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4

This cannot work. It is not possible to lower the threshold voltage. The threshold voltage can only be increased by having a voltage between backgate (bulk) and source. If there was some way to lower the threshold voltage of a MOSFET through a "circuit trick" then for sure I would know and use that trick. I've been dealing with CMOS circuits for more than ...


1

You could add one more LED and thus reducing the voltage drop across the resistor. Or choose another led, with a bigger forward voltage ex: 3.5V - 3.9V. Because the power loss is directly proportional to the voltage drop across de resistor. P = U x I In your case P = 2.1V x 0.3A P = 0.63W with a bigger forward voltage (Led with forward voltage of 3.9v) ...


1

The basic problems with your circuit are extreme sensitivity to DC bias point of the upper FET, interaction between DC bias point and gain of the input transistor and lower FET, and low AC gain. I put your circuit into LTspice and made a few changes to improve DC stability and AC gain:- To remove the interaction between M1 and Q1 biasing I AC coupled the ...


2

Class A Complementary FETs will always be a struggle with 1 BJT driver low open loop gain, thus >10% THD even with negative feedback since it is very quadratic. only get 16Vpp from a 24V supply. loss of low-frequency response with a single supply requiring large Cap may have poor characteristics. poor supply ripple rejection (hum etc) too many pots that are ...


1

You already included all you need to look for. The gate output voltage of the MIC5018 is specified as min. 11.4V @ V[supply]=4.5V, typically above 15V @ 5V. You want a FET with V[th] preferredly lower than (11.4V-5V)=6.4V (e.g. logic-level MOSFET), or at least lower than 15V-5V=10V, current capability 20A at given V[gs] of 6.4...10V, min. 48V V[ds] rating. ...


1

If higher resistance is not an issue, I'd still look into the fairly reduced current limit. At 3V, the FET will enter the linear region at about 1A of drain current. At this point the "RdsON" if the term is still valid will sky rocket, with potentially destructive power losses. Mind that your FET is 0.9 x 0.9 mm^2 !!! Moreover, appart the 3.3V ON voltage I'...


0

Quite the old topic, but one important thing is missing : Half bridge voltage may force you NOT to use charge pumps. There is no practical charge pumps that allow to supply 20V to a high side switch with 400V ground offset. Charge pumps are more of a low voltage thing (I've seen them up to 60V bus voltage with few tricks to avoid reverse currents) but ...


1

I think there is no ready ADS template for push pull type configuration. Either you need to use single ended template and double the impedance or design your own load pull setup for this type of configuration.


0

Another (I think better) way to decrease the inrush current is to PWM the motor voltage, starting at 0V and increasing it slowly. The PWM frequency should be high enough so the motor cannot 'keep up'. A not audible PWM frequency of >50 kHz will most times do. A way to determined how slow to increase the PWMed voltage is to look at the rise time of voltage ...


1

1) since you're only dissipating about 0.2 W I would say this is more than OK assuming that this PCB isn't in a very hot environment. It is possible to calculate the thermal resistance (to ambient) using tools such as those on TI's website. 2) Possible: yes but I would not choose to do that! What you are then doing is temporarily giving the MOSFET a higher ...


2

With this arrangement D4 clamps negative Vgs and the other diodes provides a load impedance above Vgs(th) if less than 1.8V , then you can drive it with a current source with pull up or a voltage source or even a capacitor with diode negative clamping with PWM if you thought this was an advantage for isolated DC driving of the gate at some different ground ...


0

The complementary emitter follower gives much more current gain using the secondary supply and inversely lower source impedance V/I=Zout from the opto current-source CTR ( equivalent to low hFE) suitable for the TLP250 to drive a MOSFET and IGBT rather than the open collector type Optos.


3

In such cases, it is useful to redraw the circuit: as @Oldfart says in their comment, we have the following situation simulate this circuit – Schematic created using CircuitLab The source of the power MOSFET is connected to the ground via three of the four diodes \$D_1, D_2, D_3, D_4\$ packaged as a couple of double diodes: the resistors \$R_{21}\$ ...


1

Mosfet Driver TLP 250 . This image found by a quick search on the internet.


2

Your simulation has some problems. First problem is the FET is wired directly across a voltage source which has zero output impedance, so no matter how well it is turned on it won't affect the voltage. Second problem is there is no return path for the Gate drive. The entire circuit is floating, so it simply jumps up and down in time with the Gate drive and ...


0

Since Vbe =0 it cannot operate. Since Vce= 0 it cannot operate. To invert input voltage with respect to output collector voltage (which is always the case for relative changes in voltage being both positive with respect to emitter or 0V, when DC biased properly. Furthermore, the NPN collect Vce must be >= Vce(sat)for some linear operation. Thus the 10k ...


1

It really depends on what all the terminals are connected to. You'd probably want to connect it to the body because it has the most surface area (or in this case drain). The body is usually the best thermal pathway as there are only small leads attached to the source and gate. Here is a pic of a mosfet that I dencapsulated (all it takes is a bit of warm ...


0

The source did, in fact, need to be connected to ground although the 10k resistor was drawing too much current. Switched to a 180k resistor and everything started working as expected.


1

Also, I would like you to take a look at my unfinished code and ask if you could suggest anything. The not-equal comparison will not work well with a real analog signal because of noise gear10us = analogRead(A1); delayMicroseconds(10) if (gear10us != analogRead(A1)) { // ... } Suppose you stop your motor, and your sensor is staring at the same thing. ...


1

I have 2 images above. The first one is Ic vs Vce which i captured in a lab at university. The second I took off the internet of a FET because i didnt have one. In the FET one you can see that all the curves split off before they come flat. Giving a R = V/I for those curves will give different resistances. However, in my BJT one, you can see that before the ...


1

FETs are often used as Ideal Diodes when RdsOn is << lower than the load R unlike Schottky diode which have a low threshold but higher Rs than Silicon. So adding a diode in series would degrade the typical performance for losses as a switch. We can compare cost, size, power, speed , voltage drop, max current to compare a power Schottky power diode ...


0

Regardless of the polarity of charge movement, the substrate needs to be controlled. A rectifying-contact blocks that control in at least one polarity.


0

MOSFETs have very large input capacitance (several thousand pF). To drive this effectively you will need to run the driver stage at higher current with a lower collector resistor. Although feeding the collector resistor from the drain of the FET gives DC feedback to stabilize the operating point it will have the disadvantage of limiting the output swing as ...


1

Most but not all MOSFET have a low PTC which permits current sharing easily,. ALL CMOS logic has a PTC effect on Ron as well. All BJT’s and IGBT have a NTC effect which requires a small series R (high power) to share current. It does this by the added resistance so that the NTC effect never causes thermal runaway with the rise in current with voltage yet ...


5

So does this concept apply to parallel mosfets? Yes, it is quite common to do this in high current DC DC converters. The other nice benefit is you get double (or whatever number your paralleling) the heat dissipation while lowering the resistance because of the additional devices.


16

Yes. It does. Note that in general you can't blindly parallel transistors. You can parallel MOSFETs without special measures since as they get hotter they conduct less well which distributes the load more or less evenly in spite of individual component differences. Positive temperature coefficient. BJTs conduct BETTER as they get hotter so the BJT that ...


1

It looks like I found a solution highly dependent on the device parameters. Low leakage parts are the critical factor. Take careful selection of Ids @ Vt and Vgs Two inversions are done: The 1st must have a gate threshold of << 1/3 to 1/2 of Vbat and not 2 to 4V. This is what causes decay on the 1st time constant. hi-side mechanical switch, low-...


1

6 siemens is 6 amps per volt in drain current.


0

All right, a wrong MOSFET type it was (regarding the voltages involved). A NDP6020P P-channel MOSFET arrived. I put that one into the circuit instead and now everything works as intended. Only in late evening, the LED comes on and goes off already in the very early morning. The P-channel MOSFET does not heat up detectably at all. I may now up the charger ...


0

How it is self biased in every stage determines the gm of each stage which multiplies greatly from current mirror to complementary common source to drain follower then this open loop gain is limited by the feedback resistor ratio required to stabilize the DC Operating point. When the voltage bias and Vt is improperly matched, you may end up with no gain and ...


3

A MOSFET gate looks like a capacitor. The resistor will slow down the MOSFET turn-on and turn-off, and it would protect a delicate driver pin (unlikely, but possible), but in the end the thing will still switch. In that circuit the resistor is to slow down the MOSFET switching, to reduce the speed that the drain voltage changes, which in turn reduces EMI ...


0

Hold OFF MCU power till minimum threshold reached on PS capacitor. It's a challenge to make predictable power switches for an application such as this where the voltage rises very slowly. Brownout detectors within MCUs are designed to hold the device reset when the power drops to a critical level. However what happens at very low voltages is unpredictable ...


2

The small signal output resistance depends on W because \$I_D\$ depends on it. After all, you have: $$R_o=\dfrac{1}{{\lambda}I_D} $$ With \$I_D\$ being the current in the saturation region for the MOSFET. In saturation, $$I_D=\dfrac{1}{2}\text{K}\frac{W}{L}(V_{GS}-V_T)^2$$ So yeah, if you increase W, the \$R_o\$ does decrease because \$I_D\$ is in its ...


1

I think this'll work. Don't start out trying it with expensive hardware. R4 and C1 are guesses -- the smaller you make R4 the quicker it'll switch and the bigger the current pulse on the HV line. If you're going to use it over a wide span of HV voltages (you cited 50V to 150V) then you'll need D1 -- choose something with low capacitance, but that'll still ...


0

Your device has a minimum high input voltage of just 2.0 V, so you can trigger it directly with your 5 V signal:


3

Two things for you to check. 1) Ensure that the power supply rise time is FAST. Generally don't switch the power supply on with the inverter connected, the inverter can get into a linear mode that prevents fast switching of the MOSFETS and leads to destruction. Connect the inverter to an already powered and enabled supply. 2) Check your inductors are not ...


0

Looking at the board pictured and assuming those are dual n-channel chips, it likely works something like this: simulate this circuit – Schematic created using CircuitLab At least, I think this is how it would work? It doesn't really seem suitable for 3A to me... R1 and R2 form a voltage divider on M1s Gate. R3 is feed back. The only formula you ...


1

What you show is correct, a p-channel power MOSFET in series with the + output, source to the battery manager, and drain to the load (the LED driver). The question is, what is the amount of daylight when the solar panel output voltage drops MOSFET's gate voltage to the point where it no longer is fully enhanced and starts to "turn off". It might be lighter ...


12

The only mistake is using the NFET as a high side switch when it should be a low side switch with Vs=0V then with Vgs>=10V you pull down the load cathode and series R from the supply with the drain. So transistors used as switches (FETs and BJT’s) are always inverting. Vgs is chosen from the specs or as a rule of thumb Vgs>2.5 x Vt(max) the threshold of ...


7

TL;DR: I was driving my MOSFET gate at the supply voltage I was switching/ so I didn't maintain a high enough VGS. I'd failed to consider that Vs is not 0V once the MOSFET is on. Per the accepted answer the simplest solution to that is putting the MOSFET on the low (ground) side so that VS ~= 0, making VGS easy to keep high. See corrected (but warning, ...


0

I think you are asking the wrong question. Is there really any point in doing this while operating in the linear region? From your question I'm guessing you want to control something like a DC motor, hence my answer will be directed at that. A more appropriate way to control a DC motor, specially if you are using an arduino (which I assume you are since ...


0

A typical N-Channel MOSFET behaves as if a capacitor exists between the gate and source terminals. Usually, when referring to "gate voltage", one is usually talking about the gate-to-source voltage in the context of 3-pin MOSFETs. This "capacitor", when charged, turns on the transistor. Your confusion lies probably in the fact that MOSFETs are actually ...


0

The channel isn't completely cut off due to the strength of the electric field between the drain and the channel. We use this region because the current is no longer dependent on Vds and we can amplify waveforms without distortion as long as we make care to stay in saturation. I mean strictly speaking you're right the depletion region around the drain in ...


3

The IRF530 FET has a high gate threshold, and will not reliably turn on with 3.3V gate to source. Use a logic-level FET that's specified for 3.3V gate-source, or use another EMH1T2R in place of M1.


0

That will do, but you'll also have to take care of the pins, by deleting or commenting out 50 in: .SUBCKT FDV301N 20 10 30 50 You could also make that a parameter, by transforming Vtemp into: Vtemp 50 0 {temperature} (or some other naming, but try to avoid temp, it's a keyword in SPICE) and adding: .param temperature=25 With this, whenever you place ...


1

Enh. FETs have a (Vgs-Vt)^2/Ids impedance while diodes have a log Vf/If slope with a bulk Ri at rated current that approximates towards a linear curve above Vf@If max. Apparently FETs made with this desired sub-threshold characteristic will behave like diodes in the low current region below gate threshold. But you can't bias just any Enh. mode FET to ...


0

To answer your question, with a straight up ideal inductor, it can possibly result in zero current only during switching because the current will be delayed. To achieve this you need a saturable element, when the switch turns on, this component will hold DC voltage until it saturates, this time until saturation will be the delay between current and voltage. ...


2

Does this mean there will be relatively lossless turn on? From the point of view of the power supply feeding the converter, energy will be delivered to the load and also stored in the inductor. The energy delivered to the load is somewhat delayed by the inductor charging up so, for a short period of time less energy is delivered to the load and this may or ...


0

I have contacted ON Semiconductor support about ECH8695R. Their response is: The ECH8695R has specified RDSon 13.3mOhm @ 2.5 V. This measurement symbol is RDSon so it is for single MOSFET in that package So I assume this applies for other devices as well.


0

It raises the gain of the overall circuit as often one MOSFET can't increase the gain dramatically. Also taking a quick look at your drawing did you forget the source resistors? It'll separate the FETs from ground and alter your resistances. For example the vgs will be nonzero so the dependent current sources will have to be taken into consideration. Google ...


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