Drive a 12 Ohm, 50mH Transformer Coil with 12V @ 1A

In summary, a power converter can be created by connecting a transformer to a capacitor and a load resistor.
  • #1
BHY-BK
18
1
I have a signal generator that is used to generate 12kHz square waves. I want to use my signal generator to drive a transformer primary coil. My coil has 12 ohms resistance, 50mH, and will have 12V applied to it @ 1A.

Can I buy a circuit to drive the coil from the signal generator pulse or do I have to make one?
What is the correct terminology for such a circuit?
 
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  • #2
Do this:
levitatorcoildriver.gif

You can get fancy to drive it with faster edges, or use a power fet instead of a 2n3055, or do a search on coil driver and look at the images for one you like
 
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  • #3
BHY-BK said:
I have a signal generator that is used to generate 12kHz square waves. I want to use my signal generator to drive a transformer primary coil. My coil has 12 ohms resistance, 50mH, and will have 12V applied to it @ 1A.
12 V square wave ? Does it go from 0V to +12V or from -6 to +6 ?
What's the part number of your transformer ? Does it have a small air gap in the core ?
 
  • #4
Sorry, it's 0V to +12V, transformer is one a friend made for me, it has a small gap in the core but it's not a flyback. It's a step up transformer with a 1:5 turns ratio.
 
  • #5
meBigGuy said:
Do this:
levitatorcoildriver.gif

You can get fancy to drive it with faster edges, or use a power fet instead of a 2n3055, or do a search on coil driver and look at the images for one you like
Will my 5V signal from my signal generator be able to drive the 2N3055?
Dumb question I know, it's just been a while since I played around with electronics and I've forgotten a lot.
 
  • #6
Probably. Do you have a model number?
Looking at the specs for an MJE3055 in a TO-220 package, it is rated for 0.6W without a heatsink. That's borderline for 1 amp. I'd use the to-220 and add a small heatsink.
Like this http://www.mouser.com/ProductDetail/Aavid-Thermalloy/6237BG/?qs=sGAEpiMZZMttgyDkZ5WiuqcEpFN0QqoQfmtMCSJ6UoA%3d

For 1A, you probably want at least 100ma base current which would be at around 4volts with a 300 ohm resistor. Can your function generator drive 100ma? If it has a 50 ohm output, then no problem.

If there is not enough drive you can go to a 2 stage circuit, or a darlington, or a mosfet. Regardless, a TO-220 package with a clip-on heatsink will be the most robust solution.

Don't forget the diode --- the transistor will blow without it.
 
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  • #7
meBigGuy said:
Probably. Do you have a model number?
Looking at the specs for an MJE3055 in a TO-220 package, it is rated for 0.6W without a heatsink. That's borderline for 1 amp. I'd use the to-220 and add a small heatsink.
Like this http://www.mouser.com/ProductDetail/Aavid-Thermalloy/6237BG/?qs=sGAEpiMZZMttgyDkZ5WiuqcEpFN0QqoQfmtMCSJ6UoA%3d

For 1A, you probably want at least 100ma base current which would be at around 4volts with a 300 ohm resistor. Can your function generator drive 100ma? If it has a 50 ohm output, then no problem.

If there is not enough drive you can go to a 2 stage circuit, or a darlington, or a mosfet. Regardless, a TO-220 package with a clip-on heatsink will be the most robust solution.

Don't forget the diode --- the transistor will blow without it.
Cool, thanks for the suggestions, I'll use them. The function generator is 50 ohms! Any idea how fast the rise and fall time will be?
Would adding a buffer like a 4050 reduce the rise and fall time?
 
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  • #8
Again, it depends on your generators output. Since you are switching an inductor, the turn-on/off time may not have much effect overall. You can speed up the transistor switching with a very small cap across the 300 ohm, but again, it won't change much. If the 4050 is faster than your generator, then that will help.

If speed is really an issue, then there may be better solutions than the 3055, like a power mosfet and a hefty mosfet driver (mosfets have huge gate capacitance)
There is probably some power mosfet you can connect directly to the generator and drive it without a heatsink or a series resistor.

I found some mosfets in this thread. Maybe that's a better way to go.
https://www.physicsforums.com/threads/electromagnet-and-mosfets.831892/#post-5224970
 
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  • #9
What's connected to the secondary of this transformer ?
 
  • #11
Connected to the secondary of the transformer is capacitor and resistive load of about 300 ohms
 

1. How do I determine the power rating for a 12 Ohm, 50mH transformer coil?

The power rating for a transformer coil can be determined by multiplying the voltage (12V) by the current (1A) to get a power output of 12 watts. This value is the maximum amount of power that can be safely delivered by the transformer coil.

2. What is the function of a transformer coil?

A transformer coil is used to transfer electrical energy from one circuit to another through the process of electromagnetic induction. It can step up or step down the voltage and current levels depending on the number of turns in the coil.

3. How do I calculate the number of turns in a transformer coil?

The number of turns in a transformer coil can be calculated using the formula: N = (V x 10^8)/(4.44 x f x B x A), where N is the number of turns, V is the voltage, f is the frequency, B is the magnetic flux density, and A is the cross-sectional area of the core.

4. What is the significance of the 50mH inductance in the transformer coil?

The 50mH inductance in the transformer coil refers to the amount of inductance present in the coil, which affects its ability to store and transfer energy. A higher inductance value results in a slower rate of change in current and a larger voltage drop across the coil.

5. What precautions should be taken when using a transformer coil?

When using a transformer coil, it is important to make sure that the input voltage and current do not exceed the rated values to avoid damaging the coil. It is also important to properly insulate and secure the coil to prevent electrical hazards. Additionally, it is recommended to use a fuse or circuit breaker in the circuit to protect against overloading the coil.

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