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Since the transistors are labelled C B E in tiny letters, I'm assuming they are bipolar NPN transisitors and not MOSFETs. I'm also assuming that my reading of your picture is correct, but I've not checked all the connections. To get the transistors to conduct, you will need to connect their bases to a positive voltage and not ground. And for bipolar ...


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First, redraw the schematic so we can see what is going on: This makes it clear that the internal switch is not in circuit and is being used to control the external switch (the PNP in the datasheet). The design procedure: Choose your current limit resistor - this is in the datasheet. The next step is a bit iterative: Choose a transistor that can ...


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Your transistor gate is being blown by static electricity generated in the flowing water stream. The 4069 does not blow out because it has protection circuity on its inputs while the gate of your transistor is a little "exposed". You can add protection circuitry as follows. Series resistor from input to mosfet gate bigger is safer but will reduce sensitivity ...


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Power v. Temperature correlation... What I think you're missing is that the transistor of that size and power rating is not likely to reach 150deg C at only a 5W output. That is, you will not simultaneously have both of those conditions true. Therefore, your subtraction is not a true statement. You think: $$T_{case}(5W) + R_{system}(5W) = 150^oC$$ when ...


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That looks to me to be a BULK trigate profile, not SOI. You should get the paper: "Multigate Transistors: Pushing Moore's Law to the limit" by J. P. Colinge (I'm pulling that from memory). It's mostly a marketing piece, but it has pictures of trigate profiles. (I know nothing of TCAD)


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You are probably going to be ok because the base voltage will be only one diode drop above ground, and the base-collector junction, with the similar voltage drop, will have a small voltage, if any, due to the reverse bias and the current is limited by your base resistor. However, best practice would be to consider an opto-isolator here and keep V+ ...


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From what I understand, the voltage you've labeled V+ will remain on, even when +5 disappears. This is an issue, because current will be able to flow through the base-collector junction of the transistor, and then through the ESD diodes of the micro, trying to power it. It may or may not be enough current to cause damage or turn on the micro, but it's ...


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For the terminology... 'Pulling' a node to a voltage or state ('on', 'off' etc.) means drawing a current from it so that it goes that way. The state the node was 'pulled' in can be (or is expected to be) over-rided by another component. A good example is a switch connected to ground in series with a 'pull-up' resistor connected to, say, 5 V. When on, the ...


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1) When Q2 is on, it has very low Vce, 'pulling' the collector of Q2 to near ground. When Q2 is off, R3 holds Q3 in an "off" state, because it causes the base voltage of Q3 to be almost identical to the emitter of Q3. When Vbe < 0.7V the transistor will be OFF OFF = no current 2) Which current? Q2 sinks current from R3 when it's on. The "Load" sinks ...


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The technique required to solve the circuit you provided, and determine why Q1 and Q2 are not both conducting is to make an assumption about which transistor is conducting at the beginning of the analysis. Assume Q1 is conducting and Q2 is not (based on the 5V applied to the joint base resistor). If Q1 is conducting then we can write a KVL equation from the ...


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First of all 'B' or beta, which is the measure of a bjt transistors gain in a known circuit, is not a factor here based on your question. The emitters are tied together and referenced to ground via a 1K ohm resistor. A little research into NPN and PNP bjt transistors show they have a base-emitter turn-on voltage (forward) of about .55 to .65 vdc for the NPN ...


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For Q2 (PNP) to conduct its b-e diode must be forward biased so its base voltage must be what with respect to its emitter voltage? Same question for Q1 (NPN), now note that the b-e voltages are the same because the emitters are connected together.... Regards, Dan.


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It is basically a multivibrator circuit. It rely on inherent imperfections on the circuit elements to aid its startup. This type of circuit will NOT stimulate properly in circuit simulators unless deliberate imbalance is introduced in at least one of the circuit elements. What is meant by imbalance is the minor difference in one of the transistor gain, a ...


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For Muhammad, or anyone else learning EE, Might I suggest "The Art Of Electronics 3rd edition" Prices vary, but it is THE best reference when learning electronics theory. Also, your question is vague. A Darlington is a BJT, just internally configured with two BJT's for greater current gain from the base signal. MOSFET's are different animals and VERY ...


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Your approach misses what's going on, and over-concentrates on minutiae. Basically you can't see the wood for the trees. A better approach would be to say a) assuming inputs are roughly at ground potential b) then means roughly 24v is across R135 (ignore 0.7v in comparison to 24v), so about 200uA is flowing in R135 c) which for equal base voltages will ...


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In theory: yes Is it practical ? No All circuits are analog Analog circuits have an infinite number of levels. Digital 2-level electronics is an abstraction of the behavior of analog circuits. This is done because: transistors are on or off, this saves power (current only flows during transitions) this is simple, simple logic descriptions are easy to ...


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This technique is used for MLC (multi-level cell) Flash, 1G and 10G Ethernet, and almost nowhere else. The reason is that discriminating among the 4 levels is complex, whereas the simple 2-level system produces gates in which every trasistor is either 'on' or 'off'.


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For the common collector configuration shown in your figure, transistor Q1 cannot saturate due to (a) the voltage developed across resistor R3 which causes negative feedback which limits \$V_{BE}\$, and (b) the "hallmark" of BJT saturation is \$V_{E} < V_{B} > V_{C}\$, and by inspection the condition \$V_{B} > V_{C}\$ cannot happen.


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If by "turn on completely" you mean "produce 1 volt across R3", the answer is "none". As in "not going to happen." The best you'll get is about 0.3 volts, which corresponds to a 0.7 volt drop across the base-emitter junction. If you want 0.9 volts or better across R3, you'll need to change your circuit to simulate this circuit – Schematic ...


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There are two basic problems with your circuit. The transistor needs at least 0.6V to turn on. Any audio signal with a peak voltage below this won't light the LED at all. You can solve this problem by using a voltage divider which raises the DC Base voltage up close to 0.6V. The audio source will probably have a capacitor on its output to remove any ...


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Yes, you can simply use a rectifier circuit and a smoothing/filter capacitor. You can adjust the value of the resistor and capacitor to suit your particular conditions. simulate this circuit – Schematic created using CircuitLab


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As Olin mentioned in his answer, the LED will be off if there is no audio output. The best compromise I can think of involves inserting a 1N4007 diode in series with a speaker (+) output. The end of the diode with a white line goes to a 1K ohm 5 watt resistor, then to the (+) wire of a 470uF 150vdc capacitor. From this connection a 10K 5W resistor connects ...


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The output of an audio amplifier is a voltage that varies with the sound content, so the LED will flash with the sound.


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Ultimately, no, you can't keep the LED on steady. Think about it. When there is no sound at all, there is no power available to run the LED from. You can harvest power from the audio signal and store any excess that isn't needed to immediately run the LED. When the audio level gets too low to provide enough power to run the LED, you then use some of the ...


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You have the general idea down. PNP transistors used to switch 5V to the anodes of each channel. This inverts logic, which is simple to fix in code. Your only problem is that you are missing a base resistor between each GPIO and the transistor base. Without it, you can fry the transistor. At I'm assuming 20mA per diode based on your existing resistors, so ...


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Why don't you use the 2N2222 to hold the 555 oscillator capacitor discharged then the 555 can never oscillate until the base drive to the 2N2222 is removed and the transistor open circuits. See pin 7 - it's called the discharge pin. Connect collector to pin 7 and emitter to GND then you can force a permanent discharge by driving base current into the BJT.


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simulate this circuit – Schematic created using CircuitLab Using a transistor to switch a subcircuit of higher voltage / higher current is very common in electronics. The biggest thing to take away from this is NPN transistors (which the 2n2222 is) "connects your circuit to ground" which PNP's connect your circuit to the positive supply. Trying to ...


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HINTS Choose R1's value so that When no light shines on LDR1 the voltage across R1 is less than M1's gate-to-source threshold voltage \$V_{TH}\$. When light shines on LDR1, the voltage across R1 is greater than M1's Miller Plateau voltage, but does not exceed M1's maximum allowed gate-source voltage \$V_{GS(MAX)}\$. The Miller plateau voltage is usually ...


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I see 2 possible problems, one of them already pointed out by The Photon. They both have a common theme and that is the base of 'Q' is being overdriven by either the voltage being turned up too high, so the BE voltage of Q is locked at about .7 volts, or the simulated tach signal is overdriving the base, causing the same saturation effect. Try increasing the ...


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The floating cathode of the opto-coupler LED makes no sense. You got something wrong in tracing the board. Either there is a trace underneath something you can't see, or the board has inner layers. Unless this board has a negative supply, ground connected to the emitter and anode doesn't make any sense either. Bottom line is your reverse-engineered ...


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What semiconductor means[?] The word "semiconductor" is often used like this to mean a chip or semiconductor device. what is VDS \$V_{ds}\$ is the voltage between the drain and source of a MOSFET. In this context it means the maximum \$V_{ds}\$ that can be applied when the MOSFET is not conducting without damaging the FET. can i just use a ...


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The optocoupler isolates the two sides of the circuit from each other, so the MCU is protected from whatever happens with the relay. The two circuits differ only in the extra transistor (Q3). Such a transistor is used when you want to use a small curent to switch a larger current; this is necessary if the optocoupler cannot pass a current that is large ...


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The capacitor is charged when you press the switch. The capacitor is discharged through the B-E junction of the transistor and RT. The current gain of the transistor means that the current flowing out of the capacitor is a tiny fraction (typically 1/100) of the current flowing through RT. It also means that any bias current required by the 555 inputs is ...


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When the button is pressed the capacitor is charged instantaneously to 8.74 V. It will then turn on the transistor and a current will flow through RT, the value of this current is (VC-0.6)/RT where VC is the voltage in the capacitor. When the button is released the transistor will be kept ON by the capacitor and a small fraction of the previous current \$\...


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This is very similar to a simple current mirror. simulate this circuit – Schematic created using CircuitLab Notice how Q2's collector and base are shorted together ? This is effectively making Q2 a diode. Which makes this circuit just like your circuit, except, rather than a diode, this circuit is using a transistor as a diode. Collector ...


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The 2 circuits on the right are called common emitter. By adding a emitter resistor here as well you can increase the linearity and stability (while decreasing the gain). The 2 circuits on the left are called common collector. I made simple table for you to compare the properties of both circuits.


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In scenario 1, take the PNP on the left: To fully turn the switch on, \$V_{in}\$ would need to go below ground to get at least 0.6V \$V_{BE}\$ which would require some source of negative voltage. [Update] In response to the comment for the PNP in scenario 1, it takes at least 0.6V to turn on a transistor base-emitter junction; with this scenario, the ...


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The two circuits on the left are the emitter followers. As the name suggest their response actually much depends on what is happening on their emitters. Imagine the two circuits on the left are used as the switch to control the load. As the load current getting higher and higher, there will be higher voltage drop at the emitter thus effectively reduce the ...


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The two circuits on the left are emitter followers. The BJTs operate in forward active mode (assuming \$V_{in}<V_{cc}\$). The two circuits on the right are common-emitter switching circuits. The BJTs may operate in saturated mode. When operating in saturated mode, the power consumption of the BJT can be substantially lower than when in active mode. This ...


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The key is that this is a tutorial circuit aimed at a specific aspect of h bridge design. There is no suggestion in the tutorial that this would usually be suitable for real world use. It would more-or-less work as is in very limited special cases , which is essentially what the tutorial says. In the tutorial it is used to demonstrate the basic switching ...


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There are also non-metallic conductors, including (solid) graphite and conductive polymers, (liquid) solutions of salts, and all (gas) plasmas. The liquids and gases probably don't qualify as part of the answer. Semi-conductors such as those based on silicon and germanium may qualify although they may be more susceptible to corrosion. Note that thermal and ...


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Gold ... In applications where this is a requirement you can order gold flash over your copper. Although in reality there will be some corrosion in very harsh environments. ENIG ... Electroless Nickel Immersion Gold is lower cost, Nickel as a barrier, gold to protect the Nickel.


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Perhaps the author chose to ignore the fact your first circuit will not work with supply voltages exceeding about 1.2V or perhaps he or she did not understand. Who knows? There is no shortage of bad circuits on the internet. When the supply voltage exceeds two Vbe drops, both base-emitter junctions conduct and the transistors will be destroyed if enough ...


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It appears to be a BS250 P-MOSFET. Here's one with similar markings at bg-electronics.de


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"Imagine a world where 2n2222s made in the 70s have different pinouts then one made in the 80s." Such a world exists, today is the third time I have received a single batch of new TO92 package transistors (2sc815 in this case) - one out of 10 follows the original 1970s NEC datasheet pin configuration ECB whilst the other nine are EBC which accords with a ...


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It is similar to RTL or PMOS, which uses a single type of BJT or MOSFET except that vacuum tubes are more like depletion mode n-channel MOSFETs. You can find many schematics online if you look. As far as wasted power in the load goes, the tube filaments use a lot of power too (several watts for a dual in many cases) the load dissipation was not that huge in ...


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I think this can be explained only after considering the effect of load line. As it is evident from the figure shown, a change in base current causes a negligible change in collector current in the saturation region. Whereas in active region , it causes a linear change. Hence collector current is the one which gets saturated upon changes in base current. ...


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So my question is: Is it possible to implement two different kinds of switching in vacuum tubes? Yes, it is possible. For example, you could build a tube consisting of one cathode, two anodes, one acceleration grid (like in a triode) and one deflection grid (like in a old display/CRT). The acceleration grid is used to establish electron flow between ...


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Much of the history of semiconductor IC logic was done with only N-channel or P-channel transistors. It is only in more recent generations that designers had the luxury of having access to both types. Certainly logic can be designed with only one type of switch. As it was back in the original days of real "firebottle" vacuum tubes (BrEnglish: "valves") and ...


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As you've realized, driving a gate with 100 volts is not a great idea. Neither is trying to switch 100 volts with an optocoupler with a maximum collector-emitter voltage of 70 volts (see page 3 of the data sheet). Input-output isolation voltage is NOT output operating voltage. Dwayne Reid has also pointed out the folly of your schematic due to what's called ...



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