# need help on rectifier using zero point switching

Discussion in 'The Projects Forum' started by leonard, Apr 10, 2006.

1. ### leonard Thread Starter Member

Nov 29, 2005
14
0
Hi everyone,

I built a huge constant current rectifier using zero point swtching controller, but the result is not as I expected. The complete diagram (in 2 pages, fig. 1 and 2) can be found at :

HTTP://usb06net.multiply.com and the components list is in attachment

The load is a series of lead-acid otomotive batteries. The desired charge curent is set by rotating the 2K-B POT. The charger will now maintain the preset charge current. The max of I is 60 A-DC.

Problems :

1. Why the system not works efficiently? At I = 30 A-DC, T1 requires 60 A-AC from source, while other branded system only needs 33 A-AC.

2. Why the system unstable? The preset charge current varying more than (+/-) 0.5%, while other branded system only varies (+/-) 0.2% max.

3. At the 2K-B POT in full-clockwise, the charger produces I more than 60 A-DC, meanwhile the T1 rating is 60 A-AC.

Any help would be much appreciated.
Please attach some figure/drawing if necessary.

Best Regards,
leonard

2. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
4
I think we need to establish how, at which points and with what instruments you did the current measurements and with how many load batteries in series. The currents you mentioned in point 1. would result in _huge_ dissipated power and generate substantial heat.

What is the ratio of the T1 P:S?

3. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
As a controls engineer and a pilot I would hardly consider a controlled variable within (+/-)0.5% of it's commanded value to be unstable.

Unstable is an extreme oscillation that it causes the firmament to rise up and smite unworthy pilots and their airframes.

4. ### leonard Thread Starter Member

Nov 29, 2005
14
0
hallo n9352527, thanks for the reply.

a. they are 2 ways to measure the current : 1. using dc clamp amperemeter, clamps at the output cable that goes to the batteries, and 2. using digital DC voltmeter. This digital DC voltmeter measures the volttage produces by the Shunt and since the shunt is known, 200 A-50 mV, the voltage is proportional to the current that flows to the batteries.

b. the amount of battery is vary, betwen 5 to 30 pcs, in serial and paralel formation and yielding of 5 yo 10 ampres current flow to each battrey. So far, with that low of charge current, no excessive heat detected at the batteries.

c. the ratio of T1 P:S is 1:1

best regards.

5. ### leonard Thread Starter Member

Nov 29, 2005
14
0
Dear Papabravo, thanks for the replay.

dear sir, do you have any idea how to make my circuit stable?

Thanks.

Bset regards

6. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I consider your system to be stable already. What makes you think improvement is possible?

If you can measure a difference between your circuit and another one, you might want to analyze the components used and or the circuit topology in the unit with superior performance.

From my location I cannot access the link in your original post.

7. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
4
What I was trying to get was the power loss calculated from your measurements. That was why I asked how many load batteries were connected when you did your measurements. Since you didn't reply how many batteries were connected when you measured the currents, then here goes some wild approximations:

Loss = 60A*380V - 30A*(output voltage/depends on how many batteries in series)

Assuming that you get the current values for commercial system (33A) from datasheet and not from measurement, loss of ~27A at 380V is approx. 10KW. A substantial power which would heat a large room easily in winter!

Assuming that you don't cool your circuit with a water/oil filled radiator or a heatsink the size of a cupboard, then something must be wrong with the measurements.

From what you've mentioned above, I suspect it was your instruments. That was why I asked what instruments you were using. Did you use true RMS meters? If you have chopped rectified sinewave, ordinary meter would not give correct measurements.

The gist is, check what you measured first, if the values seemed to be out of proportions (like that huge power loss you have), then those values most probably were wrong.

8. ### leonard Thread Starter Member

Nov 29, 2005
14
0

my previous question (#1) means :

1. a branded unit, bult up from factory requires only 33 A-AC from 330V-AC power source to delivery 30 A-DC to the load.

2. my unit requires 60 A-AC from 330 V-AC power source, to delivery the same amoumt current, 30 A-DC.

The current measurement for above two point taken at the same load. How many battery conected to the unit is not a matter.

My concern is at input section. I think the controller can not drive the 3 pcs SCRs correctly, so a large part of input power is not delivery to the load

best regards,
leonard

9. ### leonard Thread Starter Member

Nov 29, 2005
14
0
Dear Papabravo,

1. at the low/small charge current, let us say, 0.5A, the variation of (+/-) 0.5% means a lot.

2. is it possible to improve the system? I think yes. I mean, to compare to another branded unit, it is possible to do that.

3. i will let you know my new url soon i sign up a new one.

best regards,
leonard

10. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
4
The number of batteries connected does matter to calculate the power loss (see my previous post). You seem to miss my point, which is the power loss that I estimated (not enough data for actual calculation) from your measurements seems way too high and impossible (around 10KW? based on incomplete data). What I am trying to underline is if your measurements show power loss in that region and your circuit did not heat the room up to boiling point then surely there must be something wrong with the measurements.

After several replies, I still don't have enough informations to estimate the actual power loss (V*I at input and output), compare your circuit with commercial unit (both figures through measurements? same measurement methods? commercial unit figure from datasheet?) or tell whether the measurements were accurate or not and hence whether there is actually a problem or not.

Please please take a moment to read the previous posts, think and approach the problem logically and provide us with complete informations.

11. ### leonard Thread Starter Member

Nov 29, 2005
14
0
dear n9352527,

12. ### leonard Thread Starter Member

Nov 29, 2005
14
0
Dear Sir,

please klik : HTTP://PHOTOS.YAHOO.COM/MAKKUOKX1 to acces my original post.

Thanks.

13. ### n9352527 AAC Fanatic!

Oct 14, 2005
1,198
4
Okay, lets estimate the power loss for each configuration (assuming 12V batteries were used):

1. 3 batteries in series, output power would be 30A*36V = ~ 1KW.

2. 6 batteries in series, output power would be 30A*72V = ~ 2KW.

3. 30 batteries in series, output power would be 30A*360V = ~ 10KW.

Now, did each of this configuration draw 60A from mains supply? Lets just assume that configuration 3 did, then the power loss would be 60A*360VAC = ~ 20KW. We have ~ 10KW loss for configuration 3. Where did this much of power go? It should all be converted to heat.

Next questions would be, did your circuit get hot? If it did, how hot? Did you use any heatsink or similar heatsinking system? How big is the heatsinking and does it seems capable of radiating 10KW of power? Do you appreciate how huge is 10KW of power?

These questions would lead to... do you think your circuit really dissipating 10KW of power? Or do you think there's something wrong with your current measurements?