| Topic: |
Science > Physics |
| User: |
"pavan" |
| Date: |
24 Oct 2006 05:18:18 AM |
| Object: |
Autotransformer Design?? |
Dear Reader,
First of all let me thank you for taking interest
in lending a helping hand for my issue.
The issue is that i ve to design one autotransformer with secondary
tapped from the primary and is kept at a voltage equal to 106V. The
primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
rated voltage which is 110V. so the taps will be
121,132,143,154,165,176 and 187V tap. the load on the secondary will be
around 2.1K.
i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
turns, I-Current. The secondary load current is 43mA. N2 is currently
3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
first tap. so from the above expression I1 can be found, and then N1
can be found. this is repeated for remaining taps. This is how i ve
done the designing. Is this OK or any other method has to be adopted.
Please help me.
Thanking You
Regards
Pavan
.
|
|
| User: "Androcles" |
|
| Title: Re: Autotransformer Design?? |
24 Oct 2006 07:26:13 AM |
|
|
"pavan" <pavannaren@gmail.com> wrote in message
news:1161685098.760087.77820@h48g2000cwc.googlegroups.com...
| Dear Reader,
| First of all let me thank you for taking interest
| in lending a helping hand for my issue.
|
| The issue is that i ve to design one autotransformer with secondary
| tapped from the primary and is kept at a voltage equal to 106V. The
| primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
| rated voltage which is 110V. so the taps will be
| 121,132,143,154,165,176 and 187V tap. the load on the secondary will be
|
| around 2.1K.
|
|
| i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
| turns, I-Current. The secondary load current is 43mA. N2 is currently
| 3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
| first tap. so from the above expression I1 can be found, and then N1
| can be found. this is repeated for remaining taps. This is how i ve
| done the designing. Is this OK or any other method has to be adopted.
|
|
| Please help me.
|
|
| Thanking You
|
|
| Regards
| Pavan
This appears to be a homework question, Mr Designer.
Never mind, let's see what we can do.
Let us label the taps thus:
T0
T1
T2
T3
T4
T5
T6
Now let us put the number of turns against each tap.
T0 - 0
T1 - 3600
T2 - 3600*110% = 3960
But this is an autotransformer, so we have 3600 turns between T0 and T1,
needing 360 turns between T1 and T2 for a total of 3960 turns
between T0 and T2.
T0 0
T1 3600
T2 3600*110% = 3600 + 360 = 3960
T3 = ?
T4 = ?
T5 = ?
T6 = ?
I'll let you fill in the values.
WHAT IF we applied 110V to T2?
T0 0 0V
T1 3600 ?
T2 3960 110V
T3 = ?
T4 = ?
T5 = ?
T6 = ?
Can you see that we'd have 100V at T1?
But we have a problem, 100V is too low, we need 106V.
We can apply the input voltage of 110V at T6
and draw a lesser voltage of 106V from one of the taps T1-T5
T0 0 0V
T1 3600 ?
T2 3960 ?
T3 = ? ?
T4 = ? ?
T5 = ? ?
T6 = ? 110 V
OR
we can apply the input voltage of 110V at T5
and draw a lesser voltage from one of the taps T1 to T4
T0 0 0V
T1 3600 ?
T2 3960 ?
T3 = ? ?
T4 = ? ?
T5 = ? 110V
T6 = ? ?
If you've understood the principle, you can find 106 V.
Androcles
.
|
|
|
| User: "jpolasek" |
|
| Title: Re: Autotransformer Design?? |
24 Oct 2006 09:52:13 AM |
|
|
Androcles wrote:
"pavan" <pavannaren@gmail.com> wrote in message
news:1161685098.760087.77820@h48g2000cwc.googlegroups.com...
| Dear Reader,
| First of all let me thank you for taking interest
| in lending a helping hand for my issue.
|
| The issue is that i ve to design one autotransformer with secondary
| tapped from the primary and is kept at a voltage equal to 106V. The
| primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
| rated voltage which is 110V. so the taps will be
| 121,132,143,154,165,176 and 187V tap. the load on the secondary will be
|
| around 2.1K.
|
|
| i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
| turns, I-Current. The secondary load current is 43mA. N2 is currently
| 3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
| first tap. so from the above expression I1 can be found, and then N1
| can be found. this is repeated for remaining taps. This is how i ve
| done the designing. Is this OK or any other method has to be adopted.
|
|
| Please help me.
|
|
| Thanking You
|
|
| Regards
| Pavan
snip
You're making this too tough.
You specify that the output must be V2 = 106V and N2 = 3600 turns.
Divide to get the core density:
106V/3600turns = 0.0294V/turn or
34 turns/volt
as a design constant. Use these values to find your input tap.
If your input voltage is 110 you have to tap in at
110x34 = 3736 turns or in general
Ntap = 34 x input volts.
Two suggestions:
1. You have far too many turns on the transformer. At 60 cps you can
run as low as 1 volt per turn so your secondary could be,
conservatively, 125 turns instead of 3600. (At 50 cps you need 1.2
turns per volt. The saturation of the iron depends on the integral of
volt-seconds, not current).
2. You'll have a hard time getting closer than 10% with 10% taps, so
why not use a second secondary tap 5% apart so you can get to 5%?
You don't have to consider current in your design. You'll get your 4.5
watts OK.
John Polasek
BTW: I am posting a 2nd time via Google since my 1st posting was
stalled somewhere.
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Autotransformer Design?? |
24 Oct 2006 10:56:38 AM |
|
|
"jpolasek" <jpolasek@cfl.rr.com> wrote in message
news:1161701533.604580.36810@e3g2000cwe.googlegroups.com...
|
| Androcles wrote:
| > "pavan" <pavannaren@gmail.com> wrote in message
| > news:1161685098.760087.77820@h48g2000cwc.googlegroups.com...
| > | Dear Reader,
| > | First of all let me thank you for taking interest
| > | in lending a helping hand for my issue.
| > |
| > | The issue is that i ve to design one autotransformer with secondary
| > | tapped from the primary and is kept at a voltage equal to 106V. The
| > | primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
| > | rated voltage which is 110V. so the taps will be
| > | 121,132,143,154,165,176 and 187V tap. the load on the secondary will
be
| > |
| > | around 2.1K.
| > |
| > |
| > | i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
| > | turns, I-Current. The secondary load current is 43mA. N2 is currently
| > | 3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
| > | first tap. so from the above expression I1 can be found, and then N1
| > | can be found. this is repeated for remaining taps. This is how i ve
| > | done the designing. Is this OK or any other method has to be adopted.
| > |
| > |
| > | Please help me.
| > |
| > |
| > | Thanking You
| > |
| > |
| > | Regards
| > | Pavan
| >
| >snip
|
| You're making this too tough.
[snip]
| John Polasek
| BTW: I am posting a 2nd time via Google since my 1st posting was
| stalled somewhere.
Your brain was stalled, too.
Androcles, not pavan.
.
|
|
|
|
|
| User: "pavan" |
|
| Title: Re: Autotransformer Design?? |
25 Oct 2006 02:39:28 AM |
|
|
Dear Sir,
I understand the design methodology of calculating the
turns using 110% of previous Tap. But what is practically happening is
the voltage is not the same when i apply tap voltage on the primary. It
is diminishing with tap. That is across 121V when i apply 121V, the
secondary is slightly less than 106. then when i apply 132V across the
132V tap, the secondary voltage is still lesser. This continues and
when i apply 187V across 187V tap, the seconadry voltage is only around
90 odd Volts. So this is primarily because when the load of around 2.1K
is connected, there is a volatge drop that is happening.
this has to be addressed. so for that reason i took into account the
load current of 43mA and calculated the equivalent primary current.
then using the transformation formula i calculated the number of turns,
which i ve written in the mail.
Please note this and help me.
thank You for all Your help.
regards
Pavan
Androcles wrote:
"pavan" <pavannaren@gmail.com> wrote in message
news:1161685098.760087.77820@h48g2000cwc.googlegroups.com...
| Dear Reader,
| First of all let me thank you for taking interest
| in lending a helping hand for my issue.
|
| The issue is that i ve to design one autotransformer with secondary
| tapped from the primary and is kept at a voltage equal to 106V. The
| primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
| rated voltage which is 110V. so the taps will be
| 121,132,143,154,165,176 and 187V tap. the load on the secondary will be
|
| around 2.1K.
|
|
| i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
| turns, I-Current. The secondary load current is 43mA. N2 is currently
| 3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
| first tap. so from the above expression I1 can be found, and then N1
| can be found. this is repeated for remaining taps. This is how i ve
| done the designing. Is this OK or any other method has to be adopted.
|
|
| Please help me.
|
|
| Thanking You
|
|
| Regards
| Pavan
This appears to be a homework question, Mr Designer.
Never mind, let's see what we can do.
Let us label the taps thus:
T0
T1
T2
T3
T4
T5
T6
Now let us put the number of turns against each tap.
T0 - 0
T1 - 3600
T2 - 3600*110% = 3960
But this is an autotransformer, so we have 3600 turns between T0 and T1,
needing 360 turns between T1 and T2 for a total of 3960 turns
between T0 and T2.
T0 0
T1 3600
T2 3600*110% = 3600 + 360 = 3960
T3 = ?
T4 = ?
T5 = ?
T6 = ?
I'll let you fill in the values.
WHAT IF we applied 110V to T2?
T0 0 0V
T1 3600 ?
T2 3960 110V
T3 = ?
T4 = ?
T5 = ?
T6 = ?
Can you see that we'd have 100V at T1?
But we have a problem, 100V is too low, we need 106V.
We can apply the input voltage of 110V at T6
and draw a lesser voltage of 106V from one of the taps T1-T5
T0 0 0V
T1 3600 ?
T2 3960 ?
T3 = ? ?
T4 = ? ?
T5 = ? ?
T6 = ? 110 V
OR
we can apply the input voltage of 110V at T5
and draw a lesser voltage from one of the taps T1 to T4
T0 0 0V
T1 3600 ?
T2 3960 ?
T3 = ? ?
T4 = ? ?
T5 = ? 110V
T6 = ? ?
If you've understood the principle, you can find 106 V.
Androcles
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Autotransformer Design?? |
25 Oct 2006 05:08:17 AM |
|
|
"pavan" <pavannaren@gmail.com> wrote in message
news:1161761967.972208.319150@i42g2000cwa.googlegroups.com...
| Dear Sir,
| I understand the design methodology of calculating the
| turns using 110% of previous Tap. But what is practically happening is
| the voltage is not the same when i apply tap voltage on the primary. It
| is diminishing with tap. That is across 121V when i apply 121V, the
| secondary is slightly less than 106. then when i apply 132V across the
| 132V tap, the secondary voltage is still lesser. This continues and
| when i apply 187V across 187V tap, the seconadry voltage is only around
| 90 odd Volts. So this is primarily because when the load of around 2.1K
| is connected, there is a volatge drop that is happening.
|
| this has to be addressed. so for that reason i took into account the
| load current of 43mA and calculated the equivalent primary current.
| then using the transformation formula i calculated the number of turns,
| which i ve written in the mail.
|
| Please note this and help me.
|
| thank You for all Your help.
| regards
| Pavan
Any drop in voltage as a result of load is solely due to the internal
winding resistance of the transformer.
I have no way of measuring that and you have not stated a value.
However, I will now calculate as follows.
2.1Kohms * 43 mA = 90.3 V
106-90 = 16 volts "lost" to winding resistance.
Internal resistance = 16V/43mA = 372 ohms
Wattage is 16 * 0.043 = 0.67, you should not need cooling but it will
run warm.
Resistor wattage is about 4 watts, that'll burn your fingers or
start a fire on a PCB, or at least melt solder.
If you want 106 V across 2.1K then you'll need a current of
106V/2100Kohms = 50 mA.
Now you are up to 5 watts. - That's HOT.
If you wish to reduce the internal resistance you'll need
thicker wire.
AWG tables can tell you the resistance of copper wire,
http://www.powerstream.com/Wire_Size.htm
I have no idea of the length and thickness of wire you are using
because I don't know the dimensions of the transformer,
but 40 AWG has a resistance of about 1 ohm per foot and
will carry 90 mA.
Rough check:
With a resistance of 372 ohms, that's about 370 feet of wire
and 3600 turns is about 370/3600 = 0.1 feet per turn, or 1.25
inches of wire per turn, your transformer has sides of about
1.25"/4 = 0.3" using 40 AWG copper with a thickness of 0.003".
Thicker wire will mean taking more room and hence being longer.
|
| Androcles wrote:
|
| > "pavan" <pavannaren@gmail.com> wrote in message
| > news:1161685098.760087.77820@h48g2000cwc.googlegroups.com...
| > | Dear Reader,
| > | First of all let me thank you for taking interest
| > | in lending a helping hand for my issue.
| > |
| > | The issue is that i ve to design one autotransformer with secondary
| > | tapped from the primary and is kept at a voltage equal to 106V. The
| > | primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
| > | rated voltage which is 110V. so the taps will be
| > | 121,132,143,154,165,176 and 187V tap. the load on the secondary will
be
| > |
| > | around 2.1K.
| > |
| > |
| > | i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
| > | turns, I-Current. The secondary load current is 43mA. N2 is currently
| > | 3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
| > | first tap. so from the above expression I1 can be found, and then N1
| > | can be found. this is repeated for remaining taps. This is how i ve
| > | done the designing. Is this OK or any other method has to be adopted.
| > |
| > |
| > | Please help me.
| > |
| > |
| > | Thanking You
| > |
| > |
| > | Regards
| > | Pavan
| >
| > This appears to be a homework question, Mr Designer.
| > Never mind, let's see what we can do.
| >
| > Let us label the taps thus:
| > T0
| > T1
| > T2
| > T3
| > T4
| > T5
| > T6
| >
| > Now let us put the number of turns against each tap.
| >
| > T0 - 0
| > T1 - 3600
| > T2 - 3600*110% = 3960
| >
| > But this is an autotransformer, so we have 3600 turns between T0 and T1,
| > needing 360 turns between T1 and T2 for a total of 3960 turns
| > between T0 and T2.
| >
| > T0 0
| > T1 3600
| > T2 3600*110% = 3600 + 360 = 3960
| > T3 = ?
| > T4 = ?
| > T5 = ?
| > T6 = ?
| >
| > I'll let you fill in the values.
| >
| > WHAT IF we applied 110V to T2?
| >
| > T0 0 0V
| > T1 3600 ?
| > T2 3960 110V
| > T3 = ?
| > T4 = ?
| > T5 = ?
| > T6 = ?
| >
| > Can you see that we'd have 100V at T1?
| > But we have a problem, 100V is too low, we need 106V.
| >
| > We can apply the input voltage of 110V at T6
| > and draw a lesser voltage of 106V from one of the taps T1-T5
| >
| > T0 0 0V
| > T1 3600 ?
| > T2 3960 ?
| > T3 = ? ?
| > T4 = ? ?
| > T5 = ? ?
| > T6 = ? 110 V
| >
| > OR
| > we can apply the input voltage of 110V at T5
| > and draw a lesser voltage from one of the taps T1 to T4
| >
| > T0 0 0V
| > T1 3600 ?
| > T2 3960 ?
| > T3 = ? ?
| > T4 = ? ?
| > T5 = ? 110V
| > T6 = ? ?
| >
| > If you've understood the principle, you can find 106 V.
| > Androcles
|
.
|
|
|
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: Autotransformer Design?? |
24 Oct 2006 09:01:16 AM |
|
|
On 24 Oct 2006 03:18:18 -0700, "pavan" <pavannaren@gmail.com> wrote:
Dear Reader,
First of all let me thank you for taking interest
in lending a helping hand for my issue.
The issue is that i ve to design one autotransformer with secondary
tapped from the primary and is kept at a voltage equal to 106V. The
primary has 7 taps ranging from 0 Volts to 187V in steps of 110% of
rated voltage which is 110V. so the taps will be
121,132,143,154,165,176 and 187V tap. the load on the secondary will be
around 2.1K.
i ve used the formula V2/V1=N2/N1=I1/I2, Where V-Voltage, N-No of
turns, I-Current. The secondary load current is 43mA. N2 is currently
3600 which is fixed. Since V2 is needed to be 106V, V1 is 121V for
first tap. so from the above expression I1 can be found, and then N1
can be found. this is repeated for remaining taps. This is how i ve
done the designing. Is this OK or any other method has to be adopted.
Please help me.
Thanking You
Regards
Pavan
You're making this too tough.
You specify that the output must be V2 = 106V and N2 = 3600 turns.
Divide to get the core density:
106V/3600turns = 0.0294V/turn or
34 turns/volt
as a design constant. Use these values to find your input tap.
If your input voltage is 110 you have to tap in at
110x34 = 3736 turns or in general
Ntap = 34 x input volts.
Two suggestions:
1. You have far too many turns on the transformer. At 60 cps you can
run as low as 1 volt per turn so your secondary could be,
conservatively, 125 turns instead of 3600. (At 50 cps you need 1.2
turns per volt. The saturation of the iron depends on the integral of
volt-seconds, not current).
2. You'll have a hard time getting closer than 10% with 10% taps, so
why not use a second secondary tap 5% apart so you can get to 5%?
You don't have to consider current in your design. You'll get your 4.5
watts OK.
John Polasek
.
|
|
|
|
|
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|
|
