Control transformer in reverse

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Besoeker said:
Then why mention it
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Because a member asked. I did not want to blow them off or say something to the affect of it is not your business.
I was being nice and the information provided was what I had been given at the time, as I have stated several times now.
What I confirmed in person was not consistent and I posted what I found personally.
Can we put this spat to rest now?
 
Sierrasparky said:
Because a member asked. I did not want to blow them off or say something to the affect of it is not your business.
I was being nice and the information provided was what I had been given at the time, as I have stated several times now.
What I confirmed in person was not consistent and I posted what I found personally.
Can we put this spat to rest now?
Click to expand...
Spat?
I was just trying to get to the facts so that I and others might be able help you with your query.
Water under the bridge. Go in peace.
 
Current could be an important thing if your 24 volt circuit isn't stout enough. The end load may not be too significant but you do have to energize that transformer, not sure what maximum current will be to energize it.
 
Sierrasparky said:
My research provides nothing yet from the control transformer folks.
I think I am just going to have to test it.
Click to expand...
Do you have a storage scope to capture the transient?


Sierrasparky said:
With such a small trans do I need a fuse?
Click to expand...
We always did but then the smallest control transformers we generally used were 500VA.
Another thing you might need to consider is the turns ratio when operating the transformer in reverse. The transformer is likely rated to give 24V at rated load when operated normally. Make the input the 24V side and you may not get your desired 120V out.
 
Besoeker said:
Do you have a storage scope to capture the transient?

Unfortunately not. It disappeared years ago in a move. Have not really needed one I guess till now.



We always did but then the smallest control transformers we generally used were 500VA.
Another thing you might need to consider is the turns ratio when operating the transformer in reverse. The transformer is likely rated to give 24V at rated load when operated normally. Make the input the 24V side and you may not get your desired 120V out.
Click to expand...

I will check it when I test.
 
180912-1241 EDT

Last night I ran some tests because there has been so much conjecture and incorrect information in this thread. A lot of posts have occurred since last night. As a result as result of one I have the following comment.

Transformer inrush current is highly variable for any given transformer under test. It is not a constant value every time you apply power to a transformer. It depends upon the residual flux state in the core from the last turn off, and the particular time in the cycle when voltage is applied to the transformer relative to that residual flux. See my photos P6 and P7 at http://beta-a2.com/EE-photos.html .

Experiments:

I have an old NUTONE 120 to 24 V 20 VA transformer. This is used in the following tests.

Turns ratio determination:

One useful equivalent circuit for a transformer is:
1. A pimary series internal impedance.
2. An ideal transformer (no losses) of N turns ratio.
3. A secondary series internal impedance.
4. Assume that primary and secondary internal impedances are the same when adjusted by N^2.

Using the unloaded voltage ratio we can determine the apparent turns ratio from both the primary and secondary directions. If the secondary voltage is lower than the primary voltage, then when voltage is applied to the primary for the measurement the ratio will appear to be slightly larger than its actual value, and when voltage is applied to the secondary the ratio will appear slightly lower. An average of these two will provide a good estimate of the actual ratio.

If the secondary is heavily loaded and the current ratio measured, then the result will be close to the actual ratio, and the above averaged voltage ratios.

Results with the NEWTONE:
120 excited --- 120.6/26.6 = 4.534
24 excited ---- 106.6/23.76 = 4.487
The average = 4.51

Current ratio 1.00/.22 = 4.55

Good comparison.


Exciting the 120 winding with the secondary unloaded a Kill-A-Watt read
120.6 V, 0.02 A, 1.4 W, 2.4 VA . Calculated VA = 2.6 .

Loading the secondary with 25 ohms Kill-A-Watt reads
120.6 V, 0.19 A, 23.6 W, load voltage is 21.1 V.

Calculated total internal impedance is 5.5 V change/0.17 A change = 32.4 ohma.
Reflecting this to the primary produces 32.4*4.51^2 = 32.4*20.3 = 658 ohms.

If you apply 24.0 V to the 24 winding you can expect something under24*4.51 = 108 V for the no load secondary voltage.

If the 24 V winding is the power source, then excitation currnet calculates to about 0.02*4.51 = 0.09 A. Measured excitation current at 24.0 V was 0.08 A.

To get about 20 VA from back driving the transformer with 24 V in will rrequire about 24/20 = 0.83 A plus 0.08 A = 0.91 A.

Probably the big question is how much voltage drop occurs on the 24 V supply line? Apparently this is 100 ft long. No wire size has been provided. 200 ft of #18 is about 6.4*0.2 = about 1.5 ohms. At 1 A we loose 1.5 V of our source 24 V. This might not be a great problem.

.
 
Sierrasparky said:
The control circuit powers a few proximity sensors and a control board. The label of the sensor system says up to 3w a 24v. So I believe the control power 30va @ 24v is sufficient.
Click to expand...

First, to answer your original question, see http://hammondsales.com/pdf/section1.pdf page 2
Hammond explicitly says that you can use a control transformer in reverse.

However the above quote gives me pause. You say that the sensor system says up to 3W a 24V...is this the load that the system is rated to control? Because you are asking to control a 15VA load....

-Jon
 
gar said:
180912-1241 EDT

Last night I ran some tests because there has been so much conjecture and incorrect information in this thread. A lot of posts have occurred since last night. As a result as result of one I have the following comment.

Transformer inrush current is highly variable for any given transformer under test. It is not a constant value every time you apply power to a transformer. It depends upon the residual flux state in the core from the last turn off, and the particular time in the cycle when voltage is applied to the transformer relative to that residual flux. See my photos P6 and P7 at http://beta-a2.com/EE-photos.html .

Experiments:

I have an old NUTONE 120 to 24 V 20 VA transformer. This is used in the following tests.

Turns ratio determination:

One useful equivalent circuit for a transformer is:
1. A pimary series internal impedance.
2. An ideal transformer (no losses) of N turns ratio.
3. A secondary series internal impedance.
4. Assume that primary and secondary internal impedances are the same when adjusted by N^2.

Using the unloaded voltage ratio we can determine the apparent turns ratio from both the primary and secondary directions. If the secondary voltage is lower than the primary voltage, then when voltage is applied to the primary for the measurement the ratio will appear to be slightly larger than its actual value, and when voltage is applied to the secondary the ratio will appear slightly lower. An average of these two will provide a good estimate of the actual ratio.

If the secondary is heavily loaded and the current ratio measured, then the result will be close to the actual ratio, and the above averaged voltage ratios.

Results with the NEWTONE:
120 excited --- 120.6/26.6 = 4.534
24 excited ---- 106.6/23.76 = 4.487
The average = 4.51

Current ratio 1.00/.22 = 4.55

Good comparison.


Exciting the 120 winding with the secondary unloaded a Kill-A-Watt read
120.6 V, 0.02 A, 1.4 W, 2.4 VA . Calculated VA = 2.6 .

Loading the secondary with 25 ohms Kill-A-Watt reads
120.6 V, 0.19 A, 23.6 W, load voltage is 21.1 V.

Calculated total internal impedance is 5.5 V change/0.17 A change = 32.4 ohma.
Reflecting this to the primary produces 32.4*4.51^2 = 32.4*20.3 = 658 ohms.

If you apply 24.0 V to the 24 winding you can expect something under24*4.51 = 108 V for the no load secondary voltage.

If the 24 V winding is the power source, then excitation currnet calculates to about 0.02*4.51 = 0.09 A. Measured excitation current at 24.0 V was 0.08 A.

To get about 20 VA from back driving the transformer with 24 V in will rrequire about 24/20 = 0.83 A plus 0.08 A = 0.91 A.

Probably the big question is how much voltage drop occurs on the 24 V supply line? Apparently this is 100 ft long. No wire size has been provided. 200 ft of #18 is about 6.4*0.2 = about 1.5 ohms. At 1 A we loose 1.5 V of our source 24 V. This might not be a great problem.

.
Click to expand...

winnie said:
First, to answer your original question, see http://hammondsales.com/pdf/section1.pdf page 2
Hammond explicitly says that you can use a control transformer in reverse.

However the above quote gives me pause. You say that the sensor system says up to 3W a 24V...is this the load that the system is rated to control? Because you are asking to control a 15VA load....

-Jon
Click to expand...

Some good resources , Thanks all. I'll have to do some reading. :thumbsup:
 
180914-2144 EDT

Sierrasparky:

What is the unloaded voltage at the end of the 24 V 100 ft wire?

If you put a known, but approximately 20 VA, load at the end of this wire to what does the voltage drop across the load? From this we can determine the internal impedance of the 24 V source at the end of the 100 ft.

Over what voltage range can the 120 V loads operate that are at the 100 ft end?

.
 
gar said:
180914-2144 EDT

Sierrasparky:

What is the unloaded voltage at the end of the 24 V 100 ft wire?

If you put a known, but approximately 20 VA, load at the end of this wire to what does the voltage drop across the load? From this we can determine the internal impedance of the 24 V source at the end of the 100 ft.

Over what voltage range can the 120 V loads operate that are at the 100 ft end?

.
Click to expand...

I have not performed the work yet. I tested at my shop though. I will see what it is at the end of the line with a load next week or so. I'll post the result.
 
180916-1637 EDT

Sierrasparky:

I ran a simple test as follows:

1. Voltage source was a Variac feeding a Stancor P8668 transformer rated at 117 V in 28 V 2A out. Unloaded except for the unloaded NEWTONE transformer 24 V winding. Adjusted the NEWTONE input to 24.18 V. There was no long wiring between the source and the NEWTONE 24 V winding.

2. Used a 15 W 120 V incandescent as the test load. With a Kill-A-Watt EZ this reads 0.1 A and 9,8 W at 94 V.

3. Unload voltage ratio ftrom 24 V winding to 120 V winding was 108.4/24.18 = 4.48 . Essentially the same as my previous post measurement.

4. With the 15 W bulb load on the NEWTONE 120 winding the input voltage drops to 23.57 V and the output voltage drops to 93.5 V with 0.10 A currrent thru the bulb. Calculated power to bulb is 93.5*0.10 = 9.4 W.

You need to determine the minimum voltage that your load can tolerate and produce the desired results. Also the maximum continuous voltage that will not damage the load. This information will allow you to judge whether the system will work, and how much, if any, that the source 24 V can be increased above 24 V to compensate for the various voltage drops.

.
 
gar said:
180916-1637 EDT

Sierrasparky:

I ran a simple test as follows:

1. Voltage source was a Variac feeding a Stancor P8668 transformer rated at 117 V in 28 V 2A out. Unloaded except for the unloaded NEWTONE transformer 24 V winding. Adjusted the NEWTONE input to 24.18 V. There was no long wiring between the source and the NEWTONE 24 V winding.

2. Used a 15 W 120 V incandescent as the test load. With a Kill-A-Watt EZ this reads 0.1 A and 9,8 W at 94 V.

3. Unload voltage ratio ftrom 24 V winding to 120 V winding was 108.4/24.18 = 4.48 . Essentially the same as my previous post measurement.

4. With the 15 W bulb load on the NEWTONE 120 winding the input voltage drops to 23.57 V and the output voltage drops to 93.5 V with 0.10 A currrent thru the bulb. Calculated power to bulb is 93.5*0.10 = 9.4 W.

You need to determine the minimum voltage that your load can tolerate and produce the desired results. Also the maximum continuous voltage that will not damage the load. This information will allow you to judge whether the system will work, and how much, if any, that the source 24 V can be increased above 24 V to compensate for the various voltage drops.

.
Click to expand...

Wow that awesome that you ran a test setup , thanks for your help.:thumbsup:
I was thinking about purchasing one of those new clamp on multi-meters that are True RMS and inrush sensing.
 
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