Full-wave bridge rectifier oscope reading sanity check - Page 1
Forum section: electronics / beginners

author: tegowerk
author_rank: Newbie
timestamp: February 10, 2024, 06:45:43 am
content: Hey everyone. This is my first post here, so first of all, thank you for having meI'm playing around with my signal generator (UNI-T UTG932E) and oscilloscope (Owon HDS2102S) as I'm learning electronics. I builtthis simple full-wave bridge rectifier circuitwith these parameters:- Sine wave, 60Hz, 10Vpp.- 1N4001 diodes.- 100k resistor.On paper (if I'm not mistaken) the RMS voltage across the resistor should be(Vpp/2 - 1.4)*0.707 ~= 2.54V, but both my oscope and my DMM (Brymen BM235) are reading around2.2Vwhen connected to the legs of the resistor.My gut feeling is that the discrepancy comes from imperfections in the signal generator, the tolerances of the components (especially in the forward voltage range of the diodes (datasheet here)), inductance from the cables, etc. but I'd appreciate a sanity check to make sure I'm not doing something wrong. I'm attaching a photo of the oscope screen (ignore the low battery warning 😊) and one of the circuit below (it's messy and I obstructed the light when I took the picture, sorry).When connecting the oscope and the DMM across the terminals of the generator, I get the expected RMS, peak-to-peak values, and sine curves.Somebody else suggested the ground probe of the oscope is shorting out one of the diodes (since ground is shared between the oscope and the generator), but if so, I'd like some help to understand why that happens. It would make sense to me if I had a benchtop oscilloscope that ultimately connected to mains ground, but this is a handheld. Its ground probe is only connected internally, and not to the actual ground that the signal generator shares. Or am I misunderstanding something?Also, if one of the diodes was shorted, wouldn't this be somehow visible in the graph? To my untrained eye the shape of the graph on the scope looks like a correctly-rectified 60hz sine wave to 120Hz DC pulses.

author: Benta
author_rank: Super Contributor
timestamp: February 10, 2024, 11:20:26 am
content: Welcome.You trace looks exactly as it should, but your RMS calculation is too simplistic.The issue is, that you don't have a perfect rectified sine wave, but a rectified sine wave with its "feet" cut off. You can't just subtract 1.4 V to compensate for the rectifier voltage drop, the problem is that there is a "pause" in the wave, this reduces the RMS voltage more than you think.

author: tegowerk
author_rank: Newbie
timestamp: February 10, 2024, 12:24:47 pm
content: Thank you very much for replying. That makes sense! One thing that still bothers me is the 4V peak measured by the oscilloscope and which seems too high. I'd expect it to be lower than the theoretical 3.6V, but not 10% higher, right? What could account for that?

author: Jwillis
author_rank: Super Contributor
timestamp: February 10, 2024, 12:39:42 pm
content: Your Vp-p is 10V. When calculated as RMS it becomes 3.53VVrms =1/2sqr2 * Vp-p = 0.353 * Vp-pBut your rectifier has a forward voltage drop. Each wave has to cross 2 series diodes which can have a forward voltage drop of at least 0,6V for each diode.3.53 - the forward voltage drop = 2.2V . The voltage drop is around 1.3V or 2 diodes having a 0.65 V drop each. Typical of most rectifiers.So the discrepancy is the voltage drop across the rectifier.

author: tegowerk
author_rank: Newbie
timestamp: February 10, 2024, 12:48:51 pm
content: Ahh, I was subtracting the voltage drop of the diodes from the peak andthendoing the RMS calculation, so I was multiplying the voltage drop by 0.707 as well. Oh that makes so much sense, thanks a lot!

author: MrAl
author_rank: Super Contributor
timestamp: February 10, 2024, 03:17:39 pm
content: Quote from: tegowerk on February 10, 2024, 06:45:43 amHello there,If we assume the voltage drop is constant, perhaps the average of the highest and lowest voltage, then the RMS voltage is:Vrms=sqrt((8*v*sqrt(A^2-v^2)+4*asin(v/A)*A^2-pi*A^2+8*v^2*asin(v/A)-2*pi*v^2-8*v^2)/(8*asin(v/A)-2*pi))whereA is the peak amplitude (5 volts in this case),v is the average diode voltage drop,pi is the constant 3.14159265...With A=5 and v=1.2, the RMS voltage comes out to:2.42 volts rms.Note in reality the voltage drop across the diodes is a more complicated function so this is just an estimate. To get an exact value we'd have to use the model of the diode and solve the resulting equation (not shown) numerically. It's probably not worth the effort but it can be done.Also just to note, the RMS voltage is defined as the "root of the mean of the square" and that expression is:Vrms=sqrt(mean(v^2))where the mean is:mean=integrate(v^2,t,0,Tp)/Tpwhere Tp is the period of integration.To get the above formula for Vrms, we have to use the voltage which is the sine minus the dynamic voltage drop of the diodes, then solve for the start time t1 and the end time t2 and then compute the RMS value of that time period. It's not hard to do unless we use an exponential model for the diode or even a linearized model, then the solution may have to be done numerically so we would not get a convenient expression like that above for Vrms.