Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Oh that's great, will be interesting to find out how this turns out, and if you make any RMS, average, or temperature measurements (etc>) i hope you can post those too.
In the mean time, here is one of the more interesting views on the full wave rectifier with filter capacitor, with small inductance. We were just talking about this on another web site recently.
Note that here we can calculate the ACTUAL waveform of the diode current, so there's no problem in calculating the RMS current.
The idea is to first calculate angle A from knowing the peak voltage and the DC output voltage, then calculating angle B from knowing that the volt seconds charge is equal to the voltage seconds discharge. Knowing this, we can then calculate the exact inductor waveform (from angle A to angle B). This of course allows us to calculate the RMS current.
An example with L=0.001, Vdc=150v, Vpk=120*sqrt(2), and f=50Hz, we get 18.3 amps RMS. The average current is 10.7 amps. The ratio here then is 1.71 to 1.
If we add series resistance then we need to add that to the calculations, but without that we see the basic idea.
I can probably run a quick test myself too later this month, with lighter current levels. Unfortunately i dont think i have a sizable inductor around that would be suitable for this test, so probably have to limit it to just series resistance.
Gee i almost forgot the most obvious:
Use a scope to view the current waveform, do a little math to estimate the RMS current. Big 'duh' over here
If you can take some scope shots of the steady state current at full load, we can estimate the RMS current and then you would know for sure if it will work well or not over the years.
All the bits and pieces are in that Cardboard box.
Toroidal, and heatsinks are there too along with the Bridge and LM317's and stuff. And the old Cyclons etc.
That box has been stored since 2004...tell you you truth...I am weary and worried about all that I could find inside it once I open it
I was into a bit of Porn those days
I am going to try this Weekend to pluck up the courage to just do it. And not just talk about it. If I find any Porn....you Folks will be the first I share it with.
I promise....
Now I am wondering if I said the right thing?
OK, jokes aside, that box is HEAVY. I just don't know what to expect/discover once I open it.
Not sure if already mentioned.
Most big ferrite self heating are core losses from saturation from excessive low ESR on big caps causing peak currents with a crest factor that is inverse to the ripple factor for voltage.
Thus big L current smoothing is useful to raise impedance and small series R to avoid high Q Series resonant , idle excitation currents and huge ripple currents causing self-heating in caps unless low ESR then even bigger ripple current thus small external R helps. even nichome R.(PTC)
The ripple current is 100Hz not 50. or 120 vs 60... for simulation..
If using old lead acid battery de-sulphate it for a few days low power but high current (10A) 10~30nS impulses at any rate like 30KHz to stimulate the resonance of crystals on the plates that raise ESR of the battery and kill storage capacity. The lead-suphate will breakdown, restore Specific gravity of lead acid and restore A-h capacity.
I agree with Schmitt Trigger 60% of VA is possible if well designed due to crest factor in current, active PFC can restore the capacity power to 100% , but thats another design change.
Getting back to Core losses, unless it is pre-gapped with a known Bsat, going past saturation can be detected with a scope probe shorted near the core. If it saturates from too big a load cap with low ESR, then L drops rapidly during the spikes and is prone to thermal runaway when hot. ( poof go the drivers) Sometimes they are called Eddy current losses and Saturation losses. Bsat is determined by number of primary turns and current flowing and once saturation starts Volt-seconds in does not equal Volt seconds out. due to self-heating losses in peak pulse currents.
Latest excuse from me is sealing "IT" again. I need buff tape. And none available here on a Sunday. So the box stayed closed.
Meanwhile the RITAR sits here waiting for me to make up my mind.
Oh well, she has a Ten Year service life in Standby Mode....she needs a little trickle charge from time to time to keep her happy when in storage.
She has only been sitting here for a few Days though so I am OK...
I need to get my butt into gear though...because if the lights go out....and she cannot work.....I will be cross with myself.
Picture of her here:
Yip, she is a little 65Ah Sealed Lead Acid Battery.
Her specifications are here too:
She is such a "nice" battery. No airs and graces about her. I love her already
Yes looks good. I noticed that because i would like something like that for my batteries, not trusting the thin plastic handle that comes with them.
Also, could use something for a variac too. This is really a pain though because it is cylindrical not rectangular like a car battery. Looked on the web, didnt find any good ideas except building an entire 'box' around it for easy carrying. That's not good though for a big one. A rope would be nice if there was some way to attach it nicely.
As to the RMS current of the diode bridge and transformer secondary, here is a short table (when the line frequency is 50 Hz, but see 60 Hz notes below) for comparison.
In the table, the first entry of each line is the ratio of Vdc/Vpk, which is the DC voltage divided by the peak AC voltage.
So for Vdc=153 and Vpk=170 we would have 153/170=0.90 for example, so we would use that line.
The values are shown as a fraction divided by the inductance L.
For the example of Vpk=170 and Vdc=153 and L=0.001 we have a ratio of 0.90 so the entry is:
0.0083
so we multiply that by Vpk/100 and that means 0.0083*1.7 and get:
0.01411
and now divide that by the inductance L:
0.01411/0.001=14.11 amps RMS.
The table is more for comparison than for any exact calculation however.
If we look at a couple entries like 0.95 and 0.90 we can see that the RMS current goes up by a factor of 3 when the ratio of DC voltage to peak AC line voltage does from 0.95 down to 0.90 and that's quite a bit. It goes up by almost 18 times as we go from 0.90 to 0.75 and that's really a lot.
It is also interesting that the RMS current is inversely proportional to L, so the bigger the L the lower the RMS current. An inductance twice as large would produce half the RMS current.
To repeat once more, to use the table above find the fraction Vdc/Vpk and use that entry. Multiply that by Vpk/100 and then substitute L in with the actual value of L. The result is the RMS curent of the diode BRIDGE (not each diode). Each diode will see about 1/2 of that.
So for Vpk=170 and Vdc=153 and L=1mH we get about 14 amps RMS in the bridge, which is about 7 amps RMS for each diode.
This will also be 14 amps RMS for the secondary of the transformer that powers the bridge.
Note that in practice it may be hard to figure out what the value of L is because this will include the line inductance reflected to the secondary and also the transformer inductance reflected to the secondary and any added extra inductance.
At 60 Hz the RMS current is a little lower, for example for a ratio of 0.90 we get:
0.90: 0.0069/L
instead of:
0.90: 0.0083/L
So at 60 Hz and Vpk=170 and Vdc=153 and L=0.001 the RMS current would be:
(170/100)*0.0069/0.001=11.73 amps RMS.
This is close to the value obtained at 50 Hz divided by (60/50) so that probably works for the others too, due to the larger inductance effect at the higher frequency.
It is also interesting to note that these results would also be the same for any equal ratio of Vdc/Vpk, so the result we get for 170vpk and 153vdc would be the same as what we get for 100vpk and 90vdc because the ratio is 0.90 for both of these.
Note these results hold only when the inductance is allowed to reset after each half cycle current period. If it doesnt reset then there is net DC which has to be considered also.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
