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You can use 139 NPN Epitaxial Silicon Transistor instead of any MOSFET. Because 12V motor needs 1.2A current for moving and 139 NPN transistor works on approx 2A current. And motor should be connected to collector of transistor. This is the suitable circuit for 12V dc motor. The datasheet of 139 transistor:- ...


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I hesitate a bit to answer your question because I know that some persons will heavily disagree. However, because you have asked "what happens physically inside" my answer is as follows: The bipolar transistor is a voltage controlled device - that means>: The current Ic is determined and controlled by the voltage Vbe - and not by the input current Ib. This ...


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This is grossly simplified, but I think you're asking for a very simple answer, so take it in the spirit it's offered. The key lies in the fact that for some semiconductors/junctions, a) you can talk about majority and minority charge carriers, roughly electrons and holes, depending on whether the material is p-type or n-type. b) under a broad range of ...


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Transistors make use of semiconductors. Nowadays it means pretty much exclusively silicon. You have N-type (more electrons) and P-type (less electrons). When you put differently treated pieces of silicon together, you create junctions (PN junction, etc.). You can bias (put voltage on) a junction, that makes it either more conductive, or less conductive. You ...


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You've labeled your 'thermistor' RT1 as a KTY81. That device is not actually a thermistor - it is a 'silicon temperature sensor' http://www.nxp.com/documents/data_sheet/KTY81_SER.pdf Have you actually used this particular device in your real circuit? If you've used a 'normal' NTC thermistor instead then this would explain the behaviour of your circuit ...


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A vehicle speed sensor works by monitoring the passing of a toothed wheel or some other source of magnetic interference on the transaxle of your car (though the exact location and mechanism may depend on your model of car.) Because of this, it won't put out an analog signal but a PWM with varying frequency depending on how rapidly the shaft is rotating (and ...


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As Dave says, the diagram is almost certainly wrong. The bottom of the 40k emitter resistor should be connected to -20V. The explanation is still confusing on its own, so I'll try to unpack it. First, they compute the DC emitter current. They do this by ignoring the base-emitter voltage drop, which gives ~0 V at the emitter. Then they get the current ...


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You should use a logic level gate MOSFET, not the IRFZ44N, which is not guaranteed to work. Rds(on) is specified at Vgs = 10V not 5V. Vgs(th) is for 250uA. I think your motor needs more than 250uA. It would probably sort-of work badly if you try it. Connection is correct, however you must put a diode across the motor to keep the MOSFET from avalanching ...


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They're not being biased into active operation. The rheostat is for calibration. The transistors are connected as low-leakage zener diodes so they will clamp errant input voltages that may occur when the user does something idiotic. Edit: with regard to your added question, yes there is a difference. The back-to-back series transistors act as ...


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See let us take an example for a basic analog circuit where the voice signal is amplified and sent to the loudspeaker. Now your voice signal is not a electrical signal so to process it we require a transducer which give the electrical equivalent of your voice signal. Let the output impedance of the transducer be 100k (assume). Say we connect a CC (common ...


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Yes, it's possible to make some types of computer without using synchronous digital logic. The reason an op-amp is called an operational amplifier is because it was originally the building block of the electronic analog computer. The op-amp is "Operational" in the sense of performing mathematical operations such as addition, multiplication, integration, and ...


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It depends on what you want to call a "computer". There are two kinds of logic circuits, combinatorial and sequential. Combinatorial circuits have outputs that are strictly a function of their current inputs, while sequential circuits employ feedback so that their output can be a function of both their current inputs and their past history — in other ...


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It really only applies to the TIP122 part and not the TIP120. You need to look into the data sheet at figure 4 - safe operating area and you will see that the TIP122 can take certain durations of current for certain lengths of time. Plotted on that graph is collector current versus collector-emitter voltage: - From this graph I infer that the TIP122 can ...


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Here's a guide published by On Semiconductor on how to interpret their IGBT datasheets. It says that: The pulsed collector current describes the peak collector current pulse above the rated collector current specification that can flow while remaining below the maximum junction temperature. The maximum allowable pulsed current in turn depends on ...


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If you want to simply replace the switches with something connected to an RPI or other micrcontroller then the most straight forward way is to use a relay driven by a transistor, with the gate/base driven by a microcontroller. What you are looking at is "remote" for the building intercom (housed elsewhere in your building), the speaker doubles as a ...


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Q2 will not suffer from the early effect as it is being driven by the current from Q1 so th effects will be negligible. All that will happen if the Vce increases is that the Vbe will drop slightly and so increase the voltage on Q1 by a few millivolts. The overall effect for the cascade will be greatly reduced by many orders of magnitude. Have you any ...


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The "gate" you've shown is confusing I think because while its purpose may be to have LED2 and LED1 give opposite indications, the input to the "gate" is a current and the output is a voltage. In most cases, it makes more sense to define logic gate behavior for both inputs and outputs in terms of voltage. For example: simulate this circuit – ...


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Current only flows into the base of the transistor, not out. When the switch is closed, the circuit acts like this: simulate this circuit – Schematic created using CircuitLab LED2 is shorted out. With no voltage across it, it doesn't conduct. All of R2's current goes straight to ground. With the switch open, the circuit acts like this: ...


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The purpose of a NOT gate is to have the OUTPUT (LED2) be the opposite of the input (LED1). If LED1 is lit, LED2 should be off. That being said, When SW1 is pressed, current flows through LED1 turning it on, and also providing current to turn on Q1. When Q1 is "ON" is provides a low resistance path to ground--a short--bypassing LED2. LED2 was originally on ...


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Somebody thought that this circuit is an emitter follower. I don't think so. There is no load circuit in the emitter and in fact the emitter is to be grounded. It is just a plain vanilla saturated switch in common collector mode. You are driving the base from a microprocessor or similar that only can supply small numbers of mA and the load is A. So ...


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For the record, here's one of the circuits you tried: Your problem is connecting the load to the source of the FET rather than the drain. Tie the source directly to ground, and connect the load between 5 V and the drain: simulate this circuit – Schematic created using CircuitLab Now nearly all of the supply voltage can be applied to the ...


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The ACPL-217 opto shows on page 4, dark collector current = 100nA @ 48V across the collector-emitter. So when the LED anode is disconnected (pin 1), there is no light created, and the current that can flow through the transistor is 100nA. If the PMOS has a suitable pulling input resistor to ensure it's gate voltage is correct with this 100nA load, then it ...


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Is there any risk to enable the P MOSFET if i leave the PIN 1 not connected ? With pin 1 not connected there is no current into the photodiode and hence, according to the diagram you posted: - V\$_{CE}\$ will be high (due to R\$_L\$) and deactivate the P channel MOSFET (assumes gate connected to V\$_{CE}\$ and source connected to V\$_{CC}\$).


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It's not clear why you have chosen a PNP transistor to begin with - I assume it's because you had to start with something. Throw it in the bin; get some MOSFETs, Darlington transistors or beefy NPN transistors (in order of current-conducting capacity, highest to lowest, generally speaking). The following diagram uses IRF510 MOSFETs, which are standard ...


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If you look at figure 17 of the datasheet, you can see that with no heatsink, at temperatures up to about 40 degrees celsius, the device can safely dissipate about 1.5 watts of power. Per table 3, collector-emitter saturation voltage is at most 1.4 volts at 1.25 amps. 1.5 watts divided by 1.25 volts is 1.2 amps. If your individual outputs are drawing lower ...


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Besides maximum current through one darlington transistor you also have to care about power dissipation. The datasheet tells you that the saturation voltage of one darlington is up to 1.5V at a collector current of 1.5A, leading to a power dissipation of 2.25W which is alread more than is allowed at 70°C Tamb. If you manage to cool all your pins down to 90°C ...


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A floating gate has a voltage. Any wire has a voltage. Until you put it in reference to something else, that voltage can be anything. Don't expect it to be 0V or any other known voltage. That gate also has a capacitance. Your body is a capacitance. Every time you touch that gate, you're connecting two capacitors and a charge balance will occur. This will ...


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Obviously in this particular application the transistor is used to form a constant current source. I.e. its purpose is to provide stable current through the load even if the main voltage supply (20V or 12V) is not stable. This is achived by negative feedback by the voltage across the emitter resistor given that the base voltage is constant. Therefore a ...


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Based upon the schematics you post most answers here have a good amount to them concerning resistor versus zener. However, to offer a fair and balanced view towards your specific drawings I'll add a little bit of extra information towards current sinking. What happens in your circuits is that the transistor is set up as an emitter follower. It means it ...


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A zener diode bias means that the bias voltage is almost independent of the supply voltage. This could be useful in any circuit with an unregulated power supply. By contrast a resistor divider produces a bias voltage proportional to the supply voltage. The dynamic impedance of a zener diode is small (the voltage across the diode does not vary much, even ...


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It is practically easier to find a pair of resistors of determined values than a zener diode of the reverse voltage you might need. Moreover, you have to count with slightly higher power dissipation at the diode, because the current is higher than when you use a resistive voltage divider. For AC applications (capacitively coupled) you cannot use a zener, ...


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For biasing a BJT based amplifier stage it is common to use a resistive voltage divider. Usually, this divider is chosen to be low-resistance to provide a "stiff" bias voltage. The reason is that the produced base voltage should be as independent as possible of the base current of the BJT which has very large tolerances. On the other hand, these resistors ...


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Note that the behaviour of these two circuits is not equivalent. The first circuit will be active on both transitions, while the second will be active on low to high, but not on high to low (or won't work at all without that diode).


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A MOSFET may* be more suitable - Particularly if your available drive current is limited. The current required to drive the gate is almost zero. http://www.onsemi.com/PowerSolutions/product.do?id=ATP202 Is an example of a device with suitable Vth and RdsON, it will happily switch 3A with little to no heat-sinking. Note it is only 30V maximum, generally ...


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I suggest you to use darlington transistors like TIP122, these can work out on loads around 10A with a little base current, you only need a transistor in each of your loads (if it's going to be asynchronous, otherwise you could pair them on parallel) and add a resistor at the base that can manage your base current depending on your trigger voltage. If you ...


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The page you link above has this note following the last photo: Edit: I’ve been advised that you don’t actually need a Darlington array for this, you could use the second shift register to become the ground. Shifting a 1 would disable the ground and shifting a 0 would enable ground. Or, since the Darlingtons would only be driving one LED at a ...


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Sure, just copy the circuit inside the ULN2803 using (lots of) discrete parts. For push-pull drive from the micro and LED drive you can probably leave out the diode (and the parasitic diodes with the dashed lines, of course), as well as the 3K and 7.2K resistors, so just the two transistors and the base resistor. The maximum current will be set by the ...


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A relay as recommended in the comments would be a really good way to go for this if you want to build it yourself. It is a really simple solution and requires few components. It also provides isolation if that would be needed between the computer and the DSL modem. Here is a picture of how to hook it up:


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You (may) need a high-side switch to do this, which requires two transistors. Assuming the grounds are connected together (eg. a USB cable between the modem and the computer) you can use this circuit (left), or if the grounds are not connected, the right-hand circuit. : simulate this circuit – Schematic created using CircuitLab


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I understand that your teacher wants you to use the transistor as a switch to light up the LED.You must add a resistor between the base and the positive supply or the transistor will be destroyed by too much internal heat.Other than that,it can work well if you choose the right resistors(let me know if you need help calculating their values) and LED.


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Your team needs to get software version control as soon as possible. Losing your source code is unacceptable. Even if this is not a professional setting, losing the source code wastes your team's time. It doesn't matter who is at fault, it will happen again if you don't get some basic source control in place. Git, Subversion, Mercurial, any of these are ...


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Pullups on input pins like this exist as a convenience - for instance, so they can be connected to buttons as you suggest in your question - but they don't prevent you from being able to drive them high or low just like any other input pin. As long as your logic levels are the same, you can connect your Arduino's output pin to the input pin on the sound ...


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This is most easily done if two conditions exist: 1) Your Arduino board runs from 5V (instead of 3.3V) 2) You have extra output pins on the Arduino board that aren't being used. If both of these conditions exist, simply connect the grounds of both boards to the same point and then connect each input pin of the FX board to the Arduino. These output pins ...


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Oops. I got the order wrong in my original explanation, and have edited to reflect this. My apologies. In your picture, Q5 and Q6 are being used as zener diodes. Current flows through the CB junction, which behaves as a reverse-biased diode. Since an NPN transistor has two PN junctions back-to-back, with the base grounded it looks like simulate this ...


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I would think that your circuit would produce around 5V or more at the emitter of the NPN transistor, so I am thinking there is something not quite right somewhere. However your circuit will never work the way you want it to, so I am hesitant to spend too much time thinking about it. Here is one circuit that will work the way you want. Just apply your 6V ...


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You are correct in saying that the anode voltage is Vcd (Anode pin with reference to ground) and Vgh is the gate voltage (Gate pin to ground). That is the simple part. Now, I cannot say immediately what is wrong. But a few things strike me as odd, so maybe we can work this through together. If Vab is zero, Vcd must be roughly 12V, this stands true. But, ...


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The saturation current of a PN junction, as you correctly said, depends on the cross sectional area of the junction itself. In fact, if you look at a datasheet \$ I_{CBO} \gg I_{EBO} \$, confirming your idea. Moreover, Sedra/Smith (I'm looking at the 6th edition, page 361) it says: The structure in Fig. 6.7 indicates also that the CBJ has a much larger ...


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Something is being missed here. The optotriac is quite capable of switching that load current, however the triac in the optoisolator is handling the inverter's high frequency AC differently from your 'triac'. It's designed to control mains frequency AC - 50 or 60Hz, and is too slow to turn off at the very brief zero crossings when it is fed with ~1kHz AC ...


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The datasheet for the MOC3031-M specifies that it's meant to be a trigger and not directly drive the load. As a result, I would imagine that the internal phototriac is much more sensitive than the eternal triac depicted in your second circuit and is probably why the phototriac was never turning off and your EL wire was always on. Instead, you should drive an ...


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Based on your response to my question the answer is relatively simple. A DC powered device draws DC current. A triac or triac-opto coupler stays on as long as the current through it is above its minimum holding-current. Holding current is just "stay-on current". Once it's triggered as long as the current is larger than the minimum, it will stay on. This is ...



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