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## Hot answers tagged bjt

8

There is a third reason, related to your reason #2: Because the voltage across the BJT is low, its power dissipation is also low. This point was driven home for me many years ago when I was building a variable DC power supply1 with current limit. A transistor switched on a lamp2 when the current limit was engaged. When I was testing it, I deliberately ...

6

Reason 2 is the answer. Since P = VI we have two switch states: Transistor off: I = 0 so P = 0. There is no power dissipated in the transistor. Transistor on: I is high but V is low - typically 0.2 V or so for a BJT. That means that P = 0.2 × I. This may be as low as we can get with a BJT but will be a lot better than V/2 × I/2 which would be the half-way ...

6

This should actually be only a comment CircuitLab transient analyzer shows that you are right. There's no such thing as storage time - not at least when the transistor is Cicuit Lab's version of 2N3904. Here's a circuit with 1N4148. Signal V2 is bipolar square wave 5V peak: The diode conducts to reverse direction about 10 nanoseconds. Transistor turns off ...

5

As @Brian says, in the saturation region this is true. The transistor has a dynamic resistance (for very small voltage changes) of approximately Vt/Ib. So, for example, the collector of an NPN transistor with 50uA of base current will behave approximately like a 500 ohm resistor to ground for very small voltage changes (mV or less) that are AC-coupled. ...

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You seem to have at least one short. Orcad's schematic editor has the delightful habit of adding shorts here and there, perhaps there are others.

4

Remove R4 and confirm that the motor remains off. Shorting R5 is not sufficient to be sure, there could be D-G leakage on Q4 for example. I would suspect Q4 first, damage to Q3 would likely be more of a random failure (toys are not known for using the finest space-qualified components). Semiconductors do tend to fail "on" unless they've been abused ...

4

For ideal resistors, the voltage drop across a resistor is equal to the current through the resistor times the resistor's resistance. This is Ohm's law. $$V = IR$$ This law, rearranged is $$R = \frac{V}{I}$$ For a non-linear device, there is no single value of R that relates V and I in a linear way. We can still refer to a "resistance" $R=\frac{V}{... 4 No - the "quasilinear" region of the BJT cannot be used as a (grounded) resistor (as it is the case for the FET). Two main reasons: The Ic=f(Vce) set of curves does NOT cross the origin Even in the vicinity of the origin in the 3rd quadrant, these curves are NOT symmetrical to the "normal" first quadrant (for positive values of Ic and ... 3 does it have a recovery time when turning off? Yes. The transistor will not turn off until the concentration of minority carriers in the base drops significantly. Since the transistor does not get into saturation I think it should [turn off] pretty fast. Yes. Because the transistor is not in saturation, there will be less minority carriers in the base, ... 3 If you simulate your circuit, you will get this result: As you can see, the current through the bases of the transistors is almost 40mA and the current through the 1k resistor is less than 1mA. The is because the base emitter paths of the two transistors are appearing across V1. Because of this, the full 2.5V appears across the base-emitter paths of the two ... 3 Most transistors, and especially power transistors, exhibit long base-storage times that limit maximum frequency of operation in switching applications, unless clamped, such as in this circuit with Vce clamped by Vbe ( or with a Schottky diode B-C clamp or a Baker's dual diode clamp etc). In storage time tests Vbe is switched from Vbe .0.6 at some Vce(sat)/... 3 I'm sure you've encountered the idea of the slope of a curve, before. It's the first thing they teach when learning calculus. Remember this formula? f^{'}\!\!\left(x_0\right)=\lim_{h\to 0} \frac{f\left(x_0+h\right)-f\left(x_0\right)}{h} It's just the local slope of the curve at$x_0$. The dynamic resistance is like that, except that the curve is ... 3 If I have done my math correctly, to a very good approximation, \Delta V \approx \frac{kT}{q} ln(n) = V_T ln(n) where k is Boltzman's constant T is the absolute temperature in Kelvins n is the ratio between the reverse saturation/leakage currents of the two "diodes". q is the charge of an electron$V_T$is the temperature equivalent voltage ... 3 Help really appreciated I'm going to trust you on that point Miss Mulan. Your circuit won't be behaving like an active (and saturated) BJT any more - there is too much forward bias between base and collector for this to happen and, all rules about transistor amplification are laid-to-rest and, the BJT behaves like two forward biased diodes: - Compare ... 2 C5 vs. Q1: You have biased Q1 to operate at 1.5mA emitter current, which means that the resistance looking into the emitter of Q1 is about 17 ohms. Yet at 10kHz, C5 has an reactance of 105 ohms. The rule of thumb here is that the capacitive reactance should be 1/10 the emitter resistance at the design frequency -- so you want C5 to be about 100 times as ... 2 The definition of saturation is that the base emitter and base collector junctions both are forward biased. This means that for an NPN BJT, Vbc will be positive. So the real answer is that as soon as you apply 1kV or something to the collector, the transistor will not be in saturation any more. 2 You kind of don't need too many equations. Look at the datasheet to get Vce,sat and Vbe,sat - then make those the junction voltages. Then solve for currents and voltages. So maybe Vbe is 0.7 V and Vce is 0.2 V. Then just do nodal analysis and verify that current is flowing the correct way for a BJT in saturation. EDIT: This answer is for a saturated BJT. But ... 2 If the current mirror was perfect then the 300uA current and the left-hand 294.1uA would be equal to each other. As things stand the mirror is pretty good with only 5.9uA difference between the two sides. A mirror can be improved with the addition of an extra two transistors to cancel out the effect of the extra current in the right hand side of the mirror ... 2 Somewhere around 0.5 to 1mA is fine, forced beta of 10-20. The transistor is specified at a Ic/Ib = 10, but 20 will be fine too, especially at only 1mA Ic. So the resistor would be about (3.3V - 0.7V)/0.75mA or about 3.3K$\Omega$. The hFE is specified at Vce = 1.0V and you would like it to be more like 0.1V to minimize the losses in the transistor. You ... 2 The two transistors as originally wired will short out the power supply via their respective base-emitter junctions so you should arrange for this type of modification: - This circuit can now be driven correctly and, with V2 at 2.5 volts, Q2 is off and Q1 is activated. 2 Yes, high reverse leakage might lead to the switch always being partially on, even when you try to turn it off. High reverse leakage is kind of like putting a (nonlinear) resistor in its place. Then you can see it's like a collector-base feedback biasing scheme that will could slightly bias the transistor into active mode. If you are switching at very high ... 2 Assume that the collector is connected to a potential such that the transistor is in active mode (eg. grounded). Assume junctions have 0.7V across them. The current through the 10K resistors and diodes is (20V-1.4V)/20K = 0.93mA (so far so good) So the base voltage is 10.7V above the -20V rail and therefore the emitter is at 10.0V above the -20V rail so the ... 2 20V / (10k+10k) minus diode drops is approx 1mA . This means all Si diodes will be 0.6V not 0.7 @1mA Therefore the two 10k voltage divider gives; Vb= -10V + 2x0.6. = -8.6V which controls the emitter voltage. Thus Ve = -8.6 - 0.6 = - 9.2V. Now you can compute Ie which = Ic within 1% for hFE >100. 1 Ohh, the answer is simple. Remember that BJT is a non linear device. i.e. collector current varies with base emitter voltage exponential. In DC analysis, we use that exponential equation to determine the voltages at different point of circuit wrt ground. However, sometimes we just assume that$V_{BE}\\$ is 0.6-0.7V, this simplify the calculations. However, ...

1

The essence of transistor behavior in the active region is as follows. The base-collector pn junction is reverse biased majority carriers have a difficult time crossing a reverse biased pn junction minority carriers easily cross a reverse biased pn junction unless the emitter injects minority carriers into the base, there are few minority carriers in the ...

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Please guide me where I am wrong because I am a beginner in electronics. You have miscalculated the parallel impedance of the RC network and the 10 kΩ collector resistor (R4). It's more like 9540 Ω than 8 kΩ. The mistake you have made is that you have assumed the impedances are just paralleled like parallel resistors but this isn't the case when the first ...

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I don't care who is explaining gm in transistors circuits because it is incorrect. That is why you are stumped. Common base circuit, is voltage amplifier that has the current gain (approx) as 1 but is a calculation of Alpha Trans-resistance Factor (instead of Beta) is explained: of course you looking for voltage gain, it has no trans conductance because its ...

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If the load voltage is fixed and there is an emitter resistor greater than Rbe to linearize the input impedance hFE*(Rbe+Re) then using a fixed input voltage and variable Rb will amplify conduction to appear as Rb/hFE as a common emitter amplifier. This is one way to make an Active Load . The other way is to use a fixed Rb and vary the Vin to control load ...

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Do you really want to compute the oupur resistance - including ro1 and ro2 - by hand? You can expect a rather huge expression. Here is the result from a symbolic calculator: Note that the transconductance gm is here expressed by h21/h11=FHfe/Rhie and the output resistances as ro=1/Ghoe ( + Fhfe_Q1 Rhie_Q2 + Ghoe_Q1 Rhie_Q1 Rhie_Q2 + Rhie_Q2 + Rhie_Q1) ( + ...

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I don't think battery power is very suitable for this circuit. Just enough light to see- you may not need a Darlington at all, in fact you may be able to drive the LED directly from the module. There is apparently a 1K resistor in series (check that) and try just an LED to ground. A few mA into a modern bright LED produces quite a bit of light in a dark room....

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