# Tag Info

## Hot answers tagged transistors

24

These transistors are devices which are highly optimized for use in ultra-low-cost and high-volume consumer products- off-line fluorescent lamp ballasts. The anti-saturation network (for fast switching) and free-wheeling diode are integrated onto one die, minimizing the component count and silicon area required (the diagram shows two parallel 'fingers' of ...

10

The transistor boasts "high switching speed" it gets this by having the lower transistor turn on and steal base current when the main transistor approaches saturation. Thus it is a type of Baker Clamp

7

That is a strange part. At first I thought it was an integrated Sziklai pair, a type of Darlington with a lower saturation voltage. But your part is different. When the collector goes below the emitter, the first transistor will turn on and short the 2nd transistor's base to its emitter, completing its turn off much faster by sucking out its base charge. ...

5

Revised after some thoughts, "saturation clamping" makes more sense. The way how the PNP works in the circuitry (clamping) is effectively what Baker Clamp is for. Thus, the manufacture's drawing seems agreeable. It would seem to me the moment the base of the NPN goes high, the collector goes low, which in turn pulls the PNP base low, whereby it ...

3

This means if I have a 3.3V pin on the base then I need a 3.3V/0.003A = 1.1K resistor but it would probably be a good idea to lower resistance slightly to ensure proper current. I am not sure if this is the correct procedure. The base-emitter junction will drop about 0.9 volts when driving a current of around 3 mA into the base as per this: - Hence, to ...

2

Since the application is a video signal amplifier, and the signal characteristics that are to be amplified are known, it is really the best solution to simply use DC coupling for video signals. AC coupling of video signal is really a hassle, as the frequencies of interest must go down to very low frequencies and the removal of DC components mean that the ...

2

For an NPN transistor, its emitter voltage $V_E$ is 0.7V lower than its base voltage $V_B$, so that $V_E = V_B - 0.7V$. This means that the emitter follower has a gain of 1, and the amplitude of Vout is the same as the amplitude of Vin. The output is merely offset by -0.7V with repect to the input. Since the output cannot drop below 0V, the base cannot ...

2

If it’s like most motorcycles, the neutral switch has a single pin that connects to engine case ground (closes) when the selector is in neutral. It’s open otherwise. You can check this with an ohmmeter. Some motorcycles might use a 2-pin connector if the engine is in an anti-vibration mount (e.g., a newer Harley), but nevertheless would connect to ground via ...

1

Hfe is an approximation for a transistor in the linear region, it doesn't apply to a BJT in saturation. You can use it to estimate a base resistance, but the characteristic curves will show you more detail. You should not rely on the saturation voltage of a transistor to limit the current in an LED. It can vary quite a bit over temperature and from unit to ...

1

I'd put 4.3 kohms between the 4071 output and the input of the cable, and 75 ohms between the output of the cable and the transistor base. The 4.3 kohms limits the 4071 output current to ~1 mA, well within its drive capability, but still above the 300 uA you say you need. The 75 ohms terminates the cable with a matched load (at least when the transistor be ...

1

Your first two transistors are biased wrong causing them to produce severe distortion. I corrected the biasing by reducing the value of R2.

1

In MOS circuits, it doesn't really make sense to analyze an open output like that -- when the gate V is 0, the voltage at the output will be somewhat indeterminate and will generally drift (slowly) towards 0. Usually there are some other complimentary signals that will drive that high impedance node in that condition. In your circuit, the NMOS with VG=3 and ...

1

Your black negative image could not be seen so I converted it into a normal positive image with black lines on a white background. The inputs must be opposites high and low for the motor to run. If both inputs are high or low then the motor does not run. The emitter-followers reduce the output voltage.

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