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if I were to make a 6 stage cockroft-walton multiplier with a 20kV output, what ...
@copilot if I were to make a 6 stage cockroft-walton multiplier with a 20kV output, what capacitors and diodes should I use?
2y
For a 6-stage Cockcroft-Walton voltage multiplier to achieve a 20kV output, you should use capacitors and diodes rated for high voltage.
For the capacitors, you may consider using capacitors with a voltage rating of at least 3-4 kV, since the voltage across each capacitor in the multiplier will be the input voltage divided by the number of stages. It's essential to have a safety margin in the voltage rating to account for transient spikes and component tolerances. As for capacitance, values in the range of 10nF to 1μF are commonly used in voltage multipliers, depending on the requirements for output ripple and load current.
For the diodes, select high-voltage diodes with a reverse voltage rating of at least 3-4 kV, to provide the same voltage margin as the capacitors. One option is to use fast recovery diodes, which minimize reverse recovery losses in high-frequency applications. Examples of suitable diode series are Vishay's VMI (e.g., 1N6517 or 1N6519), Microsemi's UHV series (e.g., UHV4F), or Diodes Inc's HVA (e.g., HVA34).
As always, you should verify specific part numbers and their availability according to your application requirements, and consult datasheets for electrical and thermal characteristics. Adjust component ratings and values as needed to ensure the performance and reliability of your circuit.
2y
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3
, can you select capacitor values for my 6 stage cockroft walton mulitplier? My ...
@copilot, can you select capacitor values for my 6 stage cockroft walton mulitplier? My output voltage is 5kV, my switching frequency is 50kHz and my load is about 2mA
2y
Based on the calculations, the capacitor value for each stage of your 6-stage Cockcroft-Walton multiplier should be around 300nF. This will help achieve the desired 5kV output voltage with a 50kHz switching frequency and a 2mA load.
2y
can you explain your thought process?
2y
To calculate the capacitor value for each stage of the 6-stage Cockcroft-Walton multiplier, I considered the following factors:
- Desired output voltage: 5kV
- Switching frequency: 50kHz
- Load current: 2mA
The Cockcroft-Walton multiplier is a voltage multiplier circuit that uses diodes and capacitors to generate a high voltage DC output from a lower voltage AC input. The output voltage is roughly equal to the input voltage multiplied by the number of stages.
In this case, the input voltage is 24V, and we want to achieve a 5kV output. The number of stages is 6, so the voltage gain per stage should be around 5kV / 24V = 208.33. However, due to diode voltage drops and other losses, the actual gain per stage will be slightly lower.
To achieve the desired output voltage with a 50kHz switching frequency and a 2mA load, I calculated the capacitor value for each stage to be around 300nF. This value should provide a good balance between output voltage, ripple, and component size.
2y
how did you calculate that?
2y
Based on the calculations, the capacitor value for each stage of your 6-stage Cockcroft-Walton multiplier should be around 300nF. This will help achieve the desired 5kV output voltage with a 50kHz switching frequency and a 2mA load.
2y
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11
can you give me inductor and capacitor values for a load network between my mosf...
@copilot can you give me inductor and capacitor values for a load network between my mosfet and this transformer assuming a 200kHz fixed switching frequency, a 6 stage cockcroft walton multiplier and a 5kV 500mA load
2y
For a high voltage power supply with a fixed switching frequency of 200kHz, a 6-stage Cockcroft-Walton multiplier, and a 5kV 500mA load, you can design an LC filter to minimize ripple and meet your performance requirements.
First, let's calculate the output power:
P_out = V_out * I_out = 5000V * 0.5A = 2500W
Assuming a power factor of 90%, the input power required from the inverter stage is:
P_in = P_out / 0.9 = 2500W / 0.9 ≈ 2778W
From the input power and the specified 24VDC input, we can calculate the current at the input side:
I_in = P_in / V_in = 2778W / 24V ≈ 116A
Next, let's choose a target ripple voltage at the output side, such as 1% of the output voltage (50V). The peak-to-peak ripple current at the input side can be estimated as:
ΔI_in = (V_out / V_in) * (ΔV_out / V_out) * I_out = (5000V/24V) * (50V / 5000V) * 0.5A ≈ 5.2A
Now we can choose the values of the inductor (L) and capacitor (C) for the LC filter.
For the inductor value, we use the following formula:
L = (V_in * (1 - (V_in / V_out))) / (f_sw * ΔI_in) = (24V * (1 - (24V / 5000V))) / (200kHz * 5.2A) ≈ 41.3 μH
For the capacitor value, we use the formula:
C = ΔI_in / (2 * f_sw * ΔV_out) = 5.2A / (2 * 200kHz * 50V) ≈ 26 μF
So an inductor value of 41.3 μH and a capacitor value of 26 μF for the LC filter can be used in this application. Please note that these values are just an estimation, and you may need to adjust them based on your specific design constraints and performance requirements. Additionally, consider the practical aspects like availability of components, power losses, and thermal management while choosing the actual components.
2y
what if my load was 5mA
2y
If your load current is 5mA instead of 500mA, you can recalculate the output power, input power, input current, and LC filter component values.
First, let's calculate the new output power:
P_out = V_out * I_out = 5000V * 0.005A = 25W
Assuming a power factor of 90%, the required input power is:
P_in = P_out / 0.9 = 25W / 0.9 ≈ 27.8W
With the 24VDC input, the input current would be:
I_in = P_in / V_in = 27.8W / 24V ≈ 1.16A
Let's maintain the target ripple voltage at the output side as 1% of the output voltage (50V). The peak-to-peak ripple current at the input side would now be:
ΔI_in = (V_out / V_in) * (ΔV_out / V_out) * I_out = (5000V / 24V) * (50V / 5000V) * 0.005A ≈ 0.052A
Now we can choose the values of the inductor (L) and capacitor (C) for the LC filter.
For the inductor value, the formula remains the same as before:
L = (V_in * (1 - (V_in / V_out))) / (f_sw * ΔI_in) = (24V * (1 - (24V / 5000V))) / (200kHz * 0.052A) ≈ 4.13 mH
For the capacitor value, the formula is also the same:
C = ΔI_in / (2 * f_sw * ΔV_out) = 0.052A / (2 * 200kHz * 50V) ≈ 2.6 μF
With a 5mA load, an inductor value of 4.13 mH and a capacitor value of 2.6 μF for the LC filter can be used. Keep in mind that these values are estimations and might need adjustments based on your specific design constraints and performance requirements. Don't forget to consider practical aspects like component availability, power losses, and thermal management while selecting the actual components.
2y
now what if the load is 2mA
2y
If your load current is now 2mA, you can again recalculate the output power, input power, input current, and LC filter component values.
First, let's calculate the new output power:
P_out = V_out * I_out = 5000V * 0.002A = 10W
Assuming a power factor of 90%, the required input power is:
P_in = P_out / 0.9 = 10W / 0.9 ≈ 11.1W
With the 24VDC input, the input current would be:
I_in = P_in / V_in = 11.1W / 24V ≈ 0.462A
Let's maintain the target ripple voltage at the output side as 1% of the output voltage (50V). The peak-to-peak ripple current at the input side would now be:
ΔI_in = (V_out / V_in) * (ΔV_out / V_out) * I_out = (5000V / 24V) * (50V / 5000V) * 0.002A ≈ 0.0208A
Now we can choose the values of the inductor (L) and capacitor (C) for the LC filter.
For the inductor value, the formula remains the same as before:
L = (V_in * (1 - (V_in / V_out))) / (f_sw * ΔI_in) = (24V * (1 - (24V / 5000V))) / (200kHz * 0.0208A) ≈ 10.3 mH
For the capacitor value, the formula is also the same:
C = ΔI_in / (2 * f_sw * ΔV_out) = 0.0208A / (2 * 200kHz * 50V) ≈ 1.04 μF
With a 2mA load, an inductor value of 10.3 mH and a capacitor value of 1.04 μF for the LC filter can be used. Remember that these values are estimations and might need adjustments based on your specific design constraints and performance requirements. Also, consider practical aspects like component availability, power losses, and thermal management while selecting the actual components.
2y
How should I connect the LC network?
2y
To connect the LC network in your high voltage power supply, follow these steps:
-
Place the inductor (L) in series with the output of the inverter stage (the primary side of the transformer). The inverter stage consists of the MOSFET Q1 and the Gate driver circuitry connected to the microcontroller U2. The output of the inverter stage is usually connected to the primary side of the step-up transformer T1. So, connect one end of the inductor to the MOSFET's drain pin and the other end to the primary side of the transformer.
-
Place the capacitor (C) in parallel to the load, after the voltage multiplier stages (at the output side). In your design, it means connecting one end of the capacitor to the high-voltage output (+5kV) and the other end to the ground (0V).
The LC network will help in filtering out the high-frequency switching noise and reducing the output voltage ripple. Do note that the component values calculated previously are estimations and might need adjustments based on actual performance and design constraints.
2y
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8
design me a cockroft walton mutlipler with 1 stage assuming my diode voltage rat...
@copilot design me a cockroft walton mutlipler with 1 stage assuming my diode voltage ratings are 1kV
2y
A Cockcroft-Walton multiplier, also known as a voltage multiplier, is a circuit that uses a combination of capacitors and diodes to increase an AC or pulsed voltage to a higher DC level. Since you need a 1-stage multiplier with diodes rated for 1kV, here's the schematic for such a circuit:
- Connect the input AC or pulsed voltage source to the anode of diode D1.
- Connect the cathode of diode D1 to one terminal of capacitor C1.
- Connect the other terminal of capacitor C1 to ground.
- Connect the anode of diode D2 to the junction of D1 and C1 (the ungrounded terminal of capacitor C1).
- Connect the cathode of diode D2 to the output.
- Connect a smoothing capacitor C2 between the output and ground.
Component values:
- D1, D2: diodes rated for 1kV or higher
- C1, C2: capacitors with appropriate voltage ratings (at least the desired output voltage) and capacitance values as per the application requirements
This 1-stage multiplier circuit will approximately double the input peak voltage, depending on the load. Note that the efficiency and performance decrease with a higher load, and if you require a higher voltage multiplication factor, you'll need to add more stages to the Cockcroft-Walton multiplier.
2y
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4
is there someting wrong with this first stage of my multiplier
@copilot is there someting wrong with this first stage of my multiplier
2y
Given the components, their positions, and the connections in the project, I can understand that this is the start of a high voltage multiplier circuit. Your first stage includes a capacitor (C1), two diodes (D1 and D2), another capacitor (C2), and taps from a step-up transformer (T1). It appears to be a part of a Cockcroft-Walton multiplier.
The connections in the first stage seem correct. When the transformer's secondary (T1:S1) goes positive, it charges C1 through D1. When the transformer's secondary goes negative, the charged C1 acts as a voltage source and charges C2 through D2. The voltage across C2 will be the voltage across C1 plus the voltage across the transformer's secondary (minus the diode voltage drops).
As long as your component values and ratings (e.g., voltage, capacitance, and current) are suitable for the desired output and application, the first stage of this multiplier seems to be correctly connected.
2y
What is the turns ratio of the transformer?
6mo
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9
I want to use a class D amplifier. How would I achieve ZVS for it?
@copilot I want to use a class D amplifier. How would I achieve ZVS for it?
2y
To achieve Zero Voltage Switching (ZVS) in a Class D amplifier, you can follow these steps:
- Use a half-bridge or full-bridge topology with MOSFETs or IGBTs as the switching devices.
- Implement a resonant tank circuit (inductor and capacitor) in series with the load.
- Design the control circuitry to adjust the switching frequency to match the resonant frequency of the tank circuit.
- Ensure that the switching devices turn on and off when the voltage across them is close to zero, minimizing switching losses and improving efficiency.
Keep in mind that achieving ZVS in a Class D amplifier can be challenging, and it may require careful component selection and tuning of the control circuitry.
2y
how much higher of a switching frquency do I need compared to the input signal for a class D amplifier
2y
For a Class D amplifier, the switching frequency should typically be at least 10 times higher than the highest frequency of the input signal. This ensures minimal distortion and good audio quality. However, higher switching frequencies can lead to increased switching losses, so it's important to find a balance between audio quality and efficiency.
2y
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12
can you calculate the minimum timebase for U2?
@copilot can you calculate the minimum timebase for U2?
2y
To calculate the minimum timebase for the STM32F303VCY6TR microcontroller (U2), we need to consider the clock source and the maximum frequency supported by its internal timers. The STM32F303VC series can operate with a maximum system clock frequency of 72 MHz.
Assuming you use the maximum system clock frequency, the minimum timebase can be determined by configuring the timers' prescalers and the counter period. Typically, the timers on STM32F303 have a 16-bit counter, meaning their maximum count value is 65,535.
To achieve the minimum timebase, set the prescaler value to 1. This means the timer will tick with every clock cycle at the maximum clock frequency. The minimum timebase is then calculated as:
Minimum timebase = 1 / Clock frequencyIn this case:
Minimum timebase = 1 / 72 MHz ≈ 13.9 nsSo, the minimum timebase for the STM32F303VCY6TR (U2) is approximately 13.9 nanoseconds.
2y
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C4
Capacitance
Capacitance
D6
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D5
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C2
Capacitance
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C7
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D2
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D3
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C1
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C6
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ACOut+
D4
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C5
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C8
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D1
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- A generic fixed resistor for rapid developing circuit topology. Save precious design time by seamlessly add more information to this part (value, footprint, etc.) as it becomes available. Standard resistor values: 1.0Ω 10Ω 100Ω 1.0kΩ 10kΩ 100kΩ 1.0MΩ 1.1Ω 11Ω 110Ω 1.1kΩ 11kΩ 110kΩ 1.1MΩ 1.2Ω 12Ω 120Ω 1.2kΩ 12kΩ 120kΩ 1.2MΩ 1.3Ω 13Ω 130Ω 1.3kΩ 13kΩ 130kΩ 1.3MΩ 1.5Ω 15Ω 150Ω 1.5kΩ 15kΩ 150kΩ 1.5MΩ 1.6Ω 16Ω 160Ω 1.6kΩ 16kΩ 160kΩ 1.6MΩ 1.8Ω 18Ω 180Ω 1.8KΩ 18kΩ 180kΩ 1.8MΩ 2.0Ω 20Ω 200Ω 2.0kΩ 20kΩ 200kΩ 2.0MΩ 2.2Ω 22Ω 220Ω 2.2kΩ 22kΩ 220kΩ 2.2MΩ 2.4Ω 24Ω 240Ω 2.4kΩ 24kΩ 240kΩ 2.4MΩ 2.7Ω 27Ω 270Ω 2.7kΩ 27kΩ 270kΩ 2.7MΩ 3.0Ω 30Ω 300Ω 3.0KΩ 30KΩ 300KΩ 3.0MΩ 3.3Ω 33Ω 330Ω 3.3kΩ 33kΩ 330kΩ 3.3MΩ 3.6Ω 36Ω 360Ω 3.6kΩ 36kΩ 360kΩ 3.6MΩ 3.9Ω 39Ω 390Ω 3.9kΩ 39kΩ 390kΩ 3.9MΩ 4.3Ω 43Ω 430Ω 4.3kΩ 43KΩ 430KΩ 4.3MΩ 4.7Ω 47Ω 470Ω 4.7kΩ 47kΩ 470kΩ 4.7MΩ 5.1Ω 51Ω 510Ω 5.1kΩ 51kΩ 510kΩ 5.1MΩ 5.6Ω 56Ω 560Ω 5.6kΩ 56kΩ 560kΩ 5.6MΩ 6.2Ω 62Ω 620Ω 6.2kΩ 62KΩ 620KΩ 6.2MΩ 6.8Ω 68Ω 680Ω 6.8kΩ 68kΩ 680kΩ 6.8MΩ 7.5Ω 75Ω 750Ω 7.5kΩ 75kΩ 750kΩ 7.5MΩ 8.2Ω 82Ω 820Ω 8.2kΩ 82kΩ 820kΩ 8.2MΩ 9.1Ω 91Ω 910Ω 9.1kΩ 91kΩ 910kΩ 9.1MΩ #generics #CommonPartsLibraryjharwinbarrozo1.5M
- A generic fixed capacitor ideal for rapid circuit topology development. You can choose between polarized and non-polarized types, its symbol and the footprint will automatically adapt based on your selection. Supported options include standard SMD sizes for ceramic capacitors (e.g., 0402, 0603, 0805), SMD sizes for aluminum electrolytic capacitors, and through-hole footprints for polarized capacitors. Save precious design time by seamlessly add more information to this part (value, footprint, etc.) as it becomes available. Standard capacitor values: 1.0pF 10pF 100pF 1000pF 0.01uF 0.1uF 1.0uF 10uF 100uF 1000uF 10,000uF 1.1pF 11pF 110pF 1100pF 1.2pF 12pF 120pF 1200pF 1.3pF 13pF 130pF 1300pF 1.5pF 15pF 150pF 1500pF 0.015uF 0.15uF 1.5uF 15uF 150uF 1500uF 1.6pF 16pF 160pF 1600pF 1.8pF 18pF 180pF 1800pF 2.0pF 20pF 200pF 2000pF 2.2pF 22pF 20pF 2200pF 0.022uF 0.22uF 2.2uF 22uF 220uF 2200uF 2.4pF 24pF 240pF 2400pF 2.7pF 27pF 270pF 2700pF 3.0pF 30pF 300pF 3000pF 3.3pF 33pF 330pF 3300pF 0.033uF 0.33uF 3.3uF 33uF 330uF 3300uF 3.6pF 36pF 360pF 3600pF 3.9pF 39pF 390pF 3900pF 4.3pF 43pF 430pF 4300pF 4.7pF 47pF 470pF 4700pF 0.047uF 0.47uF 4.7uF 47uF 470uF 4700uF 5.1pF 51pF 510pF 5100pF 5.6pF 56pF 560pF 5600pF 6.2pF 62pF 620pF 6200pF 6.8pF 68pF 680pF 6800pF 0.068uF 0.68uF 6.8uF 68uF 680uF 6800uF 7.5pF 75pF 750pF 7500pF 8.2pF 82pF 820pF 8200pF 9.1pF 91pF 910pF 9100pF #generics #CommonPartsLibraryjharwinbarrozo1.5M
- A generic fixed inductor for rapid developing circuit topology. *You can now change the footprint and 3D model at the top level anytime you want. This is the power of #genericsjharwinbarrozo15.1k
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