8

Your multimeter and your clamp meter are both not meant to observe PWM going through an inductive devices. You need an oscilloscope with a high-bandwidth method of sensing current. I don't know your PWM frequency, but rule of thumb: estimate how fast the transition from fully on to fully off has to be, take the inverse of that time (yielding a frequency), ...


8

simulate this circuit – Schematic created using CircuitLab Figure 1. (a) A diode will conduct if it is forward biased. There is no control pin on it. It is not controllable. Figure 2. The upper waveform shows the result of the diode or the thyristor when triggered at the start of the half-cycle. The third waveform shows the result of triggering half-...


7

From the comments: ... in the case of comparator the issue is of level adjustment. As this circuitry can be repeated upto 100000 times in the industry. No. Many op-amps latch up when driven to the supply rails. Comparators are designed to be driven to the supply rails. Repeating a mistake a million times won't fix it. Figure 1. From the datasheet we can ...


5

There is no Rds(on) on an IGBT because it does not behave like a MOSFET- it has a Vce(sat). According to the SOA curve, 100A continuous is not safe even at Tc = 25°C. The limitation is thermal. Note that the Vce(sat) is not guaranteed above 80A so there is uncertainty in the maximum power dissipation at 100A. 300A is the upper limit for a single short pulse ...


5

I want to estimate the power dissipation of this IGBT SGL160N60UFD but I can't find any Rdson Well, it's an IGBT and it doesn't have a drain (d) or source (s) pin. They are actually called collector and emitter like an ordinary BJT. It does have a gate pin (like a MOSFET) of course. can this TO-264 package handle currents of 100A? From the mouser data ...


5

Use pwm from the Arduino to ramp up the power to the motor. No need for extra circuitry.


4

The Arduino's default PWM frequency is ~490 Hz. At this low frequency the motor probably won't have sufficient inductance to make the current continuous, so the rms current and voltage will be much higher than the average values that your meters read. I created a simulation in LTspice representing your situation, adjusting the motor's internal resistance, ...


4

You need to increase PWM frequency to at least a few kHz for the motor inductance to smooth out the PWM. Some types of motors have more inductance than others, so the optimum frequency will vary with motor type. Otherwise you will get large RMS current and corresponding large I^2*R losses in the motor, transistors and connecting wires. The BTS7960 is ...


4

By shifting the +Ve reference level from 0.88V to 1.68V solved my problem. This is achieved by changing the resistance R29 from 10k5 to 25k. Thanks everyone for helping.


4

To add to @Transistor's answer: The On Semiconductor LM324 datasheet (On Semiconductor) has the following specification with a 5V supply. As you can see the typical maximum output is only 3.5V. This is because the output stage of the LM324 consists of a double emitter follower giving two VBE drops. There are many newer opamps that can get within a couple of ...


3

Question What is the difference between 2N7000 and 2N7002? Answer One critical difference for me is that 2N7002 is definitely logic level trigger, 2N7000 is not so sure. References (1) 2N7000 / 2N7002 / NDS7002A N-Channel Enhancement Mode Field Effect Transistor - FairChild/On (2) 2N7002 N-channel Logic Level MOSFET - Components 101 2019jan07 (3) 2N7002 (...


3

My question, when is it preferred to use CT of 100:1 turns ratio instead of a cheaper 100:5 amps? When the meter has a full-scale of 1 A then you need a 100:1 CT so that you don't destroy the meter when the primary current goes over 20 A. When the protection circuit has a full-scale of 1 A the 100:5 CT will cause it to trip at 20 A. The VA rating of the CT ...


3

The minimum gate triggering current is the smallest current you need to give the SCR to guarantee triggering under normal working conditions. This is the amount of current the rest of your circuit needs to provide. If your circuitry provided less than this the operation could fail to function properly under some conditions. For example high-temperature, low ...


3

In general, for a circuit where the output is switched on and off, how can we analyze the power factor? In general, the power factor in such a circuit is not the power factor defined as the cosine of the angle between the current and voltage but as the real power divided by the apparent power. In this case, the apparent power is the total RMS voltage ...


2

This part is marked obsolete on D-K site but not Mouser(?), so no complete datasheet in your link. But consider for all saturated switches; $$\dfrac{V_{ce(sat)}}{I_{c(sat)}}=R_{ce}$$ 1.3V/80A typ = 16.25 mOhm. 1.6V max. =20 mOhm Max. 1.6V/160A typ = 10 mOhm typ. Not max. Lookup plot for SOA with PW50 duration @ 100A and consider margin for this “...


2

You are looking for a power sequencer. There are many brands that make them: Panamax, Furman, Pyle...


2

Use a CD4060 instead by itself - it has a built-in oscillator so you don’t need the 555. Then gate the circuit to get the one-shot behavior you’re looking for. MORE: To get an accurate time, use two CD4060s. use a 'watch crystal' with the first one to make a 2Hz signal use the second one to count down from 8Hz (more on why, below.) Here's the first part to ...


2

I expect you'll find both arrangements in both countries - they are at different stages in the power distribution system. In the top photo, wires 1, 2, and 3 are three-phase high voltage (13.2 kV in my area) The single wire running at right angles to 1, 2, 3 is a single HV phase to supply local step-down transformers to deliver 120/240 V to homes. Wire 4 ...


2

In the US picture, the wire 4 is not related to 1,2,3. As you can also see different spacing, it's not a three phase network. In Canada picture, the upper 1 is a lightning protection, connected to the earth, While 2,3,4 are L1,L2,L3 three phase network.


2

It is clear from the datasheet that the intention is to connect the two pins together. It does not say if you have to connect them together, or if they can be left unconnected if you dont need both pins for some reason. The two pins should be connected together to allow for less voltage drop at given current and to allow passing more current. Connecting both ...


2

You can use any microcontroller with 2 independent PWM interfaces. Program the hardware registers of the 2 interfaces in order to get the desired PWM period and Ton. The values that you will program in the registers will depend on the clock frequency that drives the PWM interface. Don't do it in software (delay loop). Use the hardware registers.


2

should I left the second output of the transformer empty or connect a load to it? If it's no problem, leave it open circuit or, if it's the same voltage as the other secondary, you should be able to parallel them up to reduce \$I^2R\$ losses in the secondary.


2

You will not be harmed by receiving radio signals on an antenna unless you are extremely close to the transmitter. In any case, those signals are present at your location whether you have a receiver or not. It is also doubtful if you can receive a signal at 400 GHz if your receiver is only supposed to work up to 6 GHz. It is probably generated internally by ...


1

I quite sure that is measuring problem because if you loss power of 133W on driver, it totally melt down due to power converted to heat. Only reason for that is measuring tool can not average changing voltage or current to get correct value. I suggest to use RD low pass filter before measuring as shown. Note value for component just for example, you need to ...


1

The main advantage is you get a ton of current gain (round about the square of an individual transistor of the same type). Yes, you could use two dissimilar transistors. Main disadvantage is Vbe is effectively doubled to about 1.4V, and, just two transistors (though typically you get them in one package). No, they cannot generate power "by themselves&...


1

For a high-side switch, when your input signal level is smaller than the load voltage, you generally need two transistor stages. The resistor values and transistor types are approximate, you need to analyze for your requirements. simulate this circuit – Schematic created using CircuitLab


1

You have a source follower, not high-side switch. MOSFETs are NOT controlled by the gate voltage. They are controlled by the voltage DIFFERENCE between gate and source. How is your MOSFET supposed to know what the voltage at GND is when none of its pins are connected to GND? Yet, you are applying a gate voltage relative to GND. What happens to the voltage at ...


1

The circuit as it appears now has a couple minor issues. SPST has 4 pins and appears to be a DPST or SPDT with a LED? Vgs must be 2.5 to 3x Vgs(th) max.for low RdsOn. Yet Vg=Vd=~Vs so Vgs =0 assuming V+ is common. No go. Vs must be grounded at Vbat- with Vd to low side load to pull down. (Assuming your interfaces can tolerate this in the off condition.)


1

The way you model a minority carrier device (BJT, Diode, SCR, IGBT) is via the output characteristic transfer function. \$V_{ce} = V_{ce0}+ R_0\cdot I_c\$ \$V_{ce0}\$ = on-state voltage threshold \$R_0\$ = On-state resistance approximation These figures are not provided in the datasheet but can be derived from the Ic vs Vce curve, around an operating point. ...


1

Thanks @Linkyyy It’s probably Richtek’s RT2518 (linear regulator). Datasheet here: https://www.richtek.com/assets/product_file/RT2518/DS2518-05.pdf


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