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1

Parasitic capacitance in the MOSFET between drain and source plus external parasitic capacitances mean that when the MOSFET turns on, it is trying to short out these previously charged parasitics and hence, there is always an impulse of current into the MOSFETs drain. Same applies with IGBTs and regular BJTs except internal parasitic device capacitance is ...


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After experimenting with placing a couple of capacitors at different points in the circuit, the solution I found working best is placing a 47uF ceramic capacitor between Q2E and Q2B AND a capacitor in parallel with the LED (47uF, electrolytic). It could work with smaller capacitances, but the next smallest I have is 0.1 uF, which is not enough to completely ...


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Let me address a couple things as a pilot... Why? Well, for example, if the pressure switch in the gear up part of the circuit should fail, I can still get the gear to retract by activating the override push button. This is a big no-no as a pilot. The last thing you want to do is to bypass something that is failing. If your gear doesn't retract, leave it ...


1

Terms Zero Voltage Switchingand Zero Current Switching are very unfortunate and misleading. More appropriate terms would be Zero Voltage Turn-on and Zero Current Turnoff. The idea is pretty obvious: for the switch itself it is advantageous to be turned on a tad before there will be current through it, and turned off when current is already ceased.


2

Trace at point A carries high di/dt current so it should have low inductance to the decoupling capacitor to minimize voltage spikes. \$ e = L \frac{di}{dt} \$ and in dc-dc converters di/dt is pretty high so you want to minimize L. A few amps switched in a few ns with a few nH inductance = VIN drops a lot when it switches. I've had a case where this trace ...


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As you correctly pointed out, a quasi-square-wave resonant converter colloquially called a quasi-resonant converter (QR) is a self-relaxing converter: there is no clock in the controller except for clamping purposes. The switching frequency varies with operating conditions. It is low at low line and high power, it is high at high line and low power. QR ...


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Wider traces exhibit less resistance, but more significantly less inductance. The lower resistance and inductance helps a little bit with efficiency, and a little more with ripple, but the larger gains are in EMC. A lower inductance trace radiates less, particularly in conjunction with a solid ground plane.


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The problem its caused by the fact that \$Q_{18}\$ is always ON since even when the input level fro Arduino Nano+ is high, it does not reach the 12V needed to bring \$V_{BE_{Q_{18}}}\simeq 0\$. This is due to the fact that the Arduino output reach at most +5V, while the emitter of \$Q_{18}\$ is connected with various loads (the relay coil and the series of ...


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For the N-FET to work correctly you need to make sure to tie the GND of both power systems together. In other words the GND of the Rambo (Arduino) board needs to connect to the GND of the bed heater power/switching system.


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First of all, you're using the incorrect schematic symbol for a MOSFET. They are not the same as BJTs, and we just have to guess how you have it connected. MOSFETs do not have an emitter, collector, or base, so I am going to assume you have it connected such that the source is the emitter, the drain the collector, and the gate the base. Please use the ...


1

Look for "logic level mosfet". Don't be guided by the \$V_{GS(th)}\$ value (or look for \$V_{GS(th)} < \text{about } 3.3V/2\$), but rather check the conditions for \$V_{GS(th)}\$ given for the \$R_{DS(ON)}\$ parameter. Below is a visual explanation of the text above. This is not a product recommendation: product recommendations are off-topic on this ...


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Decades ago I worked on a satellite project. Lotta good stuff to learn; most of my circuits worked first time, so they let me handle integration of the entire "box" we were building. One of my circuits was a stripped-down version of the customer-suggested relay-driver schematics; they suggested an AC_coupled positive-feedback 2-transistor circuit. I thought ...


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This figure from "Review of Switching Concepts and Power Semiconductor Devices" (no author) illustrates the power losses when a bipolar device is used as a switch: There are two considerations. One is that the average power, when multipied by the thermal resistance (junction-to-ambient if there is no heat sink, or junction-to-case if the heat sink ...


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While turning on and off, your transistor produces so-called switching losses: $$P_{SW} = f_{Sw}\cdot(E_{on} + E_{off})$$ $$f_{Sw}=\text{Switching frequency}$$ $$E_{on}=\text{Turn-on energy}$$ $$E_{off}=\text{Turn-off energy}$$ As you can see, the switching losses depend strongly on your switching frequency. If they're relevant to your application or not ...


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