# need help on rectifier using zero point switching

#### leonard

Joined Nov 29, 2005
14
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

#### n9352527

Joined Oct 14, 2005
1,198
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?

#### Papabravo

Joined Feb 24, 2006
20,582
Originally posted by leonard@Apr 10 2006, 05:39 AM
...
2. Why the system unstable? The preset charge current varying more than (+/-) 0.5%, while other branded system only varies (+/-) 0.2% max.
...
[post=16028]Quoted post[/post]​
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.

#### leonard

Joined Nov 29, 2005
14
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.

Originally posted by n9352527@Apr 10 2006, 07:21 PM
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?
[post=16034]Quoted post[/post]​

#### leonard

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

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

Thanks.

Bset regards

Originally posted by Papabravo@Apr 10 2006, 08:55 PM
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.
[post=16038]Quoted post[/post]​

#### Papabravo

Joined Feb 24, 2006
20,582
Originally posted by leonard@Apr 15 2006, 05:31 AM
Dear Papabravo, thanks for the replay.

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

Thanks.

Bset regards
[post=16196]Quoted post[/post]​
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.

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by leonard@Apr 15 2006, 10:25 AM
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.
[post=16195]Quoted post[/post]​
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.

#### leonard

Joined Nov 29, 2005
14

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

Originally posted by n9352527@Apr 18 2006, 09:47 PM
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.
[post=16286]Quoted post[/post]​

#### leonard

Joined Nov 29, 2005
14
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

Originally posted by Papabravo@Apr 17 2006, 05:38 AM
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.
[post=16230]Quoted post[/post]​

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by leonard@Apr 20 2006, 01:22 PM

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
[post=16358]Quoted post[/post]​
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.

#### leonard

Joined Nov 29, 2005
14
dear n9352527,

Originally posted by n9352527@Apr 20 2006, 07:54 PM
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.

....sorry about that. the rectifier have been tested using several battery combination. The combinations are :

1. parallel in 6 columns, each column has 3 batteries in seri. total battery in this combination is 18 pcs. since the current flows 30 A-DC from the rectifier, then each column has about 5 A-DCD flows to it. assume all the battery has the same internal R, then each battery has 5 A-DC flows to it.

2. parallel in 3 columns, each column has 6 batteries in seri. total battery in this combination is also 18 pcs, and also 30 A-DC flows from the rectifier, then each column has 10 A-DC, or each battery has 10 A-DC.

3. parallel in 2 columns, each columns has 30 pcs batteries in seri. total battery in this combination is 60 pcs, and also 30 A-DC flows from the rectifier, then each colum has 15 A-DC, or each battery has 15 A-DC.

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.

....yes. I mean, there are 2 rectifiers, one is commercial-branded and the other is mine. Using the same methode, point of measurementt, the batteies and the tools, but yielding a different result, which my rectifier is unefficient, I resume.

Please please take a moment to read the previous posts, think and approach the problem logically and provide us with complete informations.
[post=16360]Quoted post[/post]​

#### leonard

Joined Nov 29, 2005
14
Dear Sir,

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

Thanks.

Originally posted by Papabravo@Apr 17 2006, 05:38 AM
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.
[post=16230]Quoted post[/post]​

#### n9352527

Joined Oct 14, 2005
1,198
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?