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@shay yederman I have used the following configuration, the main purpose was to control a BLDC motor sensorless. Also some formulas to find out the charge needed and the capacitor value: QT = QG + (ILKCAP + ILKGS + ILKC + ILKD + IQBS) × tON + QLS I have used this simplified formula: QT = QG + (ILKCAP + ILKGS + ILKD) × tON + QLS QG = Total Gate Charge; ...


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Your VFD can't detect the pole passing with back EMF sensing (sensorless BLDC control). This issue is common for all sensorless BEMF sensing devices, even when using induction motor. 15 rpm is very low speed, so you would need an ESC with hall sensors feedback.


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WCH is the manufacturing company that first produced the CH340. They are Chinese.


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The upper opamp of circuit B works as a (positive) unity-gain amplifier - and R41 plays a special role ! In many cases, a unity gain configuration has a phase margin that is too small (step response with ringing effect or even instability). Here we have a circuit (unity gain) with a selectable loop gain which can be set to any value for ensuring a "good&...


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In the second circuit, it will work as shown — no current flows through R41 anyway — but it looks like an error. In order to balance the bias currents, R41 should be connected between Diff_In and the noninverting input, and the R40 should be the only thing connected to the inverting input. But for the same reason, there should also be a 2.5k resistor between ...


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You can indeed filter the PWM to control a current source. You can use something like the following. I will assume you just want to set the brightness and leave it on, then take the picture. If you are trying to turn on the LED as fast as the shutter, like a flash setup, this will not work. Also since you are using 2n2222, I will assume relatively low LED ...


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It is possible but not desirable. LED's work within their specifications particularly the spectrum they emit only within a current range, changing the current can change the characteristics of the emitted light. This is why PWM is used to control intensity. The correct current is supplied for a shorter time and normally your eye averages it. For your camera ...


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If you look at the data sheet, on pages 4 and 5 you'll see VOUT. Notice that, with a 5 volt supply and 600 ohm load, you cannot be certain of getting more than 3.1 volts. I am astonished that you are getting .25 volts with a 10 ohm load. This is far beyond what the data sheet suggests. Your op amp is putting out 25 mA, when its rated max is about 5 mA (10 ...


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