Passive full-wave bridge rectifier V_output > V_input ? - Page 1
Forum section: electronics / beginners

author: raff5184
author_rank: Regular Contributor
timestamp: February 19, 2018, 09:06:56 pm
content: Hi all,I realized a full-wave bridge rectifier with BAT54 diodes. My input is a 700kHz sinusoid 10V peak-to-peak, while the output open circuit is a DC voltage higher than 5V (that is more than 10/2 V) which is not expected.Where am I losing power? Is it because the input is too high?I also did a simulation in SPICE and the DC value is actually 4.8V as expectedFull-wave bridge rectifier:https://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/rectifier-circuits/BAT54 datasheet:https://www.diodes.com/assets/Datasheets/ds11005.pdf

author: newbrain
author_rank: Super Contributor
timestamp: February 19, 2018, 11:44:32 pm
content: Of course, the output voltage is supposed to be lower than the peak Vin...But:How are you measuring these voltages?Scope or DMM? Or something else?If DMM, consider that 700kHz might be well outside the bandwidth of the meter, so quite attenuated.If you are using a scope (I see you mention a 4ch 200MHz Keysight in another post) could you post screenshots of both input and output?If something else, please describe your set up!

author: IanB
author_rank: Super Contributor
timestamp: February 20, 2018, 12:24:47 am
content: Does it make a difference if you put a small load on the output of the rectifier like a 1 k resistor? There are two reasons for this: firstly that diodes need to be conducting some current to switch properly; and secondly that you should measure the output voltage across a reasonably low impedance source to avoid phantom voltages upsetting the measurement.

author: raff5184
author_rank: Regular Contributor
timestamp: February 20, 2018, 12:37:25 am
content: Quote from: newbrain on February 19, 2018, 11:44:32 pmAC voltages with a scope, DC voltages I tried both with a DSO and with a DMM, I get the same values, againacross an open circuit

author: raff5184
author_rank: Regular Contributor
timestamp: February 20, 2018, 12:40:27 am
content: Quote from: IanB on February 20, 2018, 12:24:47 amI tried with a 10 ohm resistor load and indeed the DC voltage across it becomes smaller of a couple of orders of magnitude with respect to the o.c. voltage.

author: IanB
author_rank: Super Contributor
timestamp: February 20, 2018, 01:01:23 am
content: Quote from: raff5184 on February 20, 2018, 12:40:27 amLol. I said asmallload, not a huge one

author: raff5184
author_rank: Regular Contributor
timestamp: February 20, 2018, 01:17:30 am
content: Quote from: IanB on February 20, 2018, 01:01:23 amah, you mean a large resistor that absorbs small current

author: IanB
author_rank: Super Contributor
timestamp: February 20, 2018, 01:27:15 am
content: Quote from: raff5184 on February 20, 2018, 01:17:30 amYes, like the example I gave of about 1000 ohms. With a 5 V source voltage that should only draw a few milliamps.

author: IanB
author_rank: Super Contributor
timestamp: February 20, 2018, 01:33:43 am
content: Quote from: blueskull on February 20, 2018, 01:29:09 amWell, quite. The OP is starting with a 10 V p/p AC signal and is expecting to get a rectified DC signal of a bit less than 5 V peak. The question is why might the rectified output be more than 5 V?

author: danadak
author_rank: Super Contributor
timestamp: February 20, 2018, 02:50:42 am
content: Begs the question, what V value are you measuring ? We are assumingthat you have a relatively large value cap on the output of the bridge,so that AC is filtered out.The 10V sinewave is +5V max to -5V min, so if there is no load on theCap/Bridge output then over a few sine cycles the cap charges to ~ 5V -1 diode drop, or somewhere around 4.3 - 4.8 V. Thats if cap has lowleakage, and diode is nominally .7V when conducting. All this ignoresstray C coupling across diode, and other parasitics.If using DVM, when you measure 10V sine, thats probably the average, or RMS,unless it specifically has a peak reading capability. Compound this with thefrequency response of the meter probably only around 50 - 100 Khz, so good guessyou have >> 10V sine. Oscilloscope will give you better chance reading peak V.Regards, Dana.

author: raff5184
author_rank: Regular Contributor
timestamp: February 23, 2018, 03:54:27 am
content: Hi again,so I did some measurements using different load values. And I always measure rectified voltages higher than the AC RMS values so that the efficiency of the rectifier would always be greater than 100%I'm not sure what I am doing wrong, because I measured the voltages with different DSO scopes and multimeters and I have the same values

author: IanB
author_rank: Super Contributor
timestamp: February 23, 2018, 04:00:29 am
content: Quote from: raff5184 on February 23, 2018, 03:54:27 amYou may be doing nothing wrong. If you rectify a 10 V RMS sine wave and store the rectified output in a smoothing capacitor you are supposed to see a DC voltage of about 14 V.Please give more information about exactly what your experiment looks like, what your parameters are, and what you are measuring. Unless you can give us that information we cannot help to explain what you are seeing.

author: raff5184
author_rank: Regular Contributor
timestamp: February 23, 2018, 02:29:00 pm
content: Quote from: blueskull on February 23, 2018, 03:58:56 amSure, and thank you!The wave on the scope is the AC input and the multimeter is the rectified DC

author: raff5184
author_rank: Regular Contributor
timestamp: February 23, 2018, 02:35:49 pm
content: Quote from: IanB on February 23, 2018, 04:00:29 amI only have a 0.01 uF smoothing capacitor, would it be enough to increase the voltage from 10 V RMS to 14 V DC?Like I said the input is a 10 V peak-to-peak sine wave at 700kHz. I measure this wave (correctly measured with a digital oscilloscope).Than I measure the output of the rectifier, that is the voltage drop across the smoothing capacitor (0.01uF) in parallel with a resistor. If I change the value of this resistor the input AC voltage drops as shown in the plots I attached a couple of posts above, however the DC voltage across the load still doesn't match the theoryThank you for your help