New answers tagged

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A BJT transistor's pass-thru voltage will never rise above base voltage; thus the 4.34V & also why you place it between a resistor and GND, rather than drive 1 transistor directly through the other. The resistor allows you to 'invert' and amplify the voltage change of the incoming signal. When Q14 is 'open,' R15 allows the base of Q15 to reach near the ...


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Insert base and collector resistances where appropriate, and for simple 2 wire measurements: V1 and V2 are voltage sources. AM1 and AM2 are ammeters. VM1 and VM2 are voltmeters. for minimum instrument measurement errors, the resistance of the voltmeters should be very high and the resistance of the ammeters should be very low.


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You will need to make your own bi-directional discrete H bridge (if you want to drive propellers in both directions) or a half bridge if you want only one direction. Don't get 2A transistors, get something like 8-10A rated MOSFETs and ensure your design can survive inductive loads and over-voltage spikes from the motors. Do the proper research into H ...


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With MOSFETs, you need to consider that the gain is due to the drain voltage and the change in the effective channel length (sigma). Below is an image that represents the behavior in subvt, even through it is not true to physics, in abovevt, it's the same but the bands are different due to the inverted channel and the gain is lower. If you have a fixed ...


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If the current source produces 10mA and the MOSFET is trying to take 10.0001 mA then the voltage on the drain is approximately zero. If, on the other hand the MOSFET is trying to only take 9.9999 mA then the voltage on the drain is close to Vdd. In practise, the MOSFET has an effective DS parallel resistance of several kohm so it's not quite as black and ...


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In basic terms: BJT transistors are relatively low-impedance, current-driven devices. They amplify Ibe or Ibc to a higher value Ice or Iec. FET (MOSFET, or JFET) transistors are relatively very high-impedance, potential(voltage)-driven devices. An extremely low motion of electrons (more capacitive that resistive) 'charges' carriers in a channel, allowing ...


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MOSFETs do not necessarily have lower conduction loss than BJTs- in fact in many cases (especially at low voltage) a MOSFET can do much better. For example a 5m ohm MOSFET switching 100mA (grossly overrated, but I digress) will drop only 500uV. A BJT might drop tens of mV at best. IGBTs can have competitive conduction losses with MOSFETs if they are both ...


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If I understood your question without schematics, I guess the answer is: With the resistor connected to the emmiter, when IB rises, IE rises due to the amplification. If IE rises, the voltage drop over RE rises too. Since VBE=VB-VE, VBE will lower and IB too. It works like a negative feedback. Best regards


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The VGS problem: The 2n7000 has a *VGS (Voltage Gate to Source) of 3 Volts Max. Meaning it is a low logic level Mosfet, and will have it's lowest RDS(on) (Resistance Drain to Source when On) at that voltage or above. It's great for a 3.3V signal. The IRF540 has a much higher VGS, nearly 10 Volts for lowest RDS(on). It's VGS Threshold is listed as min 2.0 ~ ...


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The DRV8816 is used quite commonly: - The power supply range is 8V to 38V and it can supply currents of up to 2.8A (normal recommended levels). There are plenty to choose from and I would recommend this method rather than building from discrete transistors. See also this for other recommendations and some pointers to devices that may not be suitable. If ...


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As the mosfet requires a negative voltage to turn on this means that my mosfet would turn on when my pwm wave falls to 0v and turn off when it goes up to 3.3v correct? So whatever I set the duty cycle in my pwm to the actual switch duty cycle be 1 minus that right? Correct. If I'm right about part 1, how likely is it that this switching circuit ...


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What does it means 100v, 10A ,100W darlington transistor? The maximum voltage that the transistor can withstand between collector and emitter is 100V. Sometimes this spec is between base and collector. The maximum current that the transistor can withstand that flows between collector and emitter is 10A. The maximum product of current flow and voltage ...


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If you read this in the datasheet that means the transistor absolute maximum rating. Take example for this N Channel, if you drive the motor with 10A then you risk to damage the transistor. For the 100W is the maximum power dissipation of the transistor.


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A simple H bridge would achieve the desired outcome. Circuit below: H bridge works by opening the opposing transistor pairs, so the motor turns one way or the other.


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If you consider an NPN BJT to be two diodes in series opposition with their anodes connected together, then the collector is the cathode of one diode, the emitter is the cathode of the other, and the base is the junction of the two anodes. For an NPN BJT, then, it means that the base must be positive with respect to the emitter and the collector must be ...


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No transistors need to be abused in the making of your device. All you need is a resistor: This causes the full LED current to be drawn all the time, whether it is on or not, but it's really simple and robust. In fact, the current drawn is slightly higher when the switch is closed and the LED off. Here is a way to use transistors to minimize the ...


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The easiest method is to remove the transistor and both resistors connected to the base of the transistor (10k, 47k). Then connect the free end of the switch to where the collector used to be. If this is an existing circuit board, remove the above components. Then install a jumper where the 47k resistor used to be and another jumper between where ...


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You want to connect the LED to the emitter instead of the collector, and the switch decides the base voltage i.e. if higher than 0.7V, the transistor canal is opened, else it is closed.


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With a transistor, you can achieve this: Give a small signal(ac) at input, and get a larger valued(higher amplitude) signal at output. But this is not all. You have to give DC supply at collector and base; emitter if required. This is called biasing the dc point. The rms power you get at the output will be less than the dc power you have supplied. If you ...


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You can try using Power BJT's. These BJT's are capable to handle lot more power than the simple BJT's.There are more number of Power BJT's available in market.They can surely handle 240V and even higher amount of voltages.Just google out for the transistor which can handle the power and voltage you expect.That's itIf you don't need good amplification,then ...


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Another, simple, but accurate way to see this is using feedback theory: The Op Amp output is simply the gain of the Op Amp (A) times the difference between the voltage at the inputs. If we call the voltage at the resistor \$V_x\$ (since we don't yet know what it is), then the output of the Op Amp is simply: $$V_o = A \cdot (V_{\text{set}}-V_x)$$ Now, we ...


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There's no limit to the number of transistors that can be used to construct a gate, but it makes the most sense to use as few as possible, so it's likely that your question is posed bass-ackwards. Here's a TTL NAND gate, in a few different flavors, from TI. Note that in its simplest implementation it takes only four transistors to get the job done. There ...


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A gate uses as many transistors as needed or as can be used with the accompanying voltage level. A 1000 input AND or OR gate would be untenable in most modern CMOS processes of the last... ever. A great way of understanding the number of mosfets (and thus inputs) is to understand how much voltage it takes to turn on and off a MOSFET and how many are ...


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I have finally found the Answer to do this - use a comparator IC (typically a IC that would generate a reset if the input goes below a certain level). Found one in TI(TPS3700) - just the right one.


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Your main problem is that, following the answer to your previous question, you made the RC/RE ratio too high. Consider what is going on: 1) R1/R2 set the base voltage at ~1 volt. 2) This sets the emitter voltage at ~0.3 volts (give or take a bit). 3) So the emitter current should be ~3 mA. 4) The collector current should be about the same as the ...


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On this question, and on your previous question, you keep getting 0.7 volts Vce. Perhaps you have swapped the base and collector leads. This would put the base emitter leads across your "emitter" resistor, resulting in 0.7 volts no matter how you bias the resistors R1 and R2. 2N3904C EDIT : If you have connected transistor properly, you can ...


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Assume that all transistors (FET, BJT etc) need a leakage diverting resistor, connected control terminal to ground. If your control circuit doesn't cause leakage under any conditions, then you don't need this resistor. Some transistors don't need such a resistor, they already have it internally. Check the data sheet of each individual transistor to see ...


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This is all possible . You can make any DC mask you want . Your power is rather low so you could just burn it up in resistors and current sinks .The volt thresholds are easily done with simple compariters like LM393.If you use mosfets for your current sink be careful .Mosfets in linear mode are for people who know what they are doing .Otherwise just use BJTs ...


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SPICE places a resistor across every p-n junction to aid convergence. The default value is 10^12 ohm (GMIN=1e-12). So, the BC junction has 10^12 ohms across it and this runs a current of about 9V/1e12 = 9 pA. This current flows in R2 (4.7k) giving a calculated 42 nV. Your SPICE may have a different GMIN, and in fact the transistor model will also have B-C ...


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Since C1 is an ideal capacitor, it has infinite resistance, which means no base current. This is supported by the 0 volts across R1. With no base current, there should be virtually no emitter or collector current, and this is supported by the 73 nV on R2. So, the emitter-base voltage is going to be determined by various leakages in the transistor, and these ...


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simulate this circuit – Schematic created using CircuitLab Figure 1. What you've probably got. There's something strange about your setup. If there is a short where the bell was (LO-Z in Figure 1) then there is a danger of the transformer overheating if the button is held in or fails short-circuit. If the bell had simply been removed there ...


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So, imagine the situation. You have your wireless receiver hooked up to drive a bell in your apartment. What happens next.... Some guy down the street activates his key fob to open his car door and your bell tinkles. Someone in the same apartment switches on their remote controlled lights and your bell tinkles. In fact anyone using the same frequency will ...


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Found the justification for the pnp transistor: Note that Serial1(TX1/RX1) should be used on the Arduino Mega. The serial port output TXD from the Roomba/Create is too weak to drive the RX serial port (Serial0) input of an Arduino properly. This is because of the USB-Serial converter on the Arduino: it also tries to drive the RX serial port input via a ...


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With the ac voltage being higher than your circuit voltage, it could easily end up damaging your components. In order to avoid this damage, you pretty much have 3 choices: Disconnect the ac power, then simply wire the switch directly to your circuit. Purchase a relay that can be triggered by 8VAC, wire the relay to the AC power & the switch, then ...


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Without seeing the robot circuitry, I'd say yeah, give it a try. The A28 doesn't have quite the gain or voltage capacity of the 2N4403, but that's probably not an issue here. Match the pins correctly, then Plug it in and see what happens.


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Contrary to what other commentators have suggested, there's nothing wrong with the circuit topology as shown. In fact, it rather cleverly sets the Q point of Q3's collecter at almost exactly 6V, half of the supply voltage. simulate this circuit – Schematic created using CircuitLab For the DC analysis, start by calculating the Thevenin ...


1

It is simply a two-stage amplifier with dc-coupling and overall negative feedback (R12). The second stage (Q3) - like the first stage - works as a common emitter stage, however, it is realized with a pnp transistor. This principle is called "DC potential shift with gain". A simple calculation of the bias conditions can show that the collector DC potential ...


2

Very briefly, the general idea is that a positive-going signal on the base of Q1 causes an increase in the current flow from its collector to its emitter. The Q1 collector current primarily comes from the base of Q3, causing an increase in the collector-emitter current flow of that transistor. This causes the voltage across R10 to increase, which is taken ...


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It's probably an MMBT2222A (a generic NPN switching transistor), assuming the markings are "1P" (upper-case P). Easy to find, and probably many other NPN types would work too, but you may as well get the proper part.


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A 2N2222A is totally inappropriate in this application. That's a small signal transistor. You need a power transistor. There are many to chose from, but something like a TIP41 is something you can probably find easily.


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A power company wants to generate electricity, but they don't want to waste it on power transmission. Power loss in a wire is given by $$P = I^2R$$ but you can also calculate power as $$ P = IV$$ where I is current, V is voltage, and R is resistance. So power companies use very high voltages, as this lets them use very low currents, which means ...


3

Have you tried finding the current through the coil? Find the current through the coil. You can either simulate it with a spice package or place a meter between the coil and transistor. Once you know the current use the equations located in this online textbook in chapter 11 Inductance and Magnetic energy (Yeah it requires math). Then find the magnetic ...


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If you know what voltage and current are, then it should be obvious why you might want to amplify them sometimes. A public address system is a common (and should have been obvious) example. A microphone puts out millivolt level signals. These are way too small to drive a speaker. Amplifying the tiny microphone signal to volts or 10s of volts allows ...


2

Ideally you would use something more efficient (and necessarily more complex) than a rudimentary slayer exciter, but to easily improve the current circuit you need just a proper power transistor that you mount to a heatsink. The 2n2222a comes in a TO-92 plastic package, can dissipate a pathetic 625 mW of heat and is only rated for a current of 800 mA. You ...


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The capacitors apply some positive feedback (positive in the sense that it's the same polarity as the original signal) to switch state suddenly. Imagine it as follows, numbering transistors from left to right: The light is on. LDR resistance is low. Q1-b is pulled low so there's no collector current and Q2 is off and its collector voltage is pulled low by ...


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You are asking a simple question which does not have a simple answer. Taking things from left to right: 1) The rectifiers must be rated for at least the output current. In this case, if you try to pull more than 5 amps, you will probably kill the diodes eventually. So the rectifiers need to be sized appropriately. Also see point 5. 2) C1 will affect the ...


1

This is a very basic series-pass linear regulator circuit. The output voltage is determined by the zener voltage of diode D2 plus the base-emitter drops of Q1 and Q2. The power dissipation in Q2 is going to be the product of the current through it and the voltage drop across it, which is the difference between the input voltage and the output voltage. ...


1

Yes, in that circuit the 3055 is the "main pass" transistor. All current to "V out" has to flow through the 3055 first, so you can't draw more power than it can handle, or you'll kill your transistor. EDIT: Also, your rectifier bridge is rated for 5A, so if the transistor's rated for 15A, the diodes will blow up first. As for efficiency, if I'm reading the ...


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It depends on the bandwidth you want the amplifier to achieve. A simple thumb rule is that a capacitor is seen as a "short-circuit" at higher frequencies and an "open-circuit" at DC. Therefore, the gain of a common-emitter is about the load connected to the collector divided by the load connected to the emitter. So, you can change the AC gain of the ...


-1

This site gives you wrong information. The bypass capacitor (CE) do not influence the DC bias condition. This capacitor "short" the emitter to ground for AC signals (removes the negative feedback from the circuit). And thanks to this the AC gain increases from \$ Av = \frac{Rc}{Re}\$ to \$ gm*Rc \approx 40*Icq*Rc\$. Where Icq is Collector current at quasient ...



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