# Tag Info

## New answers tagged transistors

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I agree with you. The first equation (including DC values) is somewhat misleading. For example: What is the difference between v_BE and v_be ? Example: V_BE=0.7 V and v_be=0.1 volts. Result: v_BE=0.8 volts. How should we interprete this value? Peak value ? Therefore, forget the first line and start with the next two equations for vbe and ic. This also ...

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For any linear two-port, we can define the h-parameters according to the equation $$\begin{pmatrix}v_1 \\ i_2\end{pmatrix} = \begin{pmatrix} h_{11} & h_{12} \\ h_{21} & h_{22} \end{pmatrix} \begin{pmatrix} i_1 \\ v_2 \end{pmatrix}$$ In the case of the BJT in the common-emitter configuration, we're taking $v_1$ to be between the base and common (...

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simulate this circuit – Schematic created using CircuitLab This is an inverting switch. R values can be changed

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Use a voltage divider to create a desired trigger level voltage and use a comparator to deliver full (3.3V for example) voltage when that voltage level is reached.

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If the base voltage is higher than the collector voltage, and the base voltage is higher than the emitter voltage, the transistor is in the "saturation mode" of operation. Current will be flowing into the base (or out of base for PNP). This is something you can enter into a search engine for better understanding. If the base voltage is higher than ...

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Does the base voltage has to always be (0.7v) in order for a bipolar NPN transistor to work? No. But a silicon small-signal BJT transistor usually has a base to emitter voltage of around 0.6 to 0.7V when it's operating. (Note that you say "base voltage", and I'm changing that to base-emitter voltage -- the voltage of the base vs. some arbitrary ...

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0.7V is not an absolute value. It varies depending on the transistor part number and even across any given lot of "identical" transistors. It's an approximation. It's also not a hard-cut-off. You don't get 0.69999V it's "off", 0.7000001V it's "on". It's a gradual range across maybe 0.5V-0.8V or so (again, transistor ...

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From Quora.com: How you spend your transistors really depends on the problem you're trying to solve. If you open up a 1 billion-transistor x86 processor, you'll see a handful of very complex CPU cores, with many transistors dedicated to decoding and scheduling instructions, predicting branches, and managing virtual memory. You'll also see many transistors ...

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The ON semiconductor device is a NVD5865NL: - The other device: - Looks like it has the Alpha & Omega Semiconductor symbol: - I thinks it's the AOD409: -

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simulate this circuit – Schematic created using CircuitLab Perhaps you were thinking somethin like this. The MCU, when outputs a high state, turns ON the MOSFET. While at low state or disconnected the R2 pulls-down the gate, to prevent spurious turn ON. You probably meant this one, but it is not gonna work, since you have an emitter follower, so the ...

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The circuit won't work regardless of the value of R2. The MOSFET will remain partially on all the time (due to R1), and possibly burn up. Even if R1 was to ground, the emitter-follower would only give you 2.6V from a 3.3V output regardless of whether the supply is 5V or 12V. You could do something like this (logic is inverted): simulate this circuit – ...

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The images you gave are IMO a kind of roundabout way of the following: simulate this circuit – Schematic created using CircuitLab This circuit is equivalent to the one shown in your figures. And calculating the Thevenin-equivalent of the boxed part will get you your answers: $$e_T = \frac{R_{B1}}{R_{B1}+R_{B2}}\cdot V_{CC}$$ $$R_T = R_{B1}//R_{B2}$$ $... 1 Given that the RAP circuit is powering other things as well, you can probably just use it to power the relay coil directly. What will happen is that if the RAP circuit in your car shuts it off suddenly, the relay coil will continue to pull current for a bit, because it's an inductor. This could cause a negative spike on the RAP circuit -- but chances are ... 1 You should put one in though it is not as critical in this case since the a manual switch is a lot hardier to voltage spikes than a transistor which is often the first thing to blow from the spike. Even so, having one will prevent the switch from arcing and wearing it out. You should put one in anyways just so the spike doesn't affect other things connected ... 1 We can tell your experience from the spelling of MOSFET but in general a MOSFET rated for 5x the load current or more will give you the low Ron resistance need to produce I^2R=Pd of heat to the switch. Preferable < 250 mW so no big heatsink is needed or Ron < = 0.25W/6*6 or <7 milliOhms, which is quite low. Slightly Higher can be tolerated with a ... 4 The valve works by attracting an iron armature into a solenoid. This means that when un-energised, there's a large airgap in the magnetic circuit. When it's energised, the airgap is much smaller. The large initial airgap means that a large current is required to develop enough force to move the armature. When the airgap is smaller, less current is needed to ... 4 Have you heard the expression "cut your coat according to your cloth"? 10kV transistors don't appear to exist, no. 1kV seems to be just about feasible with present technology and you can buy bipolar and IGBT transistors with that rating. They need appropriate cooling and sharp switching times to avoid overheating. using 1A and a 10Ah battery this ... 0 I used this circuit 4 years ago. this was working fine. Current will flow only through the LEDs when a transmission is ongoing (you see this as a "longer ON state" when transmissions are ongoing). If there is a small pause between a couple of transmissions you will see some blinking (LED turns OFF). 3 You can connect the Rx/Tx lines to something like a 74VHC123AFT (one chip for both lines), and drive the LEDs directly with the output (preferably the /Q output with a series resistor to Vcc). Try a time in the 100ms range. LED current of a couple mA should be sufficient with a good LED. The retriggerable multivibrator will cause the LED to illuminate ... 0 Your question relates to two types of regulator circuits - 'Series' and 'Shunt'. In a series regulator circuit the load voltage would be maintained by dropping the excess voltage across the series transistor. In a shunt regulator circuit the load voltage and that maintained across the shunt transistor would be one and the same. Generally, series and shunt ... 0 A pump is undefined for the load current. But for a beefy driver use a low RdsOn logic level FET. The SMD code for Y2 could be the PNP transistor SS8550 so a Pch would be compatible and for 2A and worst-case power dissipation of 200mW from I^2Ron=Pd. Choose smaller Ron () or external FET for bigger currents with a suitable heatsink. Chose Imax rating 10x ... 2 "Design a power supply with a linear regulator and bypass transistor to regulate voltage to 12 volts and supply a maximum current of 5 amperes." The solution is presented in the 78xx series datasheet on page 20. I want to be able to design a linear regulated power supply but with a bjt near the load, ... No, you want to use a BJT to boost the ... 4 Emitter-coupled Franklin oscillator with switched power supply. 1 The following oscillator should work with precision op amps (low offset voltage), as long as the values of$r$and$T$do not need to be very precise. Defining$\omega=2\pi/T$, I get$r=\frac{R_4/R_3}{2 R_2 C_2}$and$\omega^2=\frac{(R_4/R_3) (R_5/R_6)+(R_4/R_6)+(R_5/R_6)}{R_1 C_1 R_2 C_2}-r^2$. SPICE simulation seems to give the expected ... 1 I get the same result. Ammeters go in series. Voltmeters in parallel. If you check properties of voltmeter, it is a 100M$\Omega$in series with emitter. It is a simulation, not reality. The program has to come up with a number and 3.12V is it. Make the following changes and your circuit will work. You do not need two switches. Essential, as you have ... 2 The voltmeter is high-impedance, so the emitter of the transistor is not connected to ground. You should instead connect the emitter to the + terminal of the ammeter that you currently have in series with the voltmeter. That will ensure that the emitter node is shorted to the voltage source negative (ground). As for why you measure 3.12 volts - in the ... 2 If you just want to detect light then use this circuit. Reverse bias the LED into the emitter of an NPN transistor. When light hits the LED then reverse current will flow proportional to the incident light. The transistor provides current gain, boosting the LED current from say (10uA) to around 1mA. The behavior of the circuit will be this. In the dark ... 3 Quick Note The schematic you show here is from LTspice, I believe. It's default NPN BJT has$\beta=100$and$I_\text{SAT}=100\:\text{aA}$. Your "EveryCircuit" link is very unlikely to use the same default model. So LTspice probably will simulate different values. Just FYI. Nodal Analysis The nodal method is probably the easiest for solving this ... 1 Current flowing through base will be 5-0.7/50000 = 0.000086A Current flowing through the collector will IbxG = 0.000086x100=0.0086A Voltage drop on the transistor (Vcollector - Vemitter) = 5V. Voltage drop on the resistor will be 4.3V. Anything else is wrong.If you simulator gives different results then it is wrong.First trust humans then machines. 2 Starting Point Here's the basic idea, with some added parasitics worth a moment's thought: simulate this circuit – Schematic created using CircuitLab Above, I added the other collector resistor. The reason I did is because of$r_{o_{_1}}$(Early Effect parasitic.) Without the added voltage drop across$R_{C_{_1}}\\$ there's a larger voltage drop ...

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1: bipolars are cheaper So you can do larger expeiments with the same budget. 2: bipolars don't die a instant death from ESD So there's less risk of the user giving up on electronics because "nothing works as described". 3: bipolars are good enough for many tasks. V_ce(sat) < 0.1V is typical 4: most "logic" MOSFETs need ...

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In general, FETs require much higher drive voltages than BJTs. Let's say you have a circuit running on 5 volts, and you want to switch a relay. If you get a 5-volt relay, you can use a BJT and everything will work off of 5 volts on the control side. The base voltage of a BJT is only about 1 volt, so you have about 4 volts to work with in developing the base ...

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You can do a noise simulation run in LTspice to show you the spectral density of the simulated noise. Transient analysis does not include internal noise (flicker, shot and thermal noise) of the components.

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I thought the physical implementation of the "long tail" would show at least some noise as well. Is that a correct assumption? There will always be noise on any analogue output because noise is always everywhere (above absolute zero) but, the CM gain is significantly less than 1 so why would you expect to see noise if you don't have much noise on ...

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It’s usually an NPN BJT with sufficient voltage rating for the circuit and an ft in the 50MHz range because it has to oscillate at 1-2MHz at the thickness mode mechanical resonance of the disk.

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The manufacturer has scratched off the identification to make it so that it's not easy to copy or repair his product. This means you'll have to take the less easy route. Trace out the circuit - looks like a 2 layer board so should be straightforward Run the circuit while measuring various parameters, with a DMM at least, and preferably with an oscilloscope ...

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here is pratical circuit that works drawing approximate 800maLED Strip, I had BC557 laying around

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simulate this circuit – Schematic created using CircuitLab Figure 1. Make sure that you have a ground connection between the two power supplies. Without the V1- connection to the 5V- there is no return path for Q1's base current. simulate this circuit Figure 2. The circuit re-drawn in the conventional format. With the redrawn layout it is more clear ...

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You haven't explained the power supply setup. If this is powered by its own 9 V battery there should be no problem but if you are using a shared power supply with the other pedals you may have created a short-circuit. Check if the other pedals positive or negative ground. (That might cause a short on the 9 V supply.)

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I have copied the explanations below from another answer of mine to a similar question. I hope they will help understanding of this exotic analog circuit. Short explanation The output voltage VOUT is "lifted" with VGS above the input voltage VIN... but since a compensating voltage drop VR1 = VGS is subtracted from it (Fig. 1), the output voltage ...

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You want to regulate LED string current rather than output voltage. So place a resistor at the bottom of your LED string, use an amplifier to gain up the signal, and send that back to the FB pin. That's the general idea. You could just pick a resistor large enough that the LED string current created the regulation feedback voltage, but I suspect the power ...

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It is amazing how such a simple and intuitive idea is difficult to understand and it periodically raises all these questions. This made me write this detailed answer to reveal the philosophy behind the specific implementation. Short explanation The output voltage VOUT is "lifted" with VGS above the input voltage VIN... but since a compensating ...

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You can either place a negative sign in front of the measurement: Or rotate the resistor 180 degrees to swap the pins.

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If you are looking at voltage across a resistor, or current through a resistor, LTSpice assumes a 'forward direction'. This assumption may be backwards from how you are thinking about the circuit. If you get a graph that is inverted from your perspective, rotate the resistor the other way in the schematic to fix it.

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It would be better to use an n-channel MOSFET such as AO3400A. No base resistor required and it will easily handle the 900mA peak with 3.3V drive. You still need the diode across the motor, of course.

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If you had had a look at the 2N2222 datasheet you'd have seen that it says I_Cmax = 800mA, therefore it is not sufficient for 3 motors (3*300mA=900mA). Also don't forget a resistor at the base, probably 220 Ohms. The rest should be OK.

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The thing is, the scientists of today don't understand why things get hot even when specific energies are applied to achieve a specific result. It's all down to quantum energy potential to achieve the flipping of a transistor (over-loaded/under-loaded) which they can't perfect because of universal forces that act upon your chip even though it's shielded, ...

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Here's my version of a short-circuit trip circuit, using an electromagnetic relay.

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