calculating turns ratio in a transformer

Discussion in 'Homework Help' started by struglnstudnt, Sep 6, 2010.

1. struglnstudntNew Member

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A transformer has a ratings plate: 4KVA
Vp=400V
Vs=80V
Rp=1.5Ω
Rs=0.1Ω

I am aware that Vp/Vs=Np/Ns for the (ideal) transformer but how do I go about calculating the turns ratio for the above (real) transformer. Do I do the same calculation or is it different?

Help on this would be much appreciated as I am struggling finding information on the internet for this question.

2. colin55Well-Known Member

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The turns ratio is 400:80

For every 400 turns on the primary, you will put 80 turns on the secondary. All the other data is for other calculations.
Your transformer may need 1 turn per volt and this will require 400 turns and 80 turns on the secondary.
It may need 5 turns per volt 2000 turns to 400 turns
or it may need 10 turns per volt. 4,000 turns to 800 turns The answer is always the same ratio of primary to secondary turns.
In all cases you will get 80v out of the transformer.

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Hello,

The transformer has primary voltage Vp at the terminals of 400v, secondary voltage Vs of 80v with full load, and primary current with full load Ip and secondary current Is with full load.

One way to handle the transformer with primary and secondary resistances is to simply subtract the voltage dropped by the primary current and primary resistance from the primary voltage and use that as the new primary voltage, then add the voltage dropped by the secondary resistance and secondary current to the secondary voltage and call that the new secondary voltage, then take the ratio of the new secondary voltage to the new primary voltage and that will be the turns ratio required.

If we formulize this process it comes out to this:
A=(Vs+Is*Rs)/(Vp-Ip*Rp)
where
A is the turns ratio primary to secondary, and because this is a step down transformer you can take 1/A to be the turns ratio here, and
Vs is the secondary voltage, and
Vp is the primary voltage, and
Is is the secondary current, and
Ip is the primary current, and
Rp is the primary resistance, and
Rs is the secondary resistance.

Example:
We have 400v input and we want 80v output with load, and the primary resistance is 1.5 ohms and secondary resistance is 0.1 ohm, and the full load VA is 4kVA.

The primary current is Ip=4000/400.
The secondary current is Is=4000/80.
The primary resistance Rp=1.5 ohms.
The secondary resistance Rs=0.1 ohms.
The primary voltage Vp is 400 volts.
The secondary voltage Vs is 80 volts.
Plugging these into the formula:
A=(Vs+Is*Rs)/(Vp-Ip*Rp)

we come out with:
A=(80+50*0.1)/(400-10*1.5)

and after doing the math we get:
A=0.22077922

which is the primary to secondary turns ratio, and the secondary to primary turns ratio is simply 1/A which is:
1/A=4.52941176

There is one more little catch here however, and that is that there is always some primary excitation current. To get more accurate results, we'd have to look up the excitation current from the manufacturers data sheet and figure that in as an additional voltage drop on the primary side.

Colin:
The turns ratio actually is usually made a little higher to make up for the loss of the primary and secondary resistances. The formula above takes these two resistances into account.

Last edited: Sep 6, 2010

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5. unix60959New Member

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99.9% of the time ideal calculation are close enough for the device to work they way you want it to under 'real' conditions. if you want to know what the turns ratio has to be to make Vp=400V and Vs=80v then you best bet is to measure the input and output voltages and adjust the coils. Note that the resistive properties of the coils change with temperature also you voltage source of Vp is not going to be ideal either.

6. jignesh doshiNew Member

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hi MrAl,

when i used this formulla for higher rating transformer then it's working fine but when i used for some small transformer then this formulla goes wrong, so please tell me is it right???

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Hi,

What formula are you using and what kind of transformer is the "higher rating" transformer, and what kind is the "small" transformer?