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8. Capacitor Discharge Bonder

Whilst designing a replacement IR emission element [See Repairs: FT IR Source] for a FTIR chemistry analyser, I needed a bonder to weld its Kanthal A-1 [T18a] resistance wire to a solderable lead wire. EBay is flooded with cheap Chinese capacitor discharge bonders, but they are far less powerful than sellers would have you believe and those that appear to work, do so more out of creating a spark than a bond. Typically their energy is in the order of 100s of mJ.​

Deepseek AI

I brainstormed with DeepSeek AI, similar to Iron Man and JARVIS [G28] but minus the fancy virtual 3D. I did manage to teach it how to draw an NMOSFET using text alone (it took 15 iterations!), but then it made a pig's ear of the rest of the trivial circuit and I abandoned this route. 

 

DeepSeek's text description is fine with the caveat you have to question everything it says because it is often wrong, even if you state at the beginning 'do not guess, only use datasheet values' however sometimes it encounters a dichotomy in its heuristic rulebase and chooses the wrong leg. This is understandable given it is only its second release (April 2025; the first was June 2024) and in 2025, AI-assisted design is still in its infancy.

 

In addition, each session is allocated 128k resource units and the closer you are to using them up the more it overwrites what you told it to begin with, resulting in incorrect answers. A simple rule of thumb is if the vertical window scroll bar is about as high as this symbol '|' you need to start a new session because once the 128k is used up, you are no longer allowed to ask it questions. Fortunately DeepSeek will provide you with a summary that you can enter to continue where you left off. Not unlike the disposable clone Mickey 17 [G29], this design got through 5 Deepseek sessions: the answers you see presented are often the conclusion of multiple iterations correcting errors that ate up the resource.

 

DeepSeek is very capable but a novice would find it very hard to design circuitry. As an engineer, hopefully I've spotted the pitfalls. I've both learned from it as well as enjoyed the interaction. The only downside is after I end this session, it will remember nothing it has learned if I start a new one. Thankfully sessions can be saved because having never owned an electric welder, I could not guess the physical and electrical parameters and had to start from scratch, and from first principles.

Initial assessment

 

I told DeepSeek I wanted to avoid high voltages and it provided an initial design assessment:

Ideal Circuit Parameters:
Voltage: 20V - 50V (Lower voltage promotes current flow over arcing).
Capacitance: 5,000µF - 20,000µF (Large capacitance for a longer pulse).
Current Limiting Resistor: Very low, just to control the peak, e.g., 10mΩ - 50mΩ.

Electrodes: Use a flat electrode on the Kanthal side and a pointed electrode on the TCW.

The flat electrode will protect the Kanthal from vaporising.

Step-by-Step Experimental Procedure: Start low at 1 Joule.

Example: With a 10,000µF cap, V = √(2 x 1 / 0.01) ≈ 14V.

 

Implementation

As this bonder is only required for the IR element, I'm not interested in making a full project of it that would ideally be PIC based. Instead, and to save time, lab instruments will provide all voltages and timing:

Amrel PPS-1202 0-18V, 4A, 1mV rms ripple, programmable linear PSU with isolated outputs,

​Manson EP-613 0-35V, 2.5A, 5mV rms ripple, linear PSU with isolated outputs,

EG&G Princeton 9650A, 5V/10V Digital Delay Pulse Generator.

The Amrel PPS-1202 PSU will charge the capacitors,

The Manson EP-613 will power the drive electronics,

The EG&G/Princeton 9650A pulse generator will produce a 10ms timing pulse to control the weld.

Required energy

I got DeepSeek to do the hard work:

Mathematically determine using existing datasets and established theory, the minimum weld energy density in Joules to bond a 0.3mm Kanthal wire to a 0.6mm tinned copper wire (TCW).

DeepSeek​:

A complete model must include the energy to raise the temperature of both the TCW volume and the Kanthal volume to the bonding temperature.

Let's define the bonding temperature as the point where the TCW is molten (~1085°C) and the Kanthal is sufficiently hot for the molten copper-tin alloy to wet it and form a intermetallic bond.

We will use 1100°C as the target interface temperature.

 

Define the Two Volumes to be Heated:


Volume​ of a cylinder = π × r² × l
Where:

r is the radius of the TCW.

l is the length of the TCW cylinder being heated.

TCW Volume (V_tcw):

For 0.6mm diameter TCW and a 0.3mm bond width (the Kanthal diameter):

TCW radius r = 0.6/2 = 0.3mm

TCW length l = 0.3mm

V_tcw = π x (0.3 mm)² x 0.3mm ≈ 0.0848mm³

 

Kanthal Volume (V_kan):
Diameter = 0.3mm
l = 0.3mm (symmetrical)
V_kan = π x 0.15mm)² x 0.3mm ≈ 0.0212mm³


Calculate mass of TCW and Kanthal:

m = V × ρ

Where:

m is the mass of the wire volume (in grams).

V is the volume of the wire segment being heated (in mm³).

ρ is the density of the material (in g/mm³).

Mass for the TCW:

ρcu = 8.96×10−3g/mm³

mtcw = Vtcw × ρcu

Mass (m_tcw) = V_tcw x ρ_cu = 0.0848 mm³ x 8.96e-3 g/mm³ ≈ 7.60e-4g

Mass for the Kanthal:

ρkan = 7.1×10−3 g/mm³

mkan = Vkan × ρkan

Mass (m_kan) = V_kan x ρ_kan = 0.0212mm³ x 7.1e-3 g/mm³ ≈ 1.51e-4g

Calculate Energy to Heat TCW to 1100°C:

The total energy equation used was the sum of the sensible heat and latent heat required to raise the material to the bonding temperature and melt it:

 

Qtotal = (m × c × ΔT) + (m × L)

 

Where:

Qtotal is the total energy required (in Joules).

m is the mass of the material (in grams).

c is the specific heat capacity of the material (in J/g·°C).
Tinitial is my lab temperature which is usually 27°C (I like it warm).

ΔT is the temperature change, Tfinal − Tinitial (in °C).

L is the latent heat of fusion (in J/g). This term is only applied to the TCW, as it is the material that melts.

 

This was calculated separately for the TCW and Kanthal volumes, then summed:
Qtotal = Qtcw + Qkan​

Calculate Energy to Heat TCW to 1100°C:

ΔT = Tfinal − Tinitial = 1100°C - 27°C = 1073°C

 

Q_heat_tcw = m_tcw x c_cu x ΔT = (7.60e-4g) x (0.385 J/g·°C) x (1073°C) ≈ 0.314J

Q_melt_tcw = m_tcw x h_cu = (7.60e-4g) x (205J/g) ≈ 0.156J

Q_total_tcw = 0.316 J + 0.156 J = 0.472 J

Calculate Energy to Heat Kanthal to 1100°C:

ΔT = Tfinal − Tinitial = 1100°C - 27°C = 1073°C

Q_heat_kan = m_kan x c_kan x ΔT = (1.51e-4 g) x (0.46 J/g·°C) x (1073°C) ≈ 0.074J

Kanthal does not melt, so no latent heat is added.

 

Calculate Total Theoretical Minimum Energy:

Q_total = Q_total_tcw + Q_heat_kan

Q_total = 0.314J + 0.074J = 0.544J

 

Conclusion:

The mathematically determined minimum energy require to heat the combined volumes of both wires to a bonding temperature of ~1100°C when starting from 27°C ambient, is approximately 0.54J.

In other words, the minimum required energy is 0.54J.

We also need to know the maximum energy because we don't want to destroy it.

The optimum energy for a good bond is between the two. 

Maximum permitted energy

The upper limit is defined by the point of catastrophic failure: vaporizing the wires.

Let's calculate the energy required to vaporize the bonding volume.

1Energy to Vaporize the TCW Volume (0.0848mm³):

Mass (m_tcw) = 7.60e-4g (as before)

Energy to Heat from 27°C to Boiling Point (2562°C):

ΔT = 2562°C - 27°C = 2535°C

 

Q_heat_to_boil = m_tcw x c_cu x ΔT = (7.60e-4g) x (0.385 J/g·°C) x (2535°C) ≈ 0.742J

 

Latent Heat of Vaporization (L_v,Cu):

4790 J/gQ_vaporize = m_tcw x L_v,Cu = (7.60e-4g) x (4790J/g) ≈ 3.64J

 

Total to Vaporize TCW: 0.742J + 3.64J = ~4.38J

 

Energy to Heat Kanthal to its Boiling Point (~3000°C)

Mass (m_kan): 1.51e-4g (as before)

ΔT = 3000°C - 27°C = 2973°C

Q_heat_kanthal = m_kan x c_kan x ΔT = (1.51e-4g) x (0.46J/g·°C) x (2973°C) ≈ 0.207J

 

Conclusion: The Maximum Energy

The total energy to begin vaporizing the materials is dominated by the TCW.
Q_max_theoretical ≈ 4.38J + 0.21J ≈ 4.6J

 

Safe Operating Range:

Theoretical Minimum: ~0.54JT

Theoretical Maximum (Vaporization): ~4.6J

This gives a theoretical range of ~ 0.54 to < 4.6J

'Sweet spot'
 

I asked DeepSeek where it thought the sweet spot for a bond might be within the range 0.54 to 4.6J:

Based on the calculated minimum (0.54J) and maximum (4.6J) energies, the sweet spot for a reliable bond will likely be in the range of 1.5J to 2.5J.

Reasoning:

Margin over Minimum (0.54J):

The theoretical minimum has no safety factor. It assumes perfect, lossless heat transfer. In reality, significant energy is lost to conduction into the cold ends of the wires, radiation, and heating the electrodes. A practical energy needs to be 2 to 3 times the minimum to ensure the bond volume actually reaches the required temperature.

This suggests an operating point above ~1.1 - 1.6J.

Safety Margin from Maximum (4.6J):

The maximum energy is a catastrophic failure point (vaporization). A good design operates far from this cliff. A sweet spot should have a comfortable buffer, at least 40-50% below the maximum, to account for variations in hand placement, surface oxidation, and capacitor voltage tolerance.

This suggests operating below ~2.3 - 2.8J.

 

The overlap of these two derived ranges—'well above the minimum' and 'safely below the maximum' points directly to a sweet spot of 1.5J to 2.5J.

Required capacitance

The next step was to determine the required capacitance to bond the wires. I wanted to avoid high voltages, so asked DeepSeek to calculate capacitance for 1.5J and 2.5J at several low voltages:

 

C = 2E / V²

1.5J  5V 120,000µF    2.5J  5V 3200,000µF

1.5J 10V 30,000µF     2.5J 10V  50,000µF

1.5J 15V 13,300µF     2.5J 15V  22,200µF

1.5J 20V  7,500µF     2.5J 20V  12,500µF

A low ESR capacitor is essential to produce the required high peak pulse current and I would be wiring several capacitors in parallel to drop that. The best choice on eBay seemed to be Panasonic EEUFC1E222 ​2,200µF 25v 105C 12.5m x 35mm with ESR 22mΩ.

I then asked it to calculate for 12V: 27,777µF

27,777µ / 2,200µF = 12.6 capacitors.

The capacitors were on offer in packs of 4, so I bought 4 packs, total 35,200µF; ESR 22mΩ/16 = 1.4mΩ.

Energy at 12V: E = 0.5 x 0.0352 x (12)² = 2.53J.

To get exactly 2.5J from 35,200µF, charge it to 11.92V.

For 1.5J, charge to 9.23V.

I set the maximum voltage at 12V.

Destruction voltage

I asked DeepSeek to calculate the 35,200µF capacitor voltage at 4.6J:

 

E=0.5 x CV²

       _______

V =  /    E   

    √  0.5 x C

 

Where:

E = Energy in Joules (4.6J)

C = Capacitance in Farads (0.0352F)

V = Voltage in Volts

       _____________

V =  /     4.6       = 16.17V

    √  0.5 x 0.0352F

Conclusion:

To store 4.6J of energy in 35,200µF you would need to charge it to ~16.2V.

The maximum voltage of 12V gives a 26% margin of safety. 

Capacitor bank safety discharge

Although the voltage isn't dangerous, the short circuit current is significant. However even though I proved empirically the Panasonic EEUFC1E222 has a long self discharge time, the voltage is critical to maintaining the optimum bond and I didn't want to add a permanent bleed resistor because it would begin to discharge the capacitors as soon as they were charged. Instead I added a manual discharge 5A rated push switch to swap in a discharge resistor across the capacitor bank.

I had a Welwyn W21 3W 4.7Ω resistor in stock.

Deepseek ran the calcs:

Peak Current: Ipeak =  17V/4.7Ω  ≈ 3.62 A

Peak Power: Ppeak   =  17V²/4.7Ω ≈ 61.5 W

Discharge Time Constant: τ = R × C = 4.7Ω × 35,200µF ≈ 0.165s.

Discharge Profile:

Capacitor discharge time =  V(t) =  V0 x e^−T/τ

Time to 10% (~1.7V):  t ≈2.3 × τ ≈ 0.38s

Time to  5% (~0.85V): t ≈3.0 × τ ≈ 0.50s

The W21 complies with EN 140401-002 which mandates the resistor must withstand 15W for 5s.

We have a ~61.5W pulse for ~0.4s:

Capacitor Exponential Discharge Pulse Energy through 4.7Ω:

Edischarge = 0.5 x C x V² = 0.5 × 35,200µF ×17² ≈ 5.09J

CECC 40201-002 Qualification Constant Width Pulse Energy:

CECCdischarge = P × t = 15W × 5s = 75J.

Discharging the capacitor produces <7% of this.

 

If the discharge button is inadvertently pressed when the Amrel PSU is still present, no harm will come to it as the 4.7Ω resistor will limit current to a maximum of 12V/4.7Ω resistor = 2.55A, however the W21 will overheat and likely blow. A conventional fuse is the ideal solution but a polyswitch is more convenient. I asked DeepSeek which value it recommended: 

Recommendation: 1.6A Hold Current Polyswitch. Reasoning:

The 2.55A initial current is a brief surge, decaying exponentially to zero in under a second.

A 1.6A polyswitch will not trip instantly on a 2.55A surge. Its trip time depends on the overload, and for a ~160% overload (2.55/1.6), the trip time will be in the order of seconds.

The normal discharge pulse is complete in ~0.4s, well before the fuse has time to heat up and trip.
If the PSU is left on, the polyswitch will see a continuous 2.55A, which is a 160% overload. It will trip and open the circuit, protecting the W21 resistor, and then reset once the fault is cleared.

Warning LED

lastly I added a 2k resistor from the capacitor + to the anode of a red led and its cathode to the point where the discharge switch joins its 4.7R resistor. When the button is pressed, the LED will illuminate indicating the capacitors have charge and will go off when they are discharged. Although this may appear unnecessary given the fast rate of discharge, it serves to warn they are charged, without loading the bank when it is needed.

Mosfet switch

A low side mosfet switch connects the negative terminal of the capacitor bank to the negative electrode. A pulse on its gate turns the mosfet on.

 

ESD protection

However this configuration means the source of the mosfet is exposed to the world, and it could be damaged by static electricity. In addition, a capacitive discharge welder acts like a large fast inductive pulse generator and induces noise spikes on power rails. To counter this and potential ESD, I placed P6KE15CA Transient Voltage Suppressors (TVS) physically close to the mosfets, across the source to 0V and the drain to 0V. Deepseek:

The P6KE15CA is a Bi-directional TVS: it clamps both positive and negative voltage spikes equally, and its 15V standoff breakdown range is 14.3V - 15.8V (capacitor voltage will be 12V max), clamp 27.2V maximum (the mosfet is rated at 30V max). It has a peak pulse power dissipation of 600W for a 10/1000µs transient waveform with a 10A peak, defined by two parameters:

t₁ (10µs): pulse rise time, measured from 10% to 90% of the peak current (I_PP).

t₂ (1000µs): pulse duration, from 50% on the rising edge to 50% on the falling edge.

Essentialy this is the equivalent of a long, high-energy surge of the type you might expect with a capacitive welder. The energy from a fast ESD spike is less, but still needs to be clamped.

Normal Operation: The TVS diode is invisible. It has a very high impedance at voltages below ~14.3V.

ESD/Overvoltage Event: If a voltage spike tries to push Vgs beyond ±16V, the TVS diode instantly 'clamps' the voltage, shunting the destructive current away from fragile semiconductors.

Placement: Solder it as physically close as possible between the MOSFET source and 0V.

 

Mosfet selection

DeepSeek recommended using a fast 30V 240A (1050Apk) IRLB3813 NMOSFET: RDSon 1.6 mΩ max @ 10V Vgs. Its high switching current, low RDSon and relatively low 650pF typ reverse transfer capacitance, Crss, were key reasons for its selection: low Crrs allows for very fast switching with a high-current driver.

 

but for added reliability I wired two in parallel.

 

Thermal dissipation

Absolute Maximum Continuous Current (Id) of 240A per MOSFET from the IRLB3813 datasheet, giving a worst-case combined pulse current of 480A.

 

Conduction Loss (P_cond):

Formula: Pcond = I² × Rds(on) × D

I_pulse: 480A

R_{ds(on)} (total for 2x MOSFETs): 0.0008Ω

Duty Cycle (D): 0.001 (10ms pulse every 10s)

 

Calculation: Pcond=(480)² × 0.0008 × 0.001

Pcond​=230,400 × 0.0008 × 0.001=0.184W (per MOSFET)

Switching Loss (P_sw)

Formula: Esw ≈ 1² × Vds × Id × (tr+tf)

V_ds: 12V

I_d: 480A

t_r + t_f: 200ns (estimated total switching time)

Calculation:

Esw = 0.5 × 1² × 480 × (200×10−9)

Esw​ = 0.000576J (Joules per switching cycle)

Power: Psw = Esw × Switching Frequency

Psw = 0.0000576 W

 

Total Power Dissipation (P_total):

Formula: Ptotal = Pcond + Psw

 

Calculation:

Ptotal = 0.184 + 0.0000576

Ptotal​ ≈ 0.1846W (per MOSFET)

 

Heatsink Requirement Analysis:

Junction-to-Ambient Thermal Resistance (RθJA): 62°C/W

Temperature Rise: ΔT = Ptotal × RθJA = 0.1846W × 62 ≈ 11.4 °C

 

Conclusion:

A heatsink is completely unnecessary. The MOSFETs will run only slightly above ambient temperature, with a worst-case temperature rise of ~11°C.

Discharge protection

The Amrel PPS-1202 PSU charges the capacitors at 4A. This PSU has isolated outputs and sense inputs. I added a forward conducting diode between the PSU + and the capacitor bank + feeding back to PSU sense +. Similarly a reverse biased diode with its cathode to the PSU - and its anode to capacitor - which goes to the NMOSFET drains and back to PSU sense -. The diodes will protect the PSU from damage when it is turned off but there is still charge on the capacitors, and also stop the capacitors from discharging through it. I asked DeepSeek to recommend suitable diodes, suggesting 6A.

Recommendation: FR607 6A fast recovery rectifier

Datasheet (Diodes Incorporated FR607):

Average Forward Current (I_F): 6A (at Tc=75°C).

Peak Forward Surge Current (I_FSM): 150A (8.3ms).

Repetitive Peak Reverse Voltage (V_RRM): 1kV.

Reverse Leakage Current (I_R): < 10µA at 25°C.

Package: R-6.

 

The R-6 is a huge 9mm dia package but it's difficult to find a smaller one at 6A, even at low voltages; Schottky diodes are unsuitable due to their high reverse leakage current.

Gate driver

The NMOSFETs are each separately driven by a Microchip (formerly Telcom) TC4422 9Apk fast MOSFET driver. DeepSeek recommended the drivers be decoupled with ceramic X7R 100nF and 10µF:

The TC4422s are powered from the Manson EP-613 12V supply. I added two 10µF MLCs in parallel where the 12V enters the board, ensuring low ESR fast current pulse decoupling. 
 

The TC4422 datasheet recommends decoupling with 1.0µF for each 1000pF of load capacitance.

IRLB3813 input capacitance = 8420pF.

Required Bulk Capacitance = (8420pF / 1000pF) x 1.0µF = 8.42µF.
 

​Gate Driver ICs: 2 x TC4422, 9A peak, fast low side MOSFET drivers.
Local Decoupling: 10µF + 100nF MLCs (ideal for high current switching) between driver Vdd and GND.
Gate Resistors: 2.2Ω to 4.7Ω, non-inductive (carbon composition or metal film) per MOSFET gate.
Layout: starpoint for gate drive connections. Keep driver IC and capacitors 1-2 cm near MOSFET gates.

I also placed a 1.5KE15CA TVS physically close to the TC4422s and across their 12V supply.

EG&G 9850A 50Ω output

 

To terminate the 50Ω output of the EG&G 9650A pulse generator into the TC4422, I added a 47Ω pulldown resistor (because 50Ω isn't a standard E12 resistor value [E66]) together with a 100pF X7R MLC in parallel to counter localised noise that could potentially cause false triggering: 

R = 47Ω

C = 100pF = 100 × 10⁻¹²F

RC time constant τ = R × C = 47 × 100 × 10⁻¹² = 4.7 × 10⁻⁹s = 4.7ns

-3dB cutoff frequency f_c = 1 / (2π × R × C) ≈ 1 / (2π × 4.7ns) ≈ 34MHz

 

-3dB point: at 34MHz the capacitive reactance of the capacitor 1/(2πfC) is exactly equal to the resistance 47Ω. The impedance of the capacitor has now become significant compared to the resistor.

This is the frequency where the output power is halved compared to the input power.

Consequently, the output voltage is reduced to 1/√2 (~70.7%) of the input voltage.

The 100pF capacitor forms a low-pass filter with the 47Ω pull-down resistor, attenuating HF noise and potential ringing from the pulse generator's fast edges that could falsely trigger the MOSFET driver.

Gate resistor

The TC4422 is capable of a 9A peak output current. The gate resistor sits between its output and the NMOSFET gate input.

Minimum Resistor for 9A with Vdd set to 12V [Manson EP-613 PSU]:

Rg = 12V = 1.33Ω
     9A

Target Range (2.2Ω to 4.7Ω):

With 2.2Ω, the initial peak current is I = 12V/2.2Ω ≈ 5.45A

                                           

This is well within the driver's capability but provides strong control.

 

With 4.7Ω, the initial current is I = 12V/4.7Ω ≈ 2.55A

                                      

This is ideal for damping and still provides very fast switching.

The 2.2Ω to 4.7Ω range is the ideal starting point, ensuring robust performance and stability.

Resistor wattage

I asked Deepseek to calculate the maximum wattage of 2.2Ω and 4.7Ω gate resistors:

Peak Instantaneous Power (W_peak)

This is the power the instant the TC4422 switches on, when the full supply voltage is dropped across the resistor.

V_drive = 12V (from Manson PSU)

Formula: Ppeak = Vdrive²

                    R​

 

Peak Instantaneous Power:

For 2.2Ω, P = 12²/2.2  = 144²/2.2 ≈ 65.5 W

for 4.7Ω, P = 12²/4.7  = 144²/4.7 ≈ 30.6 W

Average Power (W_avg) per Pulse:

This is the total energy dissipated in the resistor during one gate charge/discharge cycle, divided by the pulse width. The energy dissipated in the resistor is approximately half of the total gate energy.

Q_g (Gate Charge) = 110nC

V_drive = 12V

Energy in Gate (E_gate):

Egate ≈ 0.5 × Qg × Vdrive = 0.5 × 110×10⁻⁹ × 12 ≈ 0.66 μJ

Energy in Resistor (E_res):

 

Eres ≈ Egate ≈ 0.66 μJ (It shares roughly half the energy).

 

Pulse Width (t_pulse) (Assume a conservative 1ms):

Formula: Pavg =  Eres/tpulse

Pavg = 0.66×10⁻⁶/1×10⁻³ = 0.66×10⁻³W = 0.66 mW

Conclusion:

The average power is negligible (~0.66mW) for both resistors.

The critical specification is the peak instantaneous power and the pulse handling capability.

2.2Ω Resistor: Must handle ~66W pulses.

4.7Ω Resistor: Must handle ~31W pulses.

 

A wirewound resistor is not suitable because it will:
​Slow down the switching speed by resisting the rapid current change (dI/dt).

Cause ringing and voltage overshoot on the gate signal, which can stress or damage the gate oxide.

You must use non-inductive, pulse-rated resistors (e.g., metal oxide film or carbon composition) rated for these high peak pulses, even though the average power is tiny; a standard 1/4W carbon film resistor would be destroyed. Requirement: Must withstand a ~65W peak pulse for ~100ns.

I asked Deepseek to recommend a suitable resistor.

It suggested Vishay PR02 2W metal film:

Analysis:

The resistor is subjected to a high-power pulse when switching the MOSFET gate. Suitability is determined not by its continuous power rating, but by its ability to absorb the pulse's energy without damage.

 

Calculate Pulse Energy:

Peak Power, Ppeak​: 65W

Pulse Duration, tpulse: 100ns

Pulse Energy, Epulse​: Ppeak × tpulse = 65 × 1×10⁻⁷ = 6.5 μJ

 

Compare to Resistor Thermal Capacity:
A 2W metal film resistor has a larger mass. Let's estimate 0.5 grams (5 x 10⁻⁴ kg) and a specific heat capacity similar to ceramic/glass/plastic (~800J/kg·°C).

The energy required to raise its temperature by just 1°C is:

E = m × c × ΔT = (5 x 10⁻⁴ kg) × (800J/kg·°C) × (1°C) = 0.4J

 

Conclusion:

The pulse energy is 6.5µJ/400mJ = 0.00001625 (or 0.001625%) of the energy needed for a 1°C rise.

This means the 6.5µJ pulse is over 61,500 times maller than the energy needed to raise the resistor's temperature by a single degree Celsius. Therefore, the Vishay PR02 0.5W resistor is perfectly suitable and will not be damaged or degraded by the 65W, 100ns gate drive pulse.

The component's thermal mass renders the brief power spike harmless.

I couldn't find reasonably priced 2.2Ω and 4.7Ω PR02s but I did find 50x 10R 5% PR02s for £4.60.
Deepseek approved of my idea to parallel them by soldering them on top of each, minimising inductance.
The large number is also beneficial in finding matching resistors out of the 5% tolerance.

Deepseek now recommended starting with 5Ω:

 

5.0Ω (2x 10Ω in parallel): Slower, more damped, most stable.

3.3Ω (3x 10Ω in parallel): A balanced middle ground.

2.0Ω (5x 10Ω in parallel): Fastest switching, highest current, highest risk of ringing.

Start with the 5Ω configuration.

Use an oscilloscope to check the gate waveform for ringing and overshoot.

If the waveform is clean and stable, try 3.3Ω to see if you can achieve faster switching without introducing instability. Only proceed to 2.0Ω if the 3.3Ω waveform remains clean.

​​​

​​Electrodes

I asked DeepSeek to recommend suitable electrodes:

Look for 2% Lanthanated Tungsten Electrodes, often used for TIG welding.

Diameter: 1.6mm (1/16"). This is a robust size that won't overheat and provides a good point for focusing current. These usually come with flat tips and you will need to grind a pointed tip on it.

 

Why Lanthanated Tungsten:

It has excellent high-temperature stability, good arc characteristics, and is non-radioactive (unlike the popular but radioactive Thoriated Tungsten).

Negative electrode:

 

A large, flat-faced electrode. This acts as the anvil. It provides a stable,. supportive backing for the workpiece.

 

DeepSeek agreed the lid of the aluminium box is a good choice: The entire weld current return path—MOSFET sources, capacitor negative, and anvil will be contained within a very small area on the lid, minimizing parasitic inductance and ensuring a sharp, powerful pulse is delivered to the workpiece.

Positive electrode:

 

1.6mm Lanthanated Tungsten; grind a small, flat tip on it ~0.5mm to 1.0mm diameter.

This small flat tip concentrates the current onto a precise spot on the wires, but the flat surface applies pressure without cutting into them.

For TIG welding electrodes, the color code for Lanthanated Tungsten is not universally standardized, but the most common conventions are:


Gold: Typically denotes 1.5% Lanthanated (EWLa-1.5)
Blue: Typically denotes 2.0% Lanthanated (EWLa-2.0)

 

For this micro-welding application, the Blue (2.0% Lanthanated) is the better choice.

 

Reasoning:
The higher lanthanum oxide content (2.0% vs 1.5%) provides slightly better electron emissivity and arc stability at lower currents. This translates to a more consistent and reliable performance for the precise, low-energy pulses in a capacitor discharge welder, giving a more forgiving and stable electrode.

Electrode positive wire

The wiring from the mosfet source to the electrodes will be two runs of 55/0.1mm CSPE hook up wire (10A, 2kV) in parallel, ensuring minimal voltage drop and a sharp current pulse to the bond point over a short distance (200mm max).

Equipment wiring​

The positive lead of the Amrel PPS1202 PSU goes to the capacitor bank +.

The negative lead of the Amrel PPS1202 PSU goes to the capacitor bank - and NMOSFET drains.

The positive electrode is connected to the capacitor bank +.

The negative electrode is connected to 0V.

Sequence of events​

 

Initially the mosfets are off and the capacitor bank negative terminal is floating. 
The PPS1202 charges the capacitor bank to its target voltage (e.g. 9.23V for 1.5J).

Once the bank is charged (DMM**) the 9650A starts a 10ms rising edge pulse to the TC4422s,
The TC4422s drive the NMOSFET gates, turning them on and shorting the capacitor bank - to 0V.
The 9650A pulse drops to 0V and the TC4422s turn off the NMOSFETs.​


**The MSO8k can be set up to generate an output pulse when the capacitor is charged.

Components

All components are low cost, leaded, and readily available in 2025:

 

FR607 6A fast recovery rectifier, R-6 x2

Panasonic EEUFC1E222 ​2,200µF 25V ESR 22mΩ 105C x16

NMOSFET IRLB3813 30V 240A (1050Apk), TO-220 x2

P6KE15CA bi-directional 15V 600W Transzorb TVS, DO-201 x3

TC4422 9A peak fast low side MOSFET driver, DIP-8 x2

100pF X7R ceramic capacitor to decouple noise on TC4422 inputs CHANGE CIRCUIT - 100pF not 10pF

100nF X7R MLC across each TLC4422 12V supply
10µF X7R MLC across each TLC4422 12V supply
10µF X7R MLC x2 in parallel where 12V enters the board  CHANGE CIRCUIT - REMOVE 15µF TANTALUM

red LED, 2mA

4.7Ω Welwyn W21 3W resistor

10R 2W Vishay PR02 non-inductive resistors for experimental gate drive x10

47Ω 1/2W resistor to terminate 50Ω pulse generator input
2kΩ 1/4W resistor for LED

10kΩ 1/4W resistors to pull down mosfet gates x2

Bourns MF-R160 1.6A polyswitch multifuse

20V 3A momentary push switch

55/0.1mm CSPE hook up wire (10A) x 0.5m

2% Lanthanated Tungsten Electrode 1.6mm dia x 150mm

BNC socket for pulse input

4mm banana coloured sockets for PSU inputs

Aluminium enclosure​

Below, LIBS 7 logbook P.20: schematic                kanthal/wire bond & gate PR02 combinations

508j01 P1140962 LIBS 6 P.20 cct for bonder 08-11-25
508j02 P1140956 LIBS 7 P.20 lead wire ring crimp & gate PR02 combos

Construction

All grounds will be star-pointed.

I'll build the circuit on 0.1" square pad board and run extra TCW along the tracks to increase capacity.

​I'll build the unit on inside of the lid of a large diecast box.

Placing the work

Method:

The pointed tungsten electrode contacts the larger lead TCW; the Kanthal is pressed against the flat aluminium anvil.

The weld forms at the interface of highest resistance and current density, which is the contact point between the two dissimilar wires. Placing the pointed electrode on the larger TCW and the flat electrode under the finer Kanthal is the correct method. Here's why:

 

Current Concentration:

The sharp point on the TCW forces the immense current to focus into a tiny area at the contact point with the Kanthal.

 

Controlled Heat Generation:

This creates an intense, localized hotspot precisely where you want the metals to fuse—at their junction.

 

Anvil Function:

The large, flat electrode under the Kanthal provides a stable, cooling backing that prevents the fragile Kanthal from being vaporized and contains the weld.

Reality

Reality is a little different: this is a tiny, tiny coil 4mm long! The space in which I have to make the bond is no more than 2mm diameter. I have hold the lead wire on top of and perpendicular to the Kanthal in a '+' fashion and I can only do that using metal as anything else would melt and foul the weld.

I need to press an angled knife edge metal former down on it with sufficient space in the middle to allow the bond. It has to be a knife edge as anything else would act as a heatsink:

A conventional steel knife blade is a pragmatic and likely effective solution for this application.

 

Analysis:

Thermal Conductivity: The primary goal is to reduce parasitic heat sinking whilst forcing the wires together. Standard steel has a significantly lower thermal conductivity of approximately 45 W/m·K.vs copper (~400W/m·K) and will indeed minimise the amount of heat drawn away from the weld zone (Kanthal A-1 is approximately 15W/m·K; this low value is intrinsic to its function as a resistance heating alloy, as it helps localise the generated heat within the wire element itself).

 

Electrical Conductivity:

Steel's electrical conductivity is lower than copper. For the brief, high-current pulse, this will result in a slightly higher voltage drop and energy loss in the anvil itself. However, given the minuscule contact area of a knife edge and the short pulse duration, this effect is likely to be negligible in practice and is highly unlikely to impact the quality of your micro-welds.

​​​

​Making the bond (weld)

Based on the wire's tiny cross-sectional area and the goal of a localized fusion weld without vaporization, the sweet spot is likely 1 - 2 Joules.

Reasoning:

Below 1J: Risk of a cold, brittle joint due to insufficient melting.

Above 2J: High risk of vaporizing the 0.3mm Kanthal wire at the weld point.

 

Test:


Start at 1J, inspect the weld (it should be a fused nugget, not a sintered powder), and increase energy in 0.2J increments until you get a strong bond. ​​

 

Make the bond using a cross-wire configuration.

No Bond? Increase energy by 0.25J - 0.5J and repeat.
Weak Bond? Increase energy slightly.
Sputtering/Splash? Energy is too high. Decrease immediately

The TCW will require less energy to melt than Kanthal, so the Kanthal is the limiting factor.
 

Reasoning:

Melting Point:
Kanthal A-1 (FeCrAl): ~1500°C
Copper (Cu):           1085°C
Solder Tin (Sn):      ~232°C
 

The tinned copper wire will soften and melt its surface coating at a much lower temperature than the Kanthal will even begin to soften.

The Welding Process:
The goal is to raise the interface to a temperature where the metals fuse. The tinned copper, with its lower melting point, will become molten first. This molten alloy wets the surface of the hot Kanthal, forming a eutectic bond.

Why Kanthal is the Limiting Factor:
You must pump enough energy into the junction to get the Kanthal hot enough to be wetted by the molten copper/tin. If the Kanthal isn't hot enough, the molten solder will just bead up on it without bonding. The entire process is gated by the temperature of the higher-melting-point material.

 

So, the statement holds: the Kanthal's need to reach bonding temperature is the limiting factor, which is why we focus the current on it via the pointed electrode on the copper side.

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