# Tag Info

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Here is an application note that does exactly what you want. Design of a Low Cost, 45W Flasher with Short Circuit Protection Using LM2902 Quad Op-Amp and CSD18534 MOSFET Texas Instruments Application Report SLVA650–April 2014. They say: This Application note presents the design of a low cost, flasher circuit with short circuit protection. The design ...

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You need to feed the transistors with some volts maybe from 3.3V of PI and you need to connect the emitter of both transistors with GND of Pi since you will put voltage in those transistors from Pi. See the first picture, then you'll understand if you haven't already.

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I see what you mean about resistor sensitivity. However, looking at your simulation - Holy Cow! I can see your problem right away, and your problem is called 10 ohms. Your collector resistors are WAY too small, and will draw 1/2 amp when the transistor is turned on. Try reworking the circuit with 1k collector resistors, and 10k base resistors. Some of the ...

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I'd use a microcontroller and do all the logic in firmware. Connect all 8 lines to a single port and read them simultaneously in a periodic interrupt. The firmware looks for the port value being not FFh (assuming buttons are wired as pulldowns). If so, it looks to see which button is pressed, declares that the winner, and locks out subsequent state ...

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You could use an emitter follower and drive it from a 5V input. It will drop one Vbe drop, about 0.7V, so the power dissipation at 150mA would be about 100mW, which should be okay for most transistors over a reasonable temperature range. Use a pull-up resistor on the base if you don't have something like an HCMOS driver. simulate this circuit – ...

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You can't make a Zener diode from a commercial MOS transistor. Zener breakdown (effect) relies on quantum tunneling of electrons trough the thin space-charge region formed in heavily dopped pn junctions [1]. In common MOSFETs, the only heavily dopped regions are the source and drain, while the bulk is usually a lightly dopped p region. Thus, making pn common ...

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This works: - simulate this circuit – Schematic created using CircuitLab As the drain voltage rises from 0V D1 is not conducting until it reaches its zener voltage (simplistic words). Then a gate-source voltage starts to develop as the drain voltage rises further. At some point (maybe 1V - 4V higher than the zener voltage) the drain starts ...

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Go to digikey.com, product listings, Discrete Semiconductors, FETs (single). Select package type TO220AB, rank by increasing Vgs(th). There are two pages of TO220 MOSFETs with gate threshold of ~1 volt or less.

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What are your power requirements? Looking at the datasheet for that MOSFET: I really depends on the voltage you will apply to the MOSFET while it is switching. This particular device has a 150W power rating, so you have plenty of room unless you are passing high current. If you are planning on switching this MOSFET with some kind of microcontroller then ...

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simulate this circuit – Schematic created using CircuitLab Use Relay.

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You'll have a tough time finding an NPN transistor that meets all the specifications of that one in the same package, but you may not need that. The particular transistor you linked to has a relatively high beta (min 240 at 500mA) and a relatively high breakdown voltage (80V) without resorting to a darlington structure. A BC639 can do 100V but it doesn't ...

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The most obvious thing you're doing wrong is using the minimum h[sub]fe[/sub] in your calculations. That is, starting from a given value of the base resistor R, you find the LED current I by solving I = h[sub]fe[/sub] x (V[sub]out[/sub] - V[sub]B[/sub])/R ,right? So using the minimum h[sub]fe[/sub] only tells you the minimum I which your resistor will ...

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The typical gain of a 2N3904 is 200-300 at a couple mA collector current (more as it warms up due to not being saturated) Even with 560K, that's half a mA or so, which will give plenty of light from a modern LED, but you should be able to see that it's not as bright as when a 10K resistor is used. Do not use the hfe for this calculation if you want the ...

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From your comment I have to say the LED turning on does not indicate saturation in this configuration. Saturation of the transistor is defined as the case when increasing the base current is not followed by a further increase in collector current (i.e. the collector current has saturated to a maximum). Actually, saturation is a state of the transistor used ...

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Firstly, you should notice that $v_{BE}$ is not the only input variable of a transistor. For instance, if you manage to keep $v_{BE}$ constant somehow, you can still change the transistor coletor current by varying $v_{ce}$ (Early effect). Now, I believe that you are confusing $v_{BE}$ with its cc component $V_{BE}$. That is, $v_{BE}=V_{BE} + ... 0 Without more context, I can't answer the question of why different expressions are used but do note that r_e = \frac{1}{g_m}||r_{\pi} so the expressions are, in fact, equivalent. To see this, recall r_{\pi}= \frac{\beta}{g_m} Thus, \frac{1}{g_m}||r_{\pi} = \frac{r_{\pi}}{\beta}||r_{\pi} = \frac{r_{\pi}}{1 + \beta} = r_e 0 This should 100% work fine the way you've drawn it, don't change a thing. Do, however, check carefully the wiring on the op-amp and other parts. There is something wired wrong. Maybe V- (pin 4) is not connected to ground or V+ (pin 8) is not actually receiving the 3.3V. Or the non-inverting input is floating around. I would expect different symptoms if ... 2 The opamp is rail-to-rail and it can run from 3.3 V, so that's OK. One thing to check is that Rl is small enough. At 1 mA, the emitter of of Q1 will be at 1 V. That means the collector should be 1.5 V minimum, preferably 2 V, for good regulation. That leaves 16 V for the load. This can't be more than 16 kΩ, else there isn't enough available ... 6 Red LEDs have a much lower voltage drop than other colors so in order to compensate, the mosfet is having to drop a significantly higher amount of voltage/power. Red LEDs usually have a drop of 1.8 V while Blue and green are up into the 2.5-3.3 range generally. This means that the Red mosfet will have to drop about an extra volt at whatever current you're ... 8 The base-emitter junction of a transistor is just a diode. When the collector is disconnected, current can flow through R1, the base-emitter junction, D1 and R2, with a current of roughly 6.5V / 25270Ω = 260µA, which is enough to see. When the collector is connected, R1 no longer limits the current to the LED, so you'll get more like 6.5V / ... 11 In such a circuit the emitter current equals the collector current PLUS the base current: Ie = Ic + Ib In normal situations the collector current is much larger than the base current, hence by good approximation Ie = Ic But in your case the collector current is zero, so the first equation degrades to Ie = Ib Which is exactly what you see: a very small ... 0 It's an MJD31, which is equivalent to a TIP31 (a popular 3A 40V NPN transistor). To see if it is causing the short you could just unsolder the Emitter lead. Inductor L3 looks like its insulation has broken down, and will also need to be replaced or rewound. 1 If you want to use BJTs, the following ought to work: simulate this circuit – Schematic created using CircuitLab You will, of course, need to do this 4 times. ETA: I got the labeling wrong on the drive levels. Tip of the hat to Dave Tweed. Q1 and Q3 take your 3-volt signals and use them to drive the bases of Q2 and Q4. The TIP series ... 2 Keep in mind that when you use NPN transistors on the "high side" of the H-bridge, you need to drive their bases to +12V, since they're functioning as emitter-followers. The emitters can't go any higher than one diode drop below the base voltage. The better approach is to use PNP transistors on the high side, so that they can be put into saturation for ... 0 You need resistors in the base leads of the transistors to limit base current, and thus collector/emitter current. Also, an NPN emitter follower can only pull things up - it can't pull things down. – Peter Bennett Thanks! I fixed the problem by adding a 10K resistor to the base. I also rearranged the transistors so that Q2 is able to pull the ... 3 If your load current is 1mA (guesswork on my part) then this will produce 1.5 volts across R3. The voltage on the base to produce this must be about 2.2 volts below Vcc (1.1 volts above 0V when Vcc is 3.3 volts). If you assume R2 is 10 kohms then R1 will have 1.1 volts across it and R2 will have 2.2 volts across it - this means a current of 220uA. This ... 1 It's not a great idea and that 3.3Megohm resistor gives you a hint why... The FET has a very high input impedance, so that 3.3Megohm resistor defines the input impedance of the amplifier. Designing a bipolar transistor stage with such a high input impedance is possible but not easy, and I believe you would need at least two transistors, possibly connected ... 2 There is no need that the left transistor in your schematic will be turning ON first. I think that the article you read may be making an assumption that the left transistor is turning ON first to explain the circuit. The setup given in the schematic is a multivibrator. The diodes in the circuit provides a positive feedback such that when ever one transistor ... 0 One of many reasons is that the Gate is at +30 volts of the first transistor, while the gate of the second transistor is at a potential of a voltage divider, somewhere below +30 volts. The R3 is just a current limiting resistor, so the Gate is experiencing +30 volts at all times. 4 Every transistor has a current gain, usually$\beta$or$h_{fe}$in the datasheet. Typical values are on the order of 100. When the transistor is not saturated, then the base current and collector current are related by this factor:  I_c = h_{fe} I_b  When the base current increases to the point where collector current can increase no more, the ... 1 To answer your question about how the CPU turns RAM signals into circuit-level operations: TL;DR Logic gates. Full answer I think an example of what you're asking could be in some ways explained with a decoder, simply because a decoder can "translate" bits of information into something else. In this case, I'll try to explain how a decoder might allow you ... 0 It's simple, really. When the transistor conducts, the current has to choose between going through the collector of Q1 to the emitter and then ground, or going through R2 and D1. The diode only allows current to flow when biased, which in practice (for the range of currents in which the LED is simultaneously visible yet not damaged) means a constant voltage ... 1 It is conventional to draw circuits so signals flow from left to right, like so: simulate this circuit – Schematic created using CircuitLab (You need a base resistor, R3, to limit base current, and a resistor from base to ground to ensure that the transistor does turn off when the switch is open) With the switch open, the base will be held at ... 1 You are correct for the most part, but what you have here is essentially two switches that are working in inverted states. If you drive the base of the transistor then you will draw about 29mA collector current from V1 and force, assuming saturation, the collector to be about 200mV (Vce sat). Current would flow through the diode if the voltage at that node ... 0 I would suggest using an off the shelf module. Googling "12V battery charger for [your type of battery]" would be a good start. I can't link you because I do not know what type of battery, I can assume lead acid though, which is good as these are fairly common circuits. As for the 1st point, connecting a 12V battery to the modem will not be damaging. The ... 0 Also, in LTspice you would model you transformers as 2 inductors and couple them using a "k statement", here is an article regarding how to model transformers in LTspice for more information 0 I use microcap and you create two inductors (say L1 and L2) then choose "k" from: - component/analog primitives/passive components/k You then get this dialogue box: - Fill out "L1 L2" and coupling and you should be fine. 2 No, BJT has less extent than MOSFET because BJT is used where we need more speed. But MOSFET is used where we want less power dissipation, low operating voltage, less noise, simple manufacturing process, low chip area and easily scalable. Now, we use both devices as BiCMOS (BJT + CMOS) which has more speed than MOSFET and low power dissipation than BJT. ... 1 Sir, in case of NMOS, we take threshold voltage Vtn ( it is positive) and in case of PMOS, we take threshold voltage Vtp (it is negative). Transistor is said to be OFF if: NMOS OFF: VgsVtp In normal case, we always take NMOS as reference (its opposite is PMOS), so we can take Vtn as Vt. During calculations in linear and saturation region for PMOS, we can ... 4 Yes you can do it with a BC548 (or another NPN transistor). You need a circuit like the following one: At the input of the circuit (i/p) you have to connect the arduino output. At the output of the circuit (o/p) you have to connect the camera pin (SHUTTER or FOCUS). To calculate Rc and Rb we need to check the Collector-Emitter Saturation Voltage in ... 0 As already stated, the transistor will be used in switching mode, thus alternating between the cut-off and saturation. My contribution, if you find it of some use, is to point its operation. I'm assuming you're using the thyristor in integral control mode (instead of phase control mode). Thus, to turn on the load (triggering the thyristor) applies a ... 0 There is no "Q" (quiescent) point in this type of circuit. This is not a linear amplifier. Instead, the switching transistor Q1 is either full on (saturated) or full off (cutoff), which is highly nonlinear. You basically need to analyze the two states separately. 0 The simplest thing is to find suitable SPICE models and do the simulation. In the beginning ignore the transofmer. I know you are hoping that there is some simpler solution but if you going to do this on the paper it would take some time. I would use LTSpice/TINA TI and try to plot amplitude/frequency characteristics. After that the quality factor is the ... 2 Your circuit is not good. The LED are in the wrong direction 10k far too high to get a reasonable LED current 10k is also too high to put the transistor in saturation Put the LEDs in the right direction. Assuming they are standard LEDs, assume 1V per LED @ 10mA, hence R2 must be ( 5 - 2 ) / 0.01 => 300 ohm, use 330 Ohm. Use ~ 1k for R1. 3 Steps to design: The collector current of the transistor is given by I_C=\frac{5V-V_{D1}-V_{D2}-V_{CEsat}}{R_2} Where$V_{D1}$and$V_{D2}$are the forward drop of LEDs.$V_{CEsat}$is around 0.2V for BC547. So first fix the value for$I_C$by selecting a value for$R_2$. The value of$I_C\\$ must be less than the maximum collector current of ...

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In the era of technology, we need smart devices (small, fast and efficient). These devices are made up of integrated circuits (IC's) which contain a no. of transistors. We need more and more transistors to make IC smarter and faster because in electronics, every circuit in an IC is made from an adder, subs-tractor, multiplier, divider, logic gates, ...

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The transistor won't turn off as fast due to the base emitter junction being saturated. I have seen this before and simply place a nmos-fet in place of the transistor. Source to GND Gate to control signal (100ohms would be more than large enough in series) Drain to LED. This should allow you to switch on and off in 10's of nanoseconds

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Assuming you can live with a certain amount of slop in your output, you don't need a transistor. simulate this circuit – Schematic created using CircuitLab Consider if the battery is missing. Then depending on the load current, the output will be in the range of 12 to 12.5 volts (Schottky diodes have lower forward drop than standard diodes). ...

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A stated earlier you have drawn an NPN transistor. I would actually recommend a FET especially if you are drawing over 50mA due to power dissipation in the transistor. If you find a PFET, connect the source to the battery and the drain to your load. Keep in mind your FET must be rated for the current you wish to draw. I would also recommend that R2 ...

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Aside from increasing raw storage capacities of RAM, cache, registers and well as adding more computing cores and wider bus widths (32 vs 64 bit, etc), it is because the CPU is increasingly complicated. CPUs are computing units made up of other computing units. A CPU instruction goes through several stages. In the old days, there was one stage, and the ...

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