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5

You have a massive power MOSFET with a lot of capacitance. If you reduce the size of the MOSFET you can go faster, or change the type to a different technology than silicon. A cheap one that should show a lot (like 5 or 10:1) improvement is the FDD1600N10ALZ. They're cheap enough, but you can run some simulations to see what the predicted performance is. ...


4

The transistor is like all things in the real world...it is not perfect. Even when we expect that no current at all should be flowing there is a tiny leakage current. Note that the current flowing when the transistor is active is about 6,000,000 times greater than the leakage current. In most practical situations the leakage current is negligible.


4

There's an easier way to find the DC Q points and R value starting the 5V for Vce1. (or UCE1 ) It is simpler because we were permitted to ignore I(R) thru T2 as it is only 0.5% more (hFE=100 * 10uA vs Ie2~2mA ). 10uA/2mA*100%=0.5%


3

Yes, it looks like the circled 100K resistor & the 10K resistor should be on the moving contact of their respective switches. Edit: Like so:


3

During a read T1 is off and T3 is on. Then T2 controls the read line from the charge on the capacitor. Since it’s gate is very high impedance, no current is drawn from C to do that. That leaves C’s charge, hence the state of the bit, unchanged. T2 is both the strength and weakness of this cell design. Yes, it makes readout nondestructive. But it also ...


2

Here's the circuit you have, drawn slightly differently: simulate this circuit – Schematic created using CircuitLab Here, you can see that if the switch isn't closed, then \$R_1\$ pulls \$Q_1\$ into an active-on saturated state, causing \$Q_2\$ to be off. However, this means the collector of \$Q_2\$ is floating (very high impedance.) If you want \$V_\...


2

There is nothing explicitly wrong with the schematic, but for more stable operation it is incomplete. Add a 1 K - 10 K resistor from the Q2 collector to GND. This will assure that Q2 leakage current has somewhere to go when Q2 is off, and you will get a firm 0.0 V reading.


2

Basically, in a FET, lower RdsON means a bigger chip which means higher Qg, higher capacitance, ie a slower FET and higher gate drive current if required to switch fast. RdsOn i sonly relevant if you use the FET as a switch, ie fully turned on. Since your circuit uses the FET in linear mode, RdsON is not a relevant selection criteria for your FET. There ...


2

It seems to me like you could just use an emitter follower circuit. If you do this with the NPN transistor then the output voltage would just be Vout=Vd-0.6 If you need a current then just scale the output resistor accordingly. If this answer isn't enough then you need to restate the problem in a more complete way.


2

Yes, it's correct. In a real design you'd want more base current to ensure the transistor is fully turned on, maybe 5% of the collector current. The collector and emitter are different (usually) on transistors. It's possible to make a transistor that is symmetrical where they are the same, but few available BJTs are made that way. JFETs, on the other hand, ...


2

There are a few problems with your layout , I can imagine from your resonance. The 10:1 probe 10pF to 15pF causes Drain resonance with your wire jumper inductance from Vcc low ESR 5V Cap to Drain @ 10nH /cm. Solution add 50 OHm series near drain for a test pin The inductance in the power to drain increases the resonant gain which can be reduced by using at ...


2

It's almost certainly an astable multivibrator, or LED flashing circuit, which is a classical circuit and one which is often a first project. Here's a tutorial on one. https://www.build-electronic-circuits.com/astable-multivibrator/ Try following the tracks on the bottom of the PCB and see if you can match it to the diagram in the tutorial I linked.


2

This circuit solution is known as (BJT) differential pair, (BJT) differential amplifier, long-tailed pair, emitter-coupled pair... It has two inputs - the T1 and T2 bases. The input voltages are applied between the bases and ground; hence the name "single-ended voltages". The circuit amplifies only the difference between them (differential mode) but does not ...


1

It is supposed to flash the LEDs alternately. Here is a likely schematic using similar part types and values. I've made it a bit asymmetric by changing R1/R2 by +/-10% from nominal (and skipped initial condition) so it will start reliably. simulate this circuit – Schematic created using CircuitLab Below is a simulation in the Circuit lab ...


1

This is your general formula 2 which is valid for all fets which have source, gate and drain The red partial derivative is the transconductance. The green one is the output conductance. To get the actual transconductance formula calculate the red partial derivative from your model equation (it seems quite simple, like an ordinary silicon mosfet, I guess ...


1

First of all, learn from a good design rather than this one. This answer will not "spoon feed" you how to learn, rather, what to learn. Your circuit has some reliable clues, that it is a terrible design and the first and most important, is that it has no "specs." or "expectations in writing" as I like to say. Typical Amp. Specs: @ +/-15V 25'C ------------...


1

Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab I used Mathematica to solve it, using the following code: Solve[{Ix == I1 + I4, I4 == I7 + I8, I2 == I5 + I3, Ix == I3 + I6, I1 == ((Vx - V2)/(R1)), I2 == ((V3 - V4)/(R2)), I3 == ((V4)/(R3)), I4 == ((Vx - V1)/(R4)), I5 == ((V4 - V5)/(R5)), I6 == ((...


1

Here's a different approach. It's the way I thought about the problem: simulate this circuit – Schematic created using CircuitLab In the problem text, they suggest that \$I_{B_1}\$ and \$I_{B_2}\$ can be ignored (\$I_{B_1}\ll I_{C_1}\$ and \$I_{B_2}\ll I_{C_2}\$) for the purposes at hand. But that's obviously wrong. If it really were true, then there ...


1

You can use a transistor as a heater, however you would be better off to add at least one resistor to control the collector current. Given that, you might want to use the resistor itself as the heater and drive the BJT or MOSFET fully on to minimize the power dissipation in the switching device. Dale makes suitable chassis-mount wire-wound resistors, and ...


1

The image of Figure 1 may help a little. Figure 1. Horowitz and Hill, The Art of Electronics. "Transistor Man" looks at the current at the base, and adjust the current at the collector so as to be a multiple of the base current.


1

No. The amount of conventional current (not electron current) entering the base and flowing out through the emitter determines whether an NPN turns on (i.e. how much current passes from the collector to the emitter). For a PNP, it is the amount of conventional current flowing into the emitter and out of the base that determines how much current passes from ...


1

You need to reverse the orientation of the LED. Connect the anode of the LED to the output of the NOT gate, then connect the cathode to the resistor and the other end of the resistor to ground. The LED will light when the output of the NOT gate is high, which is when TX is low.


1

Ignoring the ringing (Which might just be poor scope probe technique) you see basically straight line rise and fall, which says to me that the mosfet capacitance is introducing a slew rate limit. Pick a mosfet with much smaller gate capacitance's or lower the 600R resistor (or both). Or go with a BJT in place of the mosfet, will need a little base ...


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