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21

There are various choices that can be made in the design of transistors, with some tradeoffs being better for switching applications and others for "linear" applications. Switches are intended to spend most of their time fully on or fully off. The on and off states are therefore important with the response curve of the in-between states being not too ...


16

For power MOSFETs, there is a good rule of thumb indicating that the newer the part, the better it is optimized for switching applications. Originally, MOSFETs were used as pass elements in linear voltage regulators (no base current degrading the no-load losses or overall efficiency) or class AB audio amplifiers. Today, the driving force for the development ...


12

It's the direction of the arrow that tells you: N-Channel P-Channel The arrow points from P to N (like a diode) so you can tell which is which channel wise. For example in the top diagram, the arrow points to the channel, so the channel is N-type. In the bottom diagram it points away, so the channel is P-type (i.e. from P to N) To clarify, this also ...


12

You need feedback if you want an exact ratio of 12/5, an open loop transistor will not do this. I would use an opamp with a transistor amplifier. You could also look for an opamp that will output 55 mA (most won't) and leave out the transistor. You need a rail-to-rail opamp so the output will go to ground. The supply voltage needs to be a few volts higher ...


11

It's due to the way they operate - JFETs (and MOSFETs) are Field Effect Transistors, so the way they control the current is different (to a bipolar transistor - FETS are unipolar devices) The gate to source impedance is naturally very high with these components - you can think of it a bit like squeezing a hose to stop the water flow - there is no actual ...


9

The JFET has several advantages over the MOSFET. The most important are: higher gain lower noise These are the overriding factors when building preamplifiers for low-level signals, such as those from microphones. Also, since there's no thin gate oxide that can be punctured by ESD, they're a little more "rugged" in that sense. -


8

The N channel JFET needs a negative voltage on its gate with respect to source therefore this complicates the power supply regime by requiring the addition of a negative rail. Similarly, for a P channel device its gate would have to rise above Vcc to be able to control it. Here's an n channel JFET pictorially: - As you can see, the voltage on the gate has ...


8

Whilst I'm inclined to agree with @Bimpelrekkie that this is probably an X-Y problem, for those who might actually need such a device there are some options. Most MOSFETs are sold as enhancement or depletion mode (less commonly) devices, however it's possible to tune the threshold voltage to approximately zero +/- tens of mV. For example, the Advanced ...


7

It looks like those capacitors and the optional ferrite bead are only meant to attenuate high frequencies. It is fine to squash everything above 20 kHz in a audio signal because you can't hear above that anyway. The audio circuit in the phone probably does that already anyway. However, if really high frequencies get in there, then you can't rely on active ...


7

I'd like to implement this circuit using a surface-mount JFET, but frankly don't have the expertise to pick out one which is likely to work for me. I don't know how many people we tell to LOOK AT THE DATASHEET. Front page from ON semiconductor / Fairchild: - Can anyone advise me on what parameters I should be looking at in the spec sheets Without ...


6

The JFET is acting as a diode. So why not just use a diode? The main reason for using a JFET as a diode is that it has low leakage compared to ordinary diodes.


6

You could tag a voltage divider off an PNP like this simulate this circuit – Schematic created using CircuitLab Note Vcc shown above is expected to be 5V. If your logic output is at a lesser voltage you would need to recalculate R1 appropriately. Or if ground needs to really be ground, the circuit below may be better. Note the control signal ...


6

No single transistor can do what you want. Instead, you need to use multiple transistors. A simple voltage comparator will do what you want: simulate this circuit – Schematic created using CircuitLab R2 and R3 set the reference voltage for the base of Q2. As long as the voltage at the base of Q1 is less than Vref, all of the current through R1 will ...


5

The statement is wrong. Both devices are voltage-controlled. The accurate model for a bipolar transistor is the Ebers-Moll equation. Note how the independent variable in the Ebers-Moll equation is: \$V_{BE}\$. Not base current! A BJT is a transconductance device, just like a JFET. The idea that a BJT amplifies current is a feature of the approximate model ...


5

The reason is because FET devices have (almost) no current flow through their base, and as a result an incredibly high impedance. If you look at current flow through a BJT, notice that the base has current flowing in: By comparison, in a JFET device current flows only through the drain and source. I've used a MOSFET, but the principle is the same: Now in a ...


5

I can only answer the first of your two main questions. Why would you want to use a MOSFET when the MODFET supposedly has a higher carrier mobility? Cost. First, if you want to make a GaAs/AlGaAs MODFET, that means you're working with GaAs wafers rather than silicon, and these are much more costly per device. Even if you make a MODFET with varying Ge ...


5

From page 3 of the datasheet, it appears the input resistance of the codec is between 15k-30k (nominal, depending on the gain setting). This is of the same order of magnitude as your JFET amplifier's output impedance, so the signal reduction is consistent with what you have observed. Regarding the clipping, the datasheet states that the input signal must be ...


5

The circuit you posted can work. These are the results of a hastily conceived LTSpice simulation with some guesswork values for the components: Therefore there could be something wrong in the simulation parameters or the choice of the values of the components. As I said in a comment, you should post additional information for us to be able to troubleshoot ...


5

Yes, the same amplifying classes applies to FETs or other amplification devices. What defines the amplification class is the conduction angle and the presence (or absence) of switching during operation (like PWM modulation, etc), not the amplification device (transistor). However, voltage-controlled devices like FETs are generally better suited for ...


5

The op-amp circuit provides better Precision. Accuracy. Ideality of the current source. Repeatability. Immunity against power supply voltage. Immunity against temperature. The JFET circuit might exhibit (especially against 741) better AC response. The JFET also might be better at lower current (as the 741 has "quite high" input bias currents). Precision, ...


5

Any given FET (at least when at a fixed temperature) will have just one curve, somewhere between those two limits. Unfortunately, FETs are difficult to make consistently, and if you buy a number of 2Nxyz FETs, you can expect that their curves will not be identical, even if they are from the same batch. If they're from different batches, you should expect a ...


5

Build non-inverting OpAmp configuration with a gain of 12/5=2.4 (using a rail-to-rail opamp when you supply it with 12V). $$V_{out}=\left(1+\frac {R_2}{R_1} \right)V_{in}$$ Using E12 series (which I know by heart): R2 = 47k and R1 =33k yield a gain of 2.42. Using E24 or E96 may give nicer results.


4

From your description, a device like Philips BUK9MNN-65PKK might be suitable for your purposes. Dual channel, so you would need just 2 of them for your 4 channels. Inbuilt current limit and sense, junction temperature sense Vds 65 Volts Id 7.1 Amps Independent channels, minimal cross-talk The Current Sense lines will provide current / over-current info to ...


4

These metal-oxide varistor PTC current-limiter solutions are often used in USB hubs and automotive applications such as windshield motors and are very low cost. One chooses the Holding current or trip current according to the design to protect from thermal overload as the devices will generally heat up to 85'C and trip faster than the devices you need to ...


4

Because the base-emitter voltage of a BJT in its operating region will be affected by base-emitter current, and vice versa, changes to the base-emitter voltage of a given transistor will affect the collector-emitter current. On the other hand, the amount of base-emitter voltage change required to affect a given collector-emitter current change is often huge ...


4

Well, if that circuit does in some way function it's well outside of the normal operation of a 741. The common mode range of a 741 only goes to within a couple volts of the negative rail, below that the current sinks that bias the differential front end will no longer function. Even with a single-supply op-amp the circuit will not function properly because ...


4

If you are familiar with BJTs, you control current from collector to emitter (Ice) by varying the current into (NPN) or out of (PNP) the base (Ib). If you deprive the base of any current, Ice goes to zero. This is cutoff mode. So the BJT is a current controlled current source. The FET on the other hand is a voltage controlled current source. In the FET you ...


4

Are depletion mode devices being made obsolete by the development of the technology of enhancement MOSFETs? No, I wouldn't say so. Depletion mode devices, like most discrete devices, are being made obsolete by the development of integrated circuits. As time goes on, ICs are developed for progressively more specialized purposes, reducing the need for ...


4

Keep in mind a couple of things that are valid for every JFET (both N- and P-channel): For many models source and drain terminals are interchangeable, i.e. there is absolutely no difference between the two (their datasheets will often tell you this upfront). Even when S and D are not interchangeable, they work almost the same, i.e. if you exchange them you'...


4

In general \$g_m\$ in simple term is a "gain" for any transconductance amplifier. And because transconductance amplifier is nothing more then a voltage controlled current source (VCCS) the gain expression is \$g_m = \frac{I_{out}}{V_{in}}\$. For example if is \$g_m = 1\:Siemens\$ any change in the input voltage by \$1V\$ will change the output current by \$...


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