How to create a single-FET magnetic field generator circuit
- 1 Main Items
- 2 Get required software
- 3 Learn how to use CAD software
- 4 Choose a suitable magneto-optic material
- 5 Create a field generation coil
- 6 Choose a suitable MOSFET
- 7 Choose suitable bypass and filtering capacitors
- 8 Choose a suitable current-sense resistor
These main items are discussed in detail in Magnetic Field Generator Design for Magneto-Optic Switching Applications.
- Get required software (simulation and layout)
- Learn how to use CAD software
- Choose a suitable magneto-optic material (e.g. here)
- Create a field generation coil
- Choose a suitable MOSFET
- Choose suitable bypass and filtering capacitors
- Choose a suitable current-sense resistor
- Add miscellaneous passive and protection components
- Manufacture PCB
Get required software
The two softwares used in the HSSE Lab include Orcad Capture CIS Lite and Eagle PCB Designer. Downloads for these can be found here:
Learn how to use CAD software
Choose a suitable magneto-optic material
This step is specific to creating magneto-optic interferometers, but relates to the application of which the field generator circuitry will be used. There are magneto-optic materials that have low and high field requirements, and each have their own advantages and dis advantages. These materials can be purchased from Integrated Photonics, Inc.:
For the moment, let's assume the FLM Low Moment rotator crystal has been selected for use in this application.
Since the material will be cut into 1x1mm slabs, the required field must be between 0 - 225 Oe.
Create a field generation coil
As in the image shown below, we create our coil based on geometrical constraints. We will choose this particular coil to have the following properties:
- Length: 6mm
- Radius: 1.5mm
- Number of turns: 16
- Field required: 225 Oe (225 G with μr=1)
Choose a suitable MOSFET
To choose a suitable MOSFET there are many things to take into consideration. Some of the main questions to ask before searching include:
- How much current is required by the MOSFET?
- How fast does the MOSFET need to be?
- Is there a limit to the voltage I can apply on Vgs and Vds?
Determining current required by the MOSFET
For magnetic field generators, the current required by the MOSFET is usually a function of the coil used to create the magnetic field. Once this coil has been chosen and the field requirement decided, the required current can be calculated. Using the above example, we find:
The current required to create 225 G (0.0225 T) using this coil is: ~7.5 A
Determining MOSFET speed
Usually we want our MOSFET switch to be as fast as possible, and thus we must optimize the rise and fall time of the FET choice. We will always be limited by the coil inductance, however keeping everything else optimized is essential. When choosing the MOSFET, look for low gate charge (Qg). Although there are many other factors that come into play, usually low gate charge results in fast rise/fall times for the FET.
For example, seethe IRF3714z datasheet here. This MOSFET has a rise time of 13ns and a fall time of 5ns, given a gate charge of about 2.6nC of gate charge (pre + post Vth).
Determining Vgs and Vds limitations
These are things that can be found on the MOSFET datasheet (e.g. see the IRF3714z datasheet). One can, in simulation, determine the I-V characteristics for different Vgs and Vds values, demonstrated in the how-to video above.
The goal of this is to find the proper choice of Vds such that your range of Vgs values gives you the desired current levels (without unintentionally destroying the MOSFET).
Choose suitable bypass and filtering capacitors
The bypass capacitors (see basic single-fet image at top of page) are essential in creating an effective magnetic field generator. These capacitors "bypass" the DC source by acting as little energy storage tanks for the field generation circuit. When fast, high-current pulses are required by the coil, the charge is pulled from these capacitors instead of the DC supply.
To calculate the proper bypass capacitor values, the following must be known:
- Required peak current
- Required pulse width
- Required DC source voltage (Vds)
Let us assume, based on the examples above, that the following is required:
- Peak current: 7.5 A
- Pulse width: 100 μs
- DC voltage: 15 V
We first calculate the amount of charge (in C) required for the pulse:
- 7.5 [C/s] * 100 [μs] = 0.75 [mC]
We then determine the minimum capacitance required:
- 0.75 [mC] / 15 [V] = 50 [μF]
However, it is good practice to double or triple this value, to be safe and allow for flexibility:
- 50 [μF] * 3 = 150 [μF]
Two of these tantalum capacitors could be purchased to accommodate for this amount of capacitance.
Finally, small ceramic capacitors are useful for filtering noise from the power line. Purchasing two of these 0.1 μF capacitors and placing them in parallel with the bypass capacitors filters unwanted noise.
Choose a suitable current-sense resistor
The current sense resistor allows one observe the electronic pulse output and relate the measured value to the current through the coil. The choice of the resistor is dependent on the:
- Peak current expected
- MOSFET Vgs behavior
- Measurement equipment resolution
In our calculations, we determined that our coil will require approximately 7.5 A to generate the appropriate magnetic field. Thus, we must choose a current-sense resistor that is of low value, but high enough such that the voltage generated across from it will be detected by our measurement equipment. Additionally, the voltage generated across this resistor will drop our Vgs as the current increases, thus limiting total current capability.
For example, if we choose a resistor value of 1 Ω, the peak current (7.5 A) will generate 7.5 V across the resistor and thus drop Vgs by that amount. That is a lot! We must choose a smaller resistor. If we choose 0.05 Ω, the peak current will only generate 0.375 V across the resistor, which is much better. However, we must then ensure that our measurement equipment has good enough resolution to detect the 0 - 375 mV range desirably.
Finally, the resistor must have a good power rating, given the desired pulse duty cycle. Low-power resistors can handle very high currents, provided the duty cycle of the pulse is very low. The average power dissipated across the resistor must be lower than its rated power, but the instantaneous power can be much greater than the rating. For example, consider this 0.05 Ω, 25 W resistor. I would like pass a current of 150 A through this resistor. The instantaneous power would be 1.125 kW, or 45x its rated power. However, if I were to apply a pulse width of 1 μs with a 10 ms period (duty cycle of 0.01%), the average power is only 0.1125 W (using this relation).