# Tag Info

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You are mistaking two different notions of active device. The notion you are referring to is the notion from in circuit theory. In that discipline a passive resistive (a.k.a. memoryless) two-terminal device is defined as a device whose V/I characteristic curve lies entirely in the 1st and 3rd quadrant. By contrast, an active device is simply a device which ...

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It's an active device because it uses a power source to produce an output in typical operation. It's not just its input that's used to produce its output.

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how does one go about choosing the ideal transistor? The collector current ability should match with the maximum load requirement. In your example the maximum load current is 260mA. I would choose any transistor with the collector current ability of more than 350 mA. Now, I am sure load current capability is considered. Next is to check the base current ...

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You can go for any Logic Level Mosfets (I usually go for N-channel) to control such high-powered loads such as the LED strip that you were stating. Google any Logic Level N-Channel Mosfets that are available in your place. They usually have "L" in them e.g. IRL540, IRL2203N, etc. Check their datasheet to know how much current they can handle and parallel ...

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Do you care if you switch on the high side or the low side? Because if you switch on the low side, a logic-level N-channel FET will interface directly to the 3.3V GPIO of your Raspberry Pi. It will have lower IR drop and thus have less thermal dissipation than a transistor. A 2N7002 will handle ~200mA by itself. Use two or three in parallel and you have ...

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You are so very close to answering your question. You want to know how to calculate the collector current and aren't sure how to develop that value. But let's look at the schematic again: simulate this circuit – Schematic created using CircuitLab Note that I've now included some information that was included in the question that you may have missed ...

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Since you need to find the transistor's beta you should calculate what the collector and base current must be for Vi=Vo=(Vcc/2) to be true. In order for Vo to be Vcc/2 than Rc must have Vcc/2 volts dropped across it. From ohms law you should get the needed current for that to be true. Using the parameters given for Vbe you should be able to calculate the ...

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First, in normal operation, the base-collector junction is reverse-biased. This causes free carriers (electrons or holes, depending on NPN or PNP respectively) to migrate across the junction, and creates a depletion region around the junction. Second, when current carriers are injected from the emitter into the base (by adjusting the base-emitter voltage), ...

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Your switch needs to pass at least 3mA when closed and withstand 6.6V when open, and its output needs to be isolated from Ground. The obvious choice is an optocoupler. Depending on how much output current the photo-interrupter can provide, you might get away with just a sensitive optocoupler or you might need another transistor to drive it. Using a driver ...

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Figure 1. From the datasheet on the linked site. The diagram is telling us to power the sensor on 1 (+) and 3 (-) and connect the load between 1 and 4. simulate this circuit – Schematic created using CircuitLab Figure 2. The internals of the sensor. These sensors are far more complex that you seem to have imagined. The type you have chosen is a ...

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NPN photo detectors usually leave the base open, or tied with a very large resistor to the emitter. Normally the light hitting the transistor excites the base-emitter, causing base current to flow. The collector draws more current, due to the transistors current gain. So then a resistor in the collector, biased positive (VCC) relative to the emitter allows ...

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Ices is the leakage current between collector and emitter at the specified Vces of 900V. It is different at 25C (10uA) and 150C (150uA) junction temperatures. BVces is the breakdown voltage between collector and emitter. 'Breakdown' is defined as the voltage at which the leakage current rises to 250uA, at 25C. Please don't put these voltages and currents ...

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As follow up to Elliot Alderson where he says: "input capacitance of the driving inverter is relevant." You find this when adding buffers on an ASIC. Buffers on an ASIC come in different drive strength: 1x, 2x, 4x, 12x etc. This is achieved by simply putting two or three 1x buffer in parallel. Thus a 4x buffer has four times the driver strength of a 1x ...

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The propagation delay depends most significantly on the load capacitance, which is gate capacitance of any transistors that are driven by the output signal as well as the capacitance of the wiring that connects them. As you said, the propagation delay also depends on the power supply voltage. The propagation delay for an inverter is not really dependent on ...

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The notion of "resistance to which it discharges itself" is not very meaningful. It's better to think about the path of the current. When Vin goes to 1V, the left end of the MOSFET becomes the source and the $V_{GS}$ is a high voltage so the MOSFET becomes conducting. Current flows from the capacitor, through the MOSFET, and into the voltage source. In ...

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Most 2n2222 transistors have a max Ic collector current of 800mA. If each LED (only one color) drew 20mA and you had 60 (usually what a 2m strip has) of them that would be 1.2A which means you'll burn the transistor out. If your using this with a controller, then it could source amps of current which will also kill the transistor. The other problem with ...

2

Start with applying Kirchhoff's voltage law on the path shown: In this loop only base current is unknown. Find it. Next, apply kirchhoff's voltage law in this loop: Use, Emitter Current=Collector Current=beta*base current. Find voltage between Collector and Emitter. If the final result satisfies that the transistor is in active region your answer is ...

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One of the problems with BJTs, even within a single family of parts, is that their $\beta$ may vary widely between each other and also their datasheet values. Another is that their $V_\text{BE}$ can be somewhat different, as well (variations of their saturation current.) And that's true even assuming only a single temperature of operation. Add in ambient ...

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Will you make only one circuit then customize the calculation of its base resistor and operate it at only one temperature? Never replace the transistor without re-calculating for its beta. But if you make many of this circuit then you must measure the beta of each transistor and customize the appropriate base resistor for each one. Or you can buy thousands ...

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The base-emitter current is a fraction of the collector-emitter current, but a non-zero fraction. For a larger collector-emitter current you need a large base-emitter current. The RF energy might be too weak to supply the larger base current required if you were to try and maximize your collector-emitter current in a single stage. The 1 megaohm resistor is ...

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Considering an n-p-n BJT, we have Vbe = 0.7 V (approx). Saturation starts to take place when the forward current from the Collector-Base junction starts to cancel out the collector current due to the carrier flow from the Emitter-Base junction. This forward current starts getting significant from a forward bias of around 0.5-0.6 V on the collector-base ...

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First Example Text Your first example is from a micro-electronic viewpoint. It's the kind of description likely found in a textbook on micro-electronics. Here's a diagram I took from an older edition of Jacob Millman's "Microelectronics: Digital and Analog Circuits and Systems": That diagram is, of course, for the active-mode behavior. When going into ...

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For a current source the small signal impedance provided by it is ideally infinite. In case of transistor implementations, it is crucial which terminal we are "looking into" when we are defining the small signal impedance of the current source. If we use an nMOS between Vdd, and a node to which another MOS perhaps is connected, we would be measuring the ...

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Why is β ever used in this kind of analysis? In this particular bit of analysis, for normal small-signal transistors (i.e., $\beta \simeq 100$ or so), the value of $\beta$ really only enters into things in the choice of values of R1 and R2. Assuming zero base current, all you need to do is get the ratio of R1 and R2 correct. But a real base current ...

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In very simplified terms: A saturated BJT's collector acts-like a voltage source between emitter and collector with only a small, but relatively constant voltage where $10\:\text{mV} \le \quad\mid V_\text{CE}\mid\quad\le 1\:\text{V}$. The depth of saturation varies from very light (when $400\:\text{mV} \le \quad\mid V_\text{CE}\mid\quad\le 1\:\text{V}$) ...

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The only circuit element that obeys Ohm's Law is an ideal resistor. If the transistor is saturated, then we assume that the voltage across the transistor, $V_{CE}$, remains essentially constant at $V_{CESAT}$. If you increase the voltage across the resistor, by increasing Vcc, then the current will increase. If you reduce the resistance value of $R_C\... 0 Must the gate and source/drain voltages be different? Yes. The voltage from gate to source must be greater than the threshold voltage. This means that the source voltage must be significantly less than the gate voltage for an NMOS transistor. These transistors are usually used as low side switches, where the source is connected to ground and the load is ... 2 Use a normally closed relay. Reasons: The relay is isolated, just like the optocoupler The shown circuit will only work when the (voltage) potential of "NC Terminal A" is higher than the (voltage) potential at "NC Terminal B." The relay will work in both ways. The shown circuit needs to be fed as you already indicate at question 1. The relay solution doesn'... 1 Why does this driver not have a load resistor? Because it is not needed. When a motor starts, for a short time there will be a starting current / inrush current / stall current. When a motor is blocked or stalled, it will draw this stall current contineously. If Q2 can handle this current, you don't need an additional load resistor. If Q2 cannot handle ... 1 Your circuit diagram is not complete. The lamp CANNOT latch on without some path for base drive to Q3 that doesn't involve the switch. As you said yourself, a connection from the drain of Q1 to, say, the node between R4 and R6 would do the trick. Otherwise, those two resistors serve no purpose at all! 0 From what I understand, you simply want to replace the switches in the H-Bridge circuit with a photo resistor to trigger a motor while allowing it to be reversible. I used the Falstad circuit simulator to mock up this kind of circuit and I believe it may be what you are looking for. Below is just a schematic of it. simulate this circuit – Schematic ... 0 If you really want a solution with only one transistor, take a look at the circuit below. R1 and R2 values can be the same, so that these two resistors are equivalent to a 4.5V supply with some series resistance. The idea is that when Q1 is off, the LEDs are polarized between 9V on the bottom and 4.5V on top, what makes D2 turn on. When Q1 is on, then the ... 1 when I wanted an AC water sensor I used a 555 with the probe connected in place of the capacitor. the 555 frequency changes in response to water level. simulate this circuit – Schematic created using CircuitLab but you could probably do a similar sort of thing just using microcontroller ports, pull down and time how long it takes the port to change, ... 6 As designed, your circuit is an emitter follower. You're applying 3.4V to the base of an NPN, and taking power off of the emitter. The transistor will try to hold the emitter voltage at roughly$V_{be} - 0.7\mathrm{V}$, or about 2.7V. That's not nearly enough for your LED. You want something like the following. You need to choose a transistor that can ... 8 There are two potential problems in your circuit. 1. The 2kOhm base resistor is too high. By applying 3.4V through a 2K resistor, and accounting for the base-emitter voltage drop of the BJT (in the datasheet) you get a base current of:$ I_{collector} = \frac{V_{supply} - V_{be}} {R_{base}} = \frac{3.4V - 1.3V} {2kOhm} = 1.05mA\\$ In the datasheet, the ...

3

Nullbyte, I agree with you, the situation seems to be confusing because some books/articles state that the BJT would be current-controlled and some other say voltage-controlled. This is a very unsatisfying situation - I really cannot understand, why some (many) people still think (no - they only believe) that the BJT would be current-controlled. There is not ...

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It's because current is a defining factor in the relationship between the base-emitter junction and collector current and it makes the most sense to use current here. It's effectively a diode and adding a voltage supply greater than it's saturation voltage without a resistor between them is like shorting it. Sure you could use the diode equation to relate ...

1

Add a blocking capacitor with a weak bias (high value resistor) to system GND on the output side. This will produce an AC waveform centered about GND.

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Voltage does not cause current, current does not cause voltage, at least for any meaningful understanding of the word 'cause'. They both co-exist. When the base-emitter junction of a transistor is biassed, an Ib flows into the base, while a VBE exists across it. If we now measure the collector-emitter current, we find the ratio to the base current is more ...

0

You say "path from the base to the emitter equivalent to that of a diode and the potential difference is irrelevant - it is the current that controls the gate." This is completely wrong. The BJT, like all types of transistors" is a voltage-controlled current source: it is the base - emitter voltage and not the base current that controls the collector current....

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Look at the datasheet of a typical transistor, for example the P2N2222A Notice look at the specified value of the DC current gain. See how only minimum values are specified. Only for Ic = 150 mA, Vce = 10 V is there a typical value specified. For that condition the DC current gain is guaranteed to be higher than 100 and is expected to be typically around ...

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Prinzipal your selection with 74154 is a bad chip, because you can never switch off all outputs and more bad this solution makes you not free to use every Stepper separate. Better you use the SPI serial bus of the MCU and connect it to 2x 74HC595. Reset the #OE Pin 13 Connect follow Pins to the chips: 1) MOSI from MCU to SER(pin14, first 74HC595) 2) QH ...

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First, if you study the literature, there are a lot of explanations of this, at every level. The one I liked best was the one I got in a class on solid-state quantum mechanics that was aimed at engineers, but the "earlier" ones (such as the one in the ARRL Handbook, just about any year after about 1970). This phenomenon happens because the base is thin, ...

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In the context of linear voltage regulators, the pass element (transistor) is controlled by negative feedback to achieve regulation. Here's a (very simplified) linear regulator: simulate this circuit – Schematic created using CircuitLab Q1 passes output current in the direction of the arrow (and that is the reason it is called the pass transistor). ...

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The STP16NF06FP requires 10V Vgs to drive it reliably fully on. You are giving it 3.3V approximately. You either need to find a logic-level MOSFET that is rated for 3.3V or lower Vgs or make a gate driver circuit to give more voltage to the gate.

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