stepping down 300v ac

[asad1234]

I want to stepdown 300Vac 50hz to 5v(peak)ac but i have only 220v to 12vrms transformer available can someone suggest how can i use this 220v transformer with 300v. I tried using a potential divider to first divide the voltage but it doesn't works since the primary winding of transformer comes in parallel with the resistance of potential divider circuit and mess up the voltage drop any suggestions??

 

[Harald Kapp]

What is the power?
You may try a series resistor or a series capacitor (Z=1/(2*Pi*f*C) to reduce the voltage across the transformer's primary coil.

 

[asad1234]

The output of the the transformer is connected to a rectifier and is then connected to microcontroller for measuring it .Since output load is very low(low output power) i assume there will be very low input power
How can i choose the value value of series resistor?

 

[Harald Kapp]

You need to know the primary current of the transformer under load condition and at 220V. From teh voltage diference to 300V you get the required voltage drop across the series resistor. Therefore R=(300V-220V)/Iprimary.
Or use a capacitor of the same impedance to reduce loss.

 

[Merlin3189]

Where does the 300V AC come from?

 

[asad1234]

300v is the maximum output of autotransformer

 

[Merlin3189]

So what is the input to the transformer? And if 300V is the maximum output, is there a lower one?

 

[asad1234]

220v input 160v minimum output

 

[Merlin3189]

I want to stepdown 300Vac 50hz to 5v(peak)ac but i have only 220v to 12vrms transformer available can someone suggest how can i use this 220v transformer with 300v.

220v input 160v minimum output

So, you have 220V supply, you use the auto transformer to step it up to 300V,
then you ask, how to use a 220V transformer with this 300V supply!

Why not simply use the 220V transformer with the 220V supply?

That would give you 12V rather than 5V, but that would probably drive a 5V regulator if the load is not too big.

If you want to put the autotransformer into the setup, if you apply the 220V input to the 300V tap and use the 160V tap to supply the 220V transformer, then you get 6.4V out.
220 x (160/300 ) x (12/220) = 6.4

Depending on your rectifying and smoothing arrangement and the load, this could give the right input (~7.5V) into a 5V regulator.
If you need a bit more, feed the 220V into a lower tap than the 300V tap, say somewhere between 230 and 260.

 

[Merlin3189]

Oh! Sorry I've just realised that you wanted the 5V as AC, so that nullifies my whole idea.
In that case you may as well stick with a dropper resistor or capacitor.
But still use the 300V tap for the input. There's no point in stepping the Voltage up from 220 to 300 when you really want it lower. Use both in maximum step down configuration.

 

[KrisBlueNZ]

I think maybe he has a variable transformer, and he wants to monitor the output voltage. He mentioned a minimum voltage of 160 VAC and a maximum of 300 VAC.

 

[asad1234]

I think maybe he has a variable transformer, and he wants to monitor the output voltage. He mentioned a minimum voltage of 160 VAC and a maximum of 300 VAC.

Exactly i want to monitor the output of my variable transformer

 

[Merlin3189]

So what's wrong with a simple resistive potential divider?
If your meter (or sensor?) wants a maximum of 5V when the transformer is giving 300V, then add a resistor in series, which is equal to 59x the meter's (or sensor's) input resistance. If the input resistance is already large, you can reduce it with another resistor in parallel first.

 

[KrisBlueNZ]

So what's wrong with a simple resistive potential divider?

He wants to monitor the voltage using a microcontroller. I assume the microcontroller's 0V rail will be earthed, and the transformer output will have to be floating, so he needs isolation.

 

[KrisBlueNZ]

I guess your simplest option is to put a load resistor across the secondary of your 220VAC->12VAC transformer, and a resistor in series with the primary.

I'll assume a maximum dissipation in the load resistor of about 2W, so you can use a 5W resistor. The load resistance can be calculated as:
R = V^2 / P
= 12^2 / 2
= 72 ohms
Use a 68 ohm, 5W resistor. Probably wirewound.

Now you need to calculate the resistance to connect in series with the primary.
The resistance you see looking into the primary winding of the transformer will be the load resistance multiplied by the square of the turns ratio.
The turns ratio for a 220-to-12V transformer is 220/12 which is 18.333. The square of that is 336.
So the resistance looking into the primary is 68 * 336 which is 22.85 kilohms.

The resistor you connect in series with the primary needs to drop the voltage from 300V to 220V (a drop of 80V) with a load of 22.85 kilohms. The resistance you need is:
series_resistance = dropped_voltage * reflected_load_resistance / primary_voltage
= 80 * 22850 / 220
= 8.3 kilohms approx.
So you would use an 8.2k resistor.

Dissipation in the series resistor is:
P = V^2 / R
= 80^2 / 8200
= 0.8W approx.
So you would use a 2W or 3W resistor.

Because of the voltages involved, I would use a fusible resistor. I couldn't find a suitable fusible resistor on Digikey, and Mouser's search and catalogue facility is crap as usual. Digikey do have lower value fusible resistors with 0.5 watt ratings, and you could connect a few of these in series to make up the required resistance.

Alternatively, you could vary the load resistance upwards and downwards a bit, and rerun those calculations, to see if you can find some numbers that are a better fit to the available components. A spreadsheet makes these calculations easy.

One other thing. You should be using a precision rectifier (Google it) instead of a diode or diode bridge, to convert the output of the voltage transformer from AC to DC. Good luck!

 

[asad1234]

I guess your simplest option is to put a load resistor across the secondary of your 220VAC->12VAC transformer, and a resistor in series with the primary.

I'll assume a maximum dissipation in the load resistor of about 2W, so you can use a 5W resistor. The load resistance can be calculated as:
R = V^2 / P
= 12^2 / 2
= 72 ohms
Use a 68 ohm, 5W resistor. Probably wirewound.

Now you need to calculate the resistance to connect in series with the primary.
The resistance you see looking into the primary winding of the transformer will be the load resistance multiplied by the square of the turns ratio.
The turns ratio for a 220-to-12V transformer is 220/12 which is 18.333. The square of that is 336.
So the resistance looking into the primary is 68 * 336 which is 22.85 kilohms.

The resistor you connect in series with the primary needs to drop the voltage from 300V to 220V (a drop of 80V) with a load of 22.85 kilohms. The resistance you need is:
series_resistance = dropped_voltage * reflected_load_resistance / primary_voltage
= 80 * 22850 / 220
= 8.3 kilohms approx.
So you would use an 8.2k resistor.

Dissipation in the series resistor is:
P = V^2 / R
= 80^2 / 8200
= 0.8W approx.
So you would use a 2W or 3W resistor.

Because of the voltages involved, I would use a fusible resistor. I couldn't find a suitable fusible resistor on Digikey, and Mouser's search and catalogue facility is crap as usual. Digikey do have lower value fusible resistors with 0.5 watt ratings, and you could connect a few of these in series to make up the required resistance.

Alternatively, you could vary the load resistance upwards and downwards a bit, and rerun those calculations, to see if you can find some numbers that are a better fit to the available components. A spreadsheet makes these calculations easy.

One other thing. You should be using a precision rectifier (Google it) instead of a diode or diode bridge, to convert the output of the voltage transformer from AC to DC. Good luck!

Thanks man that was very thorough and helpful