# Bridge Rectifier Output ??

Discussion in 'General Electronics Discussion' started by mjc, Feb 2, 2014.

1. ### mjc

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Jan 30, 2014
Hi, Just un-clear why this happens! Have a 13vac Supply, after it passes through a Bridge Rectifier (4 diodes.- 13.0-approx.1.4v diode drop=11.6vdc) the voltage is 11.79VDC . Using the same Bridge rectifier but adding- Two 250uf, 50vdc capacitors in parallel to filter the DC output the 11.79vdc changes at the capacitors to 17.68vdc. My question: Why does this happen? and applying this voltage to a circuit would I use the 11.79 or 17.68vdc output for my calculations? Thanks in advance

2. ### duke37

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769
Jan 9, 2011
!3V AC means that the varying voltage is equivalent to 13V DC for heating say a resistor. Sometimes the voltage will be higher than 13V and sometimes lower. The voltage is calculated as the root mean square (RMS) of the waveform.

The average voltage will be 90% of the RMS voltage and this is what your meter measures, then multiplies up to get the RMS value.

The peak voltage is Vrms * sqrt(2) so 13 * 1.414 = 18.4. Knock off a couple of volts for the rectifier gives 16.4. The capacitor holds the peak voltage. I imagine that this is lower than 17.7 that you measure is because the rectifier drop will be less than what you assume if it is only supplying the minicule meter current.

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,401
2,777
Jan 21, 2010
13 volts AC is 13V RMS AC. The peak voltage is 13 * sqrt(2) = 13 * 1.414 = 18.382

You're seeing 0.7 volts less than this because the diodes drop some voltage (in this case about 0.35V each -- this will vary with current).

The output capacitors charge to the peak voltage, without them you see an average voltage. Depending on your meter you might see different figures, and you're seeing a value a little lower than the RMS voltage less the diode drop.

When you draw power from the rectified and filtered source (the 18.79V one), the capacitor will discharge between the peaks of the AC waveform creating what is called ripple and reducing the average output voltage.

edit: and of course at more than minuscule currents the diode drop will be around 0.7 volts, rising above 1V at higher currents. Because a bridge rectifier effectively has two of these in series at any point in time, the voltage drop that was measured as 0.7V in this case will rise to at least 1.4V and maybe above 2V.

Last edited: Feb 2, 2014
4. ### mjc

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Jan 30, 2014
Thanks guys , Ican see / understand Why this is happening , is it ok to use the output of 17.68vdc (after the capacitors) for the circuit and its calculations? I'm thinking Yes but just want to be sure by asking.

5. ### duke37

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Jan 9, 2011
The voltage will not be stable and will vary with the load. For calculations you will need to use a high voltage if the load is low and a lower voltage when the load is high.

Voltage regulators are used after the capacitors to give a constant voltage for voltage sensitive circuits.

6. ### mjc

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Jan 30, 2014
Hi guys, tried a voltage regulator (LM317 including resistance) to see the output of the circuit by using a breadboard. There is an approximate 7-8v drop at the output. Leaving only 8.5v to run the Led circuitry. Guess I’m just going to use the 17v output right off the filter capacitors cause the load is low (about 160ma total) and let resistors at the LED’s limit the voltage. You may want to read my previous post Parallel Led’s under Circuit Help in this forum to grasp what my circuitry is trying to do. Any other suggestions are welcome! Thanks

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,401
2,777
Jan 21, 2010
That sounds like you didn't calculate the resistors correctly or you wired it up wrong.

What was the input voltage to the regulator when your output voltage fell to 8.5V?

8. ### mjc

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Jan 30, 2014
Checked wiring and resistor values all is well. With-out the LM317 the output after the capacitors is 17.3v . With the LM317 in the circuit the output after the capacitors (to the LM317) is 9.7v at Vin of the LM317, and 8.2v at Vout. reflecting a 1.5v drop thru the regulator but a 7-9v drop at the final output.

9. ### duke37

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769
Jan 9, 2011
You have something very wrong if the rectified voltage is so low. you are very likely to have something fail if run like this.

Show us a circuit diagram of what you have made so we can comment.
You do not need a controlled voltage to drive leds unless you need a controlled light output.

10. ### mjc

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Jan 30, 2014
Hi Duke 37, I will figure out how to post the circuit diagram so you could comment:
To answer your question; I do not need a controlled light output so I won’t need to use a controlled voltage to run my LEDs the LM317 is out. Here is my problem: I am using an AC input of 13vac and rectifying it to a DC output. The rectified output is 11.7vdc without using filter capacitors which is about right considering the voltage drop of the diode rectifier. But after installing the filter capacitors (parallel 250UF caps) the output becomes 17.6 vdc (5.9vdc higher). Which voltage should I use for my calculation since it would make a difference in the series resistor sizing per string of 3 LEDs? at17.6vdc use 470ohm resistors, at 11.7vdc use 120 ohm resistors -- Im confused. Hope you can help. Scanned the circuit drawing and will figure out how to send it to this thread.
Thanks’ Mike

Last edited: Mar 3, 2014
11. ### mjc

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Jan 30, 2014
View attachment LED0001.pdf

I think I attached the scanned file in a pdf format .

Question 2- The attached circuit of 27 LEDs in parallel with each other will work according to an LED wizard from the internet. The way I need to wire them is in sections as shown on the attachment, these sections still seem to be a parallel arrangement , I feel pretty confident that this arrangement will work but what do you think ? (each section is to be a lighted deck post cap using a 12awg wire to supply them) Thanks’ Mike

12. ### mjc

13
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Jan 30, 2014
View attachment LED0001.pdf The attached circuit of 27 LEDs in parallel with each other will work according to an LED wizard from the internet. The way I need to wire them is in sections as shown on the attachment these sections still seem to be a parallel arrangement , I feel pretty confident that this arrangement will work but what do you think ? (each section is to be a lighted deck post cap using a 12awg wire between them)

Think I figured out how to send a scanned attchment of my circuit diagram its a pdf.

13. ### duke37

5,364
769
Jan 9, 2011
That circuit looks good but you have desribed it wrongly. The LEDs are not in parallel, it is the strings which are in parallel.

As to the voltage to be used in the resistance calculation, use what you have. I would try the rectified voltage without smoothing as it is simpler, cheaper and smaller. The light output will be pulsed but this may not disturb you. If you take an LED digital clock and wave it about you will see that the display is pulsed.