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It's s triple emitter follower. Whatever voltage is on the base of T2 determines what is on the emitter of T4 (minus three BE volt drops totalling somewhere around 2.5 volts). So if you have 10V on the wiper of the pot you'll have about 7.5 volts on the output. Assuming the current limit circuit isn't activated there will be 24V on the top connection of the ...

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You're using the MOSFET as a switch. When the MOSFET is ON, it will appear as a low value resistance between drain and source. This is called Rds ON. For a 2N7002 from NXP this is about 2.8 Ohm (assuming Vgs=10V, Id=0.5A and Tj=25C). With 200mA of drain current the voltage drop is 0.56V. The power dissipation of the MOSFET is 0.56V * 0.2A = 112mW. This ...

2

What I did understand: R1 T1 VD5 - they form a voltage divider which provides a voltage referrence that is applied through a voltage divider formed by R3 to the input of a common collector amplifier formed by T2, T3 and T4. This allows us to set a desired output voltage by varying the base voltage on T2. Not quite. R1, T1 and VD6 form a ...

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For selecting proper value of resistors you should have at least two option : Vce_saturation and Vbe_on. Lets assume your VCC=5 volt , Vbe_on = 0.7 volt and Vce_sat=0.2 volt. For Maximum output swing we should set Ve voltage to : Ve = (VCC - Vce_sat) / 2 Ve = (5-0.2) / 2 = 2.4 volt. So Vb = Ve + 0.7volt = 3.1 Volt. Then we need to set Rb1 and Rb2 to ...

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You make a common mistake. You should be using the NMOS transistor as a switch meaning that it is either conducting or non-conducting. When it is non-conducting, no current flows so P = V * I = 18 V * 0 = 0 W, Good, no problem. When the NMOS is conducting and the solenoid is activated current flows but the 18 V should not fall across the transistor, it ...

2

You're calculating the power dissipation of your transistor wrong. 18V * 0.2A is the power of the solenoid. The power of the NMOS is the voltage drop across the transistor times the current through it. Typically for a completely switched MOSFET you will have an ohmic value given in the datasheet as rds(on), which is the on-state resistance. So you can ...

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What you have calculated is the power dissipated by the solenoid and transistor together. The transistor will maybe drop 0.1 volts across it when passing 0.2 amps hence its power dissipation is 20 milli watts. In your case, the 2N7002 is maybe a little closer to the limit. See the graph below: - What this tells you is that if you drive the gate with 5 ...

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You need to check the data sheet for your transistors. Look for maximum rated voltages for the collector. And stay below that with your battery. Rt means input impedance (resistance, total). Its rather simplistic, and you will not get a complete range between full dry and full wet, the ratio being dependant on the value of the pot vs the emitter ...

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I used a zero cross - Triac output optoisolator : MOC43043 and isolated tape Triac : TBA06 for 6A , TBA08 for 8A, TBA10 for 10A ...

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If you have 3V to drive the relay, then you can possibly use a Solid State Relay. It seems to be just barely enough to turn it on. However I feel that if you are asking this kind of question, you shouldn't be modifying the AC wiring in your home just yet. It could possibly kill you and others either by electrical shock or fire. Please be careful and ...

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Your schematic is a weird way of showing the first step of an AC analysis. The original DC version should like this: simulate this circuit – Schematic created using CircuitLab For AC analysis, you're supposed to deactivate any DC sources. Voltage sources become short circuits, and current sources become open circuits. Your schematic shows this ...

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The capacitor across Uo the supply voltage that's labled C infinity is just a power supply bypass cap .It should have a low impedence at the lowest frequency of interest.Large values are common here.The output impedence of your amplifier is close to Rc so loading it will affect the output volts unless Rl is very high compared to Rc.

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I’ve read that in order to use a transistor as a switch the base current should be high enough to fully saturate the transistor. If the base current is too low it could destroy the transistor. Why does the transistor get destroyed with less current? A transistor can certainly get destroyed if the base current is too low - the transistor may not ...

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1) Power dissipation in the active region usually is much higher than in saturation. So most of the power is dissipated during switching time rather than in conduction (saturation) as you can see in the (last) graph below: (Source Power Electronics Handbook) This puzzle is best explained by looking at a (simplified) model of a [power] transistor (which ...

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So I spent some time researching the problem, and I came up with a possible solution: First we draw the small signal equivalent for the circuit and add a test voltage $v_{\text{d}}$: I have made the assumption that $r_{\text{ds1}}$ = $r_{\text{ds1}}$ = $r_{\text{ds}}$ because the transistors are identical. We find expressions for ...

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Assuming that by "dynamic RAM" you are referring to DRAM, it is the first time I hear about "6 transistors per DRAM bit". DRAM's bit is usually implemented in the following way (or some slight variation on it): As you can see, there is just one transistor and one capacitor here. Since capacitors in integrated circuits also have "transistor-like" ...

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I see that you got your sensor schematic from the lucaszanella website. Generally it's a bad idea to connect the transistor's base directly to a pad that anyone will be toughing with their finger. An electrostatic discharge can damage the base and short to 5V. Although 5V seems benign for most people it can be deadly for people with pacemakers. What you ...

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The gate voltage of M2 is the same as M4, equals the voltage at Iref input. M4 sits on top of M3 that has a gate voltage of Vt+Von. So the total a the Iref input = M4(Vgs) + M3(Vgs) As you stated the Vgs of all devices is Vt+Von as they are all passing the same current so substitute in the above Ref input = Vt+Von + Vt+Von = 2Vt + 2Von.

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Here is how the circuit works for each of the cases shown in the truth table for a NAND gate: Note that for either or both inputs = 0 (low), the output is 1 (high, or 3.3v). This is because the top two parallel transistors are inverting, so if the input(s) are low, the corresponding transistor(s) will be on, connecting 3.3v to the output. Meanwhile one ...

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Go with @Wouter van Ooijen's suggestion: simulate this circuit – Schematic created using CircuitLab Cut the wiring / trace at either end of the relay coil on the grill. Bridge them with a normally closed reed switch. By experiment find out how many turns of expresso wire you need to get the reed switch to change over and add, say, another 25% ...

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In VLSI the transistors with the little circle on their gates are the p-channel transistors while the ones without it are the n-channel. When n-channel transistor gate is at the GND level (and source is also at GND) it's in the cutoff state (OFF). There is no conduction between drain and source. When it's at 3.3V the current flows between drain and source ...

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I think your 'extremely simplified' diagram is what's causing the confusion. That's not a CMOS inverter, it's a pmos inverter. You'll notice that it never drives the signal to ground, which is an issue when we're talking about devices with very small current drain. The second diagram, of then nand gate, is cmos. That's complimentary metal oxide ...

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There are so many MOSFETs that will fit this bill. If you are sure that you want a low-side switch, then AO3414 is a reasonable choice, it will drop only a few mV at 50mA and is very inexpensive. You might want to consider suing a p-channel MOSFET as a high-side switch, depending on the exact application.

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You'll notice that in Figure 2 it states "Ic/Ib = 10". This means that to saturate the transistor you should drive it as if the Hfe is only 10, not the almost 100 in Figure 1. Note also that figure 1 is the typical figure not the guaranteed value. The guaranteed value in the data sheet is only 10 at 3A collector current. You need to provide more base ...

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You have measured the results of the so-called Early effect. Increasing the voltage Vce (and, thus, the voltage Vcb for a fixed Vbe) causes an increase in Ic and a (small) decrease of Ib. This is because the sum of both (Ic+Ib=Ie) must be constant (Vbe determines Ie and remains constant). Why does Ic increase and Ib decrease? Answer: Because an increased ...

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Nowhere. A BJT acts not only as a transistor as normal but also a pair of diodes, both pointing either towards or away from the base.

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It is supposed to be left unconnected. This circuit is for testing just the base-collector diode.

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In a common base circuit, the signal input source is applied to the emitter terminal. The "lump" of semiconductor that seperates base and emitter is a forward biased diode. The base is held at a constant voltage and therefore the emitter signal voltage causes a signal current to flow thru the base terminal. Irrespective of whether the circuit is common ...

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Given that $\alpha$ and $\beta$ are related by $\alpha = \frac{\beta}{1+\beta}$ as stated in the wiki article, obviously you can do your sums with either. However, which is going to be easier to use? I personally always use $\beta$, regardless of the transistor configuration. In common emitter $I_c = \beta\times I_b$, so I can say 'I need to ...

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When connecting small features to the outside world you need a metalisation step. The feature (sub μm or several μm doesn't really matter) is too small to connect to in any kind of way. So quite often those features are etched such that they lay a bit higher than any other conducting element in the vicinity. The vicinity for normal or crude devices ...

1

Two transistors are needed if the output must be actively driven both high and low, which is usually the case. If, for instance, only the lower transistor were used, then when the transistor is ON, the output is pulled low, but when it is OFF, the output voltage can be high or low or anywhere in between. As shown, either the output is pulled low (logic 0) or ...

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Rising and failing edges can be detected in several ways. One is to add an additional flip-flop which serves a a simple state machine, so that only transitions from 0 to 1 are used and transitions from 1 to 0 are ignored. Here is the equivalent for a falling edge detector: Another is to use a short delay line, such as three 74HCT04 inverters, which ...

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According to simulation results, traditional matched source termination seems to work just fine. I also can't think of anything wrong with matched termination on the other side of the transmission line (i.e. in parallel with R2), however keep in mind that the steady-state voltage will be quite low if R1 is large compared to the termination at low ...

2

You have a few choices. Make $R_2=Z_0$. This does consume more power, but it is essentially how a CML driver works. (CML reduces the logic swing to 1.0 V or less to mitigate this). Add an RC circuit in parallel with R2 to make the line match at high frequencies, while keeping the low static power dissipation when the output is low. Don't match at all. ...

1

With such high pull-up the slew rate will be slow enough that you don't need to worry about the low to high case. For the high to low case you have a common situation so just terminate with a series resistor so that the transistor resistance plus the series termination equals Z0 (50 ohm in your example). Since you don't want to dissipate the power in the ...

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The G5LA-1 5DC relay specified in that link takes a rated 70mA to operate. This is more than the GPIO can handle. Look at the circuit diagram in the link, and use all the components it shows. The series resistor R1 is needed to limit the base current, they suggest 1k will give about 3mA. The diode D1 is needed to avoid the relay turn-off spike killing the ...

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The Raspberry Pi GPIO outputs (and outputs on most other microcontrollers) can only deliver a few mA, and even a small relay will require more current than the GPIO can handle. The site you link suggests keeping the GPIO output current below 3 mA, and suggests that the relay may require 50 - 100 mA. A very small current into the base of a transistor can ...

2

Yes, you can use the heatsink as a common drain for all transistors if your heatsink is not covered with paint but: - It is mandatory to use a screws with washer in order to make a good contact. If you use the circuit outside or in a humid environment probably after few months will appear some oxid on your heatsink and the contact resistance will be bigger ...

2

Use the following graph in the TIP31 datasheet to determine the voltages you need across your transistor to saturate at your load's max current. For 2A = 2000mA load (collector current), we have a Vce(sat) of 300mV = 0.3V. Multiply 0.3V by 2A to get a 0.6W dissipation on the transistor when it's "on". That will work ok without a heatsink for a TO-220 ...

3

If you saturate that transistor, then you will pull its collector low. It will be nearly ground potential. If the base current times the transistor gain is greater than the current that gets through that resistor, then its collector should be "pulled" within millivolts of ground. An easy way to understand it, is to understand voltage dividers. Voltage ...

0

this mosfet driver is much better it have a very less rise time compared to fan http://datasheets.maximintegrated.com/en/ds/MAX15018-MAX15019.pdf the supply voltage to this ic is mosfet max gate voltage (stay 2 volt less than that) and this also have internal diode just use a boot capacitor while choosing a mosfet drivet things to consider 1)n channel ...

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The heatsink should be use with some thermal paste or a material that conducts heat, if you want to connect the drain to heatsink you cannot add thermal paste.Only with some screw could make the contact.In this case should work because the heatsink is a low resistance material.

1

You have the solenoid wired between the transistor's emitter and ground. In this configuration, the maximum voltage across the solenoid will be limited to the output voltage of the MCU minus the VBE drop of the transistor, which is about 0.65 V. A 3.3V MCU output will give you at most about 2.6V across the solenoid. Any significant load current will cause ...

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At first, the shown diagram is an equivalent circuit diagram for a transistor amplifier stage in common emitter configuration (the bottom line grounded). Note that R1 and R2 are the base bias resistors, Rs is the signal source resistance, rbe is the dynamic (differential) resistance between base and emitter, Rc is the collector resistor and RL is a load ...

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Both circuits will achieve the result of common ground, but there are some conditions. Assuming first that you are driving a low current load (e.g. <100mA), and at low frequency (i.e. not PWM), then either circuit would work just fine. If however you start introducing PWM switching frequencies, then the circuit on the left becomes problematic. Basically ...

5

Of course it's possible. However, it will take significantly more CPU resources than having a UART do all the low level timing for you. Transmitting is easier than receiving because you own the clock. You know when the leading edge of the start bit is, so you don't have to do any sub-bit timing. For receiving, you have to synchronize to the leading edge ...

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As others have said, it's a JFET, which behaves similarly to a depletion mode MOSFET. For a voltage across the device that is much bigger than than the cutoff voltage (about 2V) the current is fairly constant. Here is a typical curve from the Onsemi datasheet: So, at Vgs = 0, the current for voltages of more than a few volts only varies from about 3.5 ...

0

A few issues. 9V batteries are not that strong. A 1A solenoid will drain it quickly. A 12V solenoid may trigger at 9V, but th at's already 75% of the target voltage. Any lower and it won't trigger. A solenoid needs less current and a lower voltage to hold a state than it does to trigger. A solenoid under load will be worse. A 1k ohm resistor, with 3.3V ...

0

Try this logic-level MOSFET That keeps you from having to deal with current gain problems (Beta/hFE) if the gain is too small. (ie if the Base current is 16mA and the hFE is 25, the collector current will only be 0.4A) (If you're only driving 100mA or so, just go ahead and use the BJT though!) If you put a 200 Ohm resistor between the GPIO and Gate pin, ...

1

100 nA is the dark current. That is, with no light falling on the detector, it will put out about 100 nA. 0.8 volts is the saturation voltage. That is, if you shine a light on the transistor when it is connected like simulate this circuit – Schematic created using CircuitLab when the light is dim, VOUT will be nearly Vcc. As the intensity ...

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