old posts suddenly appear at the new posts position? This has happened to me several times, and it is not like I am dumb or something. It was listed right up with some new posts, at the point where I wind up when I get to the forum, "general electronics chat".
 

djsfantasi

Joined Apr 11, 2010
8,256
OK, now I see that it is indeed a very old thread. I would suggest that first, the dates of posts be posted at least as clearly as the date that the poster joined. And second, how do these old posts suddenly appear at the new posts position? This has happened to me several times, and it is not like I am dumb or something. It was listed right up with some new posts, at the point where I wind up when I get to the forum, "general electronics chat".
They appear as new posts when the TS finds the old post, either through a search or a link from a recent post. And then proceeds to necropost!

I’ve often thought that all posts should be auto-locked after some constant period expires after the last post.
 

MisterBill2

Joined Jan 23, 2018
11,582
They appear as new posts when the TS finds the old post, either through a search or a link from a recent post. And then proceeds to necropost!

I’ve often thought that all posts should be auto-locked after some constant period expires after the last post.
NO, not auto-locked, just adequately identified as to the last posting date, possibly with a larger font. Auto-locking is DUMB, since it may be that somebody needs to see an older thread to learn something.
 

djsfantasi

Joined Apr 11, 2010
8,256
NO, not auto-locked, just adequately identified as to the last posting date, possibly with a larger font. Auto-locking is DUMB, since it may be that somebody needs to see an older thread to learn something.
Maybe I have a semantical problem. Auto-locking as I perceive it does NOT prevent anyone from seeing an older thread nor does it prevent anyone from learning anything.

Auto-locking places the thread in a read-only state. It “locks” that thread from further responses. It can still be linked to, if a member has a follow up question. The follow up can start a new contemporaneous dialogue.

But since it is a separate follow up thread, responses are more relevant.

That is all.
 

MrAl

Joined Jun 17, 2014
8,866
You know what is ironic?
It looks like neither of the two possible answers given in the first post are right.
The average DC (if it is that) comes out higher than 2.6 volts.
Maybe this will be beneficial to someone else.
 
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