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how to determine current parameters

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rajaram04

rajaram04

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Hello sir
I want to know about
if we know about the voltage out of any stepdown transformer whoaw current parameter is not shown then how its possible to determine about that ?
Suppose we have a 6v labelled transformer but the currebt is not mentioned on it so how to determine it when have multimeters n other passive components relating electronics ??

I too want to know about the total minimum current required by an electronic circuit to work efficiently , designed homely by us .(say a square wave generator using 555 ic)

what would the maximum limit of current applicable for the same circuit ? how can we derive practically ?
Thanks
 

Transformer current

1. Check the secondary turn diameter by using Standard wire gauge (SWG)
2. Connect the transformer to supply mains and connect the secondary winding through a high wattage variable resistor. Resistor value starting from approx. double the value of your trans. secondary voltage. Measure the voltage across the resistor. Vary the resistor value to minimum with checking the voltage. When the secondary voltage falls from rated note the current. That is your safe rated current now.

Electronics Circuit

1. There is no relation between efficiency and your electronic circuit. Try to use 500mA or above 500mA. For experimental Purpose use high current rating and voltage rating should be as per your circuit. I = V/R Circuit will draw current according your on state resistance.
 

ya okk sir the info awsome but tell me how could i measure the wire diameter as the transformer is sealed completely & no wires are coming out.It has only the tap points for soldering purposes.
by breaking seals we may suffer damages here , now what's the solution ? :( :roll:

- - - Updated - - -

Sir,i too want to know whatr is actually we determine on state resistance ? we never did any practicals in acedamic time . please tell the clean procedure . . thanks
 

What is your application ?
 

Try drawing increasingly more current from the secondary, by attaching loads with lower resistance (and higher watt rating).

See when the transformer gets too warm to hold. That is the chief way to determine safe maximum current output.

To go further you can multiply each volt reading times the simultaneous current reading. The result is power in watts. Plot the results with ohms as the X axis. There will be three Y axes: volts, amps, watts.

The power graph will be a bell-shaped curve. The top of the curve shows the transformer's maximum power capability. It will have a corresponding current figure.

This has to do with the concept that maximum power is transferred when outgoing impedance matches impedance of the next stage.

You can draw more current above the maximum power. Volt output will decrease. Be on the watch for overheating at some point, depending on how long you draw greater current.
 

I think ."By monitoring the temperature and take a decision from that" It is not a good practice. Your transformer
insulation property will be lost. In smaller KVA rating transformers it is the actually happening with this test.
 

lalgpt said:
I think ."By monitoring the temperature and take a decision from that" It is not a good practice. Your transformer
insulation property will be lost. In smaller KVA rating transformers it is the actually happening with this test.
Click to expand...

Yes, damage can happen if too high current flows too quickly.

It's better to build up current gradually. Start small. Let a load stay connected for several minutes, and see how things go.

It's also wise to compare the physical size and weight with other transformers which are meant to step down mains supply, and which are labelled as to current rating.
 

BradtheRad said:
..............................................

The power graph will be a bell-shaped curve. The top of the curve shows the transformer's maximum power capability. It will have a corresponding current figure.

This has to do with the concept that maximum power is transferred when outgoing impedance matches impedance of the next stage.

You can draw more current above the maximum power. Volt output will decrease. Be on the watch for overheating at some point, depending on how long you draw greater current.
Click to expand...
You will likely be seriously overloading the transformer before you ever get to the maximum power point. 8-O Power transformers are designed to operate well below the maximum power point to have good efficiency (typically well above 90%). At the maximum power point the efficiency is only 50%.
 
crutschow said:
You will likely be seriously overloading the transformer before you ever get to the maximum power point. 8-O Power transformers are designed to operate well below the maximum power point to have good efficiency (typically well above 90%). At the maximum power point the efficiency is only 50%.
Click to expand...

Yes. I find I must defer to you. Impedance matching is an advantage in many situations but it's risky when you're talking about a power transformer.

My homemade power supply uses a 12V transformer rated at 3A. That's 36 W. It will make the transformer heat up about as much as a 36 W bulb. Yet it is still within the rating which is printed on it.

It would be a mistake to draw more than 3A indefinitely (although I have drawn more than 3A for several minutes on occasion). The transformer will maintain 12V up to 3A. However beyond that, volt level will start dropping. Carried further, it will eventually reach a region of maximum power transfer. But to go to that extent risks ruining the transformer.

I have done ohm/volt/amp/watt experiments (post #5) on several power supplies, batteries, and PV panels.

The power supplies had other components besides the transformer.

I have gone back to look at the curves generated by my power supply. Now I realize these additional components altered the load-power curve to something different than what the transformer alone would yield.

It caused the curves to show that maximum power output was when the load was drawing 24W. That's quite a bit less than the transformer's 36W rating. I made the mistake of believing it had to do with the transformer's operating parameters, whereas there were other things influencing the data.
 

BradtheRad said:
Yes, damage can happen if too high current flows too quickly.

It's better to build up current gradually. Start small. Let a load stay connected for several minutes, and see how things go.

It's also wise to compare the physical size and weight with other transformers which are meant to step down mains supply, and which are labelled as to current rating.
Click to expand...


Transformer core materials is always not be the same ( silicon steel with iron content high ). Then how the weight can compared ?.

- - - Updated - - -

BradtheRad said:
Yes. I find I must defer to you. Impedance matching is an advantage in many situations but it's risky when you're talking about a power transformer.

My homemade power supply uses a 12V transformer rated at 3A. That's 36 W. It will make the transformer heat up about as much as a 36 W bulb. Yet it is still within the rating which is printed on it.

It would be a mistake to draw more than 3A indefinitely (although I have drawn more than 3A for several minutes on occasion). The transformer will maintain 12V up to 3A. However beyond that, volt level will start dropping. Carried further, it will eventually reach a region of maximum power transfer. But to go to that extent risks ruining the transformer.

I have done ohm/volt/amp/watt experiments (post #5) on several power supplies, batteries, and PV panels.

The power supplies had other components besides the transformer.

I have gone back to look at the curves generated by my power supply. Now I realize these additional components altered the load-power curve to something different than what the transformer alone would yield.

It caused the curves to show that maximum power output was when the load was drawing 24W. That's quite a bit less than the transformer's 36W rating. I made the mistake of believing it had to do with the transformer's operating parameters, whereas there were other things influencing the data.
Click to expand...

Absolutely right BradtheRad

- - - Updated - - -

crutschow said:
You will likely be seriously overloading the transformer before you ever get to the maximum power point. 8-O Power transformers are designed to operate well below the maximum power point to have good efficiency (typically well above 90%). At the maximum power point the efficiency is only 50%.
Click to expand...

Efficiency % =Output/input * 100


Output + losses = Input power

losses = Copper loss + core loss


Then by this argument at max for 50% efficiency output = 1/2 the input is it true ?
 

lalgpt said:
Transformer core materials is always not be the same ( silicon steel with iron content high ). Then how the weight can compared ?.
........................................

Efficiency % =Output/input * 100

Output + losses = Input power

losses = Copper loss + core loss

Then by this argument at max for 50% efficiency output = 1/2 the input is it true ?
Click to expand...
All power transformers use very similar steel in their design and any changes in the steel type would have little effect on the steel density, thus transformers of similar power and voltage capability will weigh about the same. But you must compare like types together. For example a toroidal transformer would likely have a different weight then a similar rated C-core type.

At the maximum power point (well above the maximum rated transformer power) the output voltage drops to 1/2 the no-load voltage, the power dissipated in the transformer equals the load power and the efficiency is 50%, not a desirable operating condition for a power transformer. Thus power transformers are operated at well below this point to minimize the voltage drop with changes in load for good voltage regulation, and to minimize power loss for high efficiency.
 

crutschow said:
All power transformers use very similar steel in their design and any changes in the steel type would have little effect on the steel density, thus transformers of similar power and voltage capability will weigh about the same. But you must compare like types together. For example a toroidal transformer would likely have a different weight then a similar rated C-core type.

At the maximum power point (well above the maximum rated transformer power) the output voltage drops to 1/2 the no-load voltage, the power dissipated in the transformer equals the load power and the efficiency is 50%, not a desirable operating condition for a power transformer. Thus power transformers are operated at well below this point to minimize the voltage drop with changes in load for good voltage regulation, and to minimize power loss for high efficiency.
Click to expand...



At the maximum power point (example)

Input voltage 230V
Output voltage 115V

regulation (230=115)/230*100
Only 50%

Oh my *** is it true ?
 

A similar approach applies to using an AC generator.

It's good for the generator to have a low operating impedance.

However you do not want to attach a load which has low impedance. It will overheat the generator.

You want to attach a load which will permit the generator to maintain 230 VAC. This means the load should have moderate or high impedance.

It's the same with a transformer.
 

lalgpt said:
At the maximum power point (example)

Input voltage 230V
Output voltage 115V

regulation (230=115)/230*100
Only 50%

Oh my *** is it true ?
Click to expand...
Yes it's true. That's why maximum power is only used where you absolutely want the maximum power output and the efficiency is a secondary consideration, such as broadcast RF transmitters and tube audio amps. (I remember seeing some old tube high-power audio amps where the tubes were running at such high current that the plates glowed a dull red. Not exactly a high efficiency design).
 

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