# Tag Info

3

Well...the first approach seems rather messy doesn't it? Not only are you sampling at unpredictable positions in the cycle, but you are increasing the processing load by having more samples to deal with. Plus you are probably only going to be updating once per PWM cycle at most anyways...so is your latency really any better? What good is all that extra ...

3

No, it has nothing to do with correcting power factor. The bus supplying the inverter is DC. It's because there is substantial inductance and resistance between the battery and the switches that make up the 3 phase inverter, and if they were to switch without any capacitance on the local bus, there would be a substantial voltage drop and probably a lot of ...

3

Why use both high side and low Side diode-ORs ... ? It is to protect against all forms of misconnections (see below). Linear Technology explains it in their Design Note 444. It can be easily seen that removing one of both ORing diode pairs in Figure 4 will result in a short of one of the power supplies. And in Figure 5 Battery A and Battery B would have ...

3

You should reverse the connections of diode D2. The anode should be connected to Q1 and the cathode to 5V. When you turn on Q1 the current flows through it and D2 between the ground and voltage supply forming an almost short circuit.

3

It depends what level of 'design' you want to go for. If you want to design the circuits of the inverters, then it would be an interesting and long learning exercise for you, and the result, in a year or two's time, would be more expensive and almost certainly inferior to anything you could buy. If you want to choose panels, inverters etc from a catalogue, ...

2

Look in the linked data sheet and, the first diagram of a core they give is this (with my red lines on it to show what an average turn might look like): - Can you see that the diameter of that average turn could be calculated as: - $$\dfrac{E}{2} +\dfrac{D}{2}$$ Or, if the core isn't fully utilized then use a smaller length but go no smaller than a ...

2

Take a tumbler filled with tap water and add a tablespoon of table salt and mix. Put in a couple forks, separated by a couple inches with a sponge or dishrag. Put that makeshift resistor in series with your LED and it won’t likely burn out. Don’t worry about getting it backwards, most LEDs can withstand quite a bit of reverse voltage.

2

A voltmeter will well you which lead is what. Otherwise, sometimes the cord will have a small 'ridge' on the + side, but that's not for certain. An LED with a 1K or so load resistor in series can also work. But be warned that LEDs don't much like reverse polarity, you can protect it with a regular diode in parallel, wired in opposite polarity to the LED. A ...

2

Shortly, it's the effect of $V_L=L\ di/dt$. Here's the equivalent diagram of your setup (neglecting the real model of the inductor): simulate this circuit – Schematic created using CircuitLab The inductor sees a positive DC voltage for the half of the period and a negative DC voltage for the rest of the period. So, from $i=\frac{1}{L}\int{V_L\ ... 1 It is used to connect a diode in parallel to the inductor. It is needed since when the transistor is off all the energy accumulated by the inductor has to be dissipated by the equivalent capacitor of the diode. Check the link below for more informations: https://en.wikipedia.org/wiki/Flyback_diode 1 The circuit shows how one device may be powered from two different supplies without one supply "forcing" current into the other. This is useful to allow graceful failover of one supply, e.g. a dead battery, while the other is still working. The purpose of the two diodes is to separate the supplies from each other. Why is that needed? Consider if one supply ... 1 There is no co-relation between a battery's remaining capacity and it's open-circuit voltage. The discharged battery's no load voltage could be above 12 V but its internal resistance would also be high. Without the required starting current, the motor would remain stationary and present itself as a dead short across the battery terminals. The battery ... 1 Is the value of 100 ohm 1 watt correct on pin 6 of MOC3021 Figure 1. The datasheet suggests this circuit. Remember that when the MOC triac is off that there is no current through your R16 so no power will be dissipated. When the MOC turns on there will be a very brief current through R16 but Q4 will turn on almost instantly and the voltage across the ... 1 Essentially a soft-start circuit limits the output current during some time at startup. It is usually accomplished by charging a capacitor with a given constant current source (like you are doing already). If you want an even simpler solution, you could try the following (LTSpice): How it works B1 modulates the soft start's constant current source + ... 1 I'm not saying there aren't other problems with your simulation but this one is glaring: - The 1 H inductor is completely shorted out. Also try and learn how to cut and paste pictures. 1 I am building a 5 V PSU to charge my phone. I am using a 7805 regulator. This is risky. For example, if you loose the connection to the centre pin (GND) the output may go close to VIN and permanently damage your phone. When I connect to the phone the Android Charger Tester app reports a charge current of only 100 mA or less. Charge management is done ... 1 The 7800 series of regulators take a quiescent current of a few milliamps. This is probably what's burning down your 9V batteries when you're not operating the unit. Modern LDOs can be had with quiescent currents in the microamp range. When operating, your efficiency is under 70%--not bad for really low duty cycle, but if your on time is significant you ... 1 You'll have to calculate the MLT yourself, based on the core geometry, the bobbin or insulation you wind it on, and the window fill - the amount of copper wire you're putting in, so it isn't a constant for a given core. For a cylindrical wind, this is fairly easy, since it's just the length of the winding halfway out of the hollow cylinder formed by the ... 1 you should switch the live wire, not neutral or ground, switching neutral makes more of the circuit live which reduces safety. The relay is intended to switch AC, it may be suitable for DC but the maker hasn't marked that on the case of the relay. if you can find the datasheet for this relay it will say if it can switch DC reliably. Do a search on "JQC-3ff"... 1 Note: you have your notation backwards from what I believe to be convention.$b$refers to the outgoing wave from the network and$a$refers to the wave incident on the network. That being said, the following will use your notation instead. When$\Gamma_g=0\\$, what is implied is that the source impedance of your generator at port 1 is matched to the ...

1

Check the battery. If it's a lead acid battery, then cells can suddenly fail, presenting a very high resistance. With no load, it may appear normal, but adding any load on the battery will make the voltage drop instantly.

Only top voted, non community-wiki answers of a minimum length are eligible