# Tag Info

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The problem seems to be as obscure as expected. Not related to having the supply loaded or anything. I am simply using a capacitor to smooth out the signal to the base and the circuit works. Discovering further, this external power supply is intended to be earthed and wasn't. Even still, the capacitor seems to be necessary for startup to be successful. I ...

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If you were to define "saturated" as $V_{CE} > V_{GE}$ then for an 'on' gate voltage of, say, 12V it would be desaturated iff $V_{CE}$ > 12V. That would be an indication of something bad happening and it would be time to switch it off fast (but not too fast, or that itself could cause damage). In reality, that definition does not seem to be widely ...

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This relates to your other question, so this may seem a little circular. In the circuits that require the highest device density, MOSFETs are preferred for a number of reasons, mainly for their electrical characteristics such as minimal static power dissipation. But it turns out that MOSFETs are also easier to pack more tightly, because their physical ...

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It is very dependent on the circuit. See this reply on details why one would use one over the other: When is a MOSFET more appropriate as a switch than a BJT? For discrete transistors there are advantages to each. When we talk about ICs and VLSI systems the MOSFET is the device of choice because of the simpler manufacturing process and the fact that they ...

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There are multiple reasons you want to do this in your case the reasons I can think of are: 1) The relay needs 12V on the input to turn on. An Arduino GPIO only outputs 5V. A transistor in between will allow a 5V signal from the Arduino to switch a 12V signal via the transistor to the relay - thus turning it on. 2) The transistor provides protection. In ...

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Is it because the arduino can't supply enough current to trigger the coil in the relay? Yes. An Arduino (really, a ATMEGA328p or similar) can only provide 20~40mA on a single pin without significant voltage droop and possibly frying the pin. A relay coil can take significantly more. And the Arduino pin output is referenced to VIN, or 5v on a typical ...

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Usually a 'diode-connected' MOSFET is taken to be a MOSFET with the gate and drain connected, analogous to a diode-connected BJT with base and collector connected. In US6960946 a level shifter design is disclosed that uses a diode-connected MOSFET (N21) as a load. The $V_{GS}$ breakdown voltage (it's not a 'junction') must exceed the higher supply ...

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Your question is hard for me to interpret, especially talking about diode-connected transistors in the context of CMOS. But you find diode-connected BJTs (not CMOS) in every current mirror or bandgap reference, for example. Since most opamps using BJTs also need current source/sinks, they are also found there. They are found in SCRs. Etc. The MOSFET ...

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Minimizing the number of components is trivial if the transistor is expected to work as effectively as a diode in (say) a power supply circuit. Consider a bridge rectifier. It has four pins and a reverse voltage capability of possibly hundreds of volts. Most BJTs can't handle more than 10v across the base emitter junction in reverse. The bridge may be rated ...

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My advise is to forget about a conventional bridge circuit and do this with the two coils: - Transmit coil and Receive coil aligned (like two of the circles in the audi car logo). This is done to produce a null signal in the receive coil when no car is present. Transmit coil - maybe 5 turns to ten turns parallel tuned with an NP0 capacitor to generate a ...

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If that is a 100 ohm resistor in series with the speaker (not 100% clear in the diagram) and the supply voltage is nominally 6V then the maximum current through the transistor is going to be about 60mA - this could rise to 100mA on a larger supply voltage so be aware of that. In the circuit you've drawn Hfe variations between one transistor and another ...

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Look at the obvious things first: how much voltage and current must the transistor handle in the circuit? How much power must it dissipate? Any transistor that meets these specs will probably work to some degree. Then look at lower-level details such as current transfer ratio (does the circuit only work with a certain minimum gain?) and capacitance values ...

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Could it be static friction? What happens if your arduino sketch first starts the motor (maximum PWM) and then slowly decreases over several seconds? My experience is that it's very hard to run a DC motor slowly (unless you have position feedback of course, or gears).

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Your diode is in the wrong position- it should be across the motor (blocking!) not across the transistor. The purpose of the diode is to allow current that is flowing in the motor coil to continue to flow in the same direction when the transistor turns off. When the transistor turns off, the voltage at the transistor collector will rise as it was flowing ...

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I see that your question is old, has already been properly answered and that you have understood why your transistor was burning up. But I felt like complemeting the answer anyway. While this doesn't answer your question, I wanted to point out that there is this BJT H-Bridge design that prevents such situations from occurring. It gives you control over ...

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This is the way I see it, I hope it adds something useful to the discussion: SEMICONDUCTORS, DIODES AND TRANSISTORS ELECTRONS AND HOLES Let's think of a row of pennies laid out in a line, touching, across a table. Move the right hand end penny one penny's width to the right, leaving a gap. Then keep moving the penny to the left of the gap into the space. ...

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The way I write it is: $V_{in} - I_b \cdot R_b - V_{be} - I_e \cdot R_e = 0$ Knowing $I_e = (\beta+1) \cdot I_b$ and re-ordering to solve for $I_b$. So, in the active region, excluding saturation, little-re, and the Early effect: $I_b \approx \dfrac{V_{in} - V_{be}}{R_b + (\beta+1) \cdot R_e}$ But in saturation, the value of $\beta$ obviously ...

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My knowledge of transistor physics is outdated but let me try. When physicists talk of "potential" they mean energy (actually potential energy per unit charge ). Imagine ( like in the old vacuum tube days ) an electron moving between two charged plates. Basic electromagnetism says there is a constant electric field between the plates. To cross the plates ...

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When you DON'T have an emitter resistor, the voltage you put on the base pretty much defines the current into the base - it's a forward biased diode so base current starts to rise rapidly above about 0.4V and by the time the base voltage is 0.6V you might be putting 10mA into the base. The transistor will be likely saturated with collector at about 0.2V ...

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Read and re-read Dave's excellent answer. Then mentally reverse what's going on... You have a forward-biased base-emitter junction, and external circuitry connected to the base demands a current Ib, which is supplied from electrons sourced by the emitter. But when an electron enters the base region, it encounters a strong electric field pulling it towards ...

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When electrons flow through a forward-biased diode junction, such as the base-emitter junction of a transistor, it actually takes a finite amount of time for them to recombine with holes on the P side and be neutralized. In an NPN transistor, the P-type base region is constructed so as to be so narrow that most of the electrons actually pass all of the way ...

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If you look at how actual IC fabrication is done you will see that the minimum feature size is the size or the width at which a transistor or any type of material on the silicon surface can be drawn at. This is usually analogous to the transistor gate length because the properties of the transistor depend on the ratio between the width and the length of the ...

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If the pins on your microcontroller can sink enough current to drive the LED (usually between 10 and 20 ma), then you don't need any transistors at all -- just two diodes. Check the datasheet for the microcontroller and look for I/O pins maximum sink current (low voltage). It is usually either higher or the same as the maximum source current (VDD voltage). ...

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I think the problem is that the segment drivers are not push-pull, rather they go high-impedance when they are turned off. Since the darlingtons in the UDN2981 will turn on with some tens of uA it will take some time for them to turn off. Since your brightness is so high for the 'on' segments, the 'ghosting' is significant. Try a pull-down resistor on the ...

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How about if you wanted to control the LED from more that two places - how about 4. What about being able to reverse the demand from one input - how could you do that. I'm aware that this is beyond the scope of the question but it's fun to answer: - This circuit can control an LED from 4 independent places using an analogue switch(es) to invert state ...

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Left circuit: If either controller wants the LED on, it turns on. High = 'on' Right circuit: If either controller wants the LED off, it turns off. High = 'off' simulate this circuit – Schematic created using CircuitLab

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You could probably use a open collector configuration. simulate this circuit – Schematic created using CircuitLab Something similar to this. Connect the microcontroller to one input and the other source to the other. Both inputs can turn the LED on.

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The rearranged version of your schematic looks like simulate this circuit – Schematic created using CircuitLab The problems have already been mentioned by jippie, and rearranging the schematic shows that the lower four transistors (Q8,Q9,Q6,Q11) are inverted (collector<->emitter). Also note that even if the level of the bridge supply was the ...

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The circuit won't work because Q7, 5, 3 and 6 are always active unless you have 24V microcontroller outputs. On top of that Q8, 9, 10 and 11 will not be driven reliably, because the emitter voltage varies with current through the stepper's windings. Convert the 0/5V microcontroller output to 0/24V logic, able to source sufficient current to drive the power ...

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For the first answer, this barrier is caused due to the different materials used to construct the transistor. Think of each connection in the transistor as a diode. Now, the source may be of n type material(i.e. have extra free electrons) and the substrate might be p type material (i.e. have extra holes - positive charges). Now if you don't apply any ...

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Did you read the transistor rating of BC546 in the datasheet, Collector Current − Continuous IC 100 mA dc. Your mcu specs mention an absolute max or 4mA as output current (you have already exceeded that with the base resistor you have used). You need to select a more powerful transistor and a high gain Darlington would suit your needs, a device like ...

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The bipolar transistors are current amplifiers. The reason is that the base-emitter act exactly like a diode. No matter if you use a low power transistor such as the 2N2222A or test with a large TO-220, you will find +0.7 volt on the base when the emitter is 0.0 volt (for NPN). The gain of transistor in datasheets is always given as a ratio of the ...

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Impedance matching is NOT one of the major applications of a common collector circuit. Why should a transistor be able to match an impedance? Impedance matching is done with passive components. A common-collector circuit is good at providing a high input impedance (to a weak signal) and generating a low output impedance (at the emitter) - it is a power ...

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The problem here is that it is a boost. So, if you were to just turn off the LM2577 somehow, there would still be Vin on the output. You will need to use a P-channel FET to turn off the input voltage to the whole boost circuit. Something like this could work. As long as Vin is more than 5V and less than 10V the FET should switch correctly. Of ...

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In general, electron tunneling is based on the quantum mechanical concept that the "wave function" of the electron — a function that describes the probability of finding the electron in any particular position — does not end abruptly at the surface of a conductor (or semiconductor), but can extend all of the way through an otherwise impassable ...

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Because the source has N-type doping and the substrate has P-type doping, and together they form a P-N junction, or diode. Even with no external potential applied, such a junction forms a depletion layer. An enhancement-mode MOSFET turns on by using the electrical field from the gate to shift charge carriers so that the substrate immediately below (the ...

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If you open the ground connection on the 7805, the output will equal the input minus a couple of volts. Probably not what you want, and it could damage the load. You could use a high-side p-channel MOSFET switch in front of the 7805 but it's easier to just use a regulator with an enable input as others have suggested. Here is a typical circuit from ...

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Checking the datasheet of the 7805, the quiescent current is indicated as 5mA, so maybe it's not worth bothering (depending on your battery capacity), also given the cheap price of the 7805 ? What output current do you need ? For example, if 50mA is enough, there is also the LM2936, which is supposed to have a very low quiescent current. It is however more ...

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If your ampere requirements are less than 300mA maybe this could work for you: - It's 5V, low quiescent current and can be turned off/on with a control line.

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A schematic would help. TLDNR; An analog switch might work. First, let's look at what you are doing here. I presume you are also connecting the ground of your external 3.3V supply to the emitter as well (and thus also to the "0-volt" pin on your PSU's connector). This will work because the two supplies are isolated. Generally speaking, you really do ...

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This is a perfect application to use a logic level N-Channel MOSFET. This will eliminate the heavy base current requirement coming from the Arduino and make it easier to find a part that has a low RDSON specification.

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While the transistor itself can tolerate 600mA, figure 3 on the datasheet shows that after about 85mA, VCE(sat) increases sharply. If we use something near the last given value on the chart, about 5V or so, we get 1.75W at 350mA. This much power can easily fry a TO-92 given the right circumstances. And the real voltage is likely to be much higher, based on ...

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The diode is in the best position, and is of an appropriate type. It conducts when the input is negative, the same as the transistor base conducting when the input is positive. The 47K resistor is about 1/10 of a normal RS-232 load. One could also block the voltage, but then a -100V spike (ESD say) could break down the 1N4148 and break down the E-B ...

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I am not familiar with the SOP and canonical terms since I did not study electronics in English, but this seems about Boole logic stuff. May I suggest you took a look there : http://sce.umkc.edu/~hieberm/281_new/lectures/forms-of-bool-expressions/forms-of-exprs.html and there : How to convert an expression from SOP to POS and back in Boolean Algebra? for ...

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This is an AC analysis of signals and all the clutter of DC components is removed - input and output capacitors are replaced with short circuits and power supplies are also replaced with short circuits. It's not a DC analysis or any other analysis - it's an analysis of signal amplification and AC impedances with a minimalist approach diagramatically. This ...

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Seeing this diagram out of context, it's hard to say, but frequently in AC circuit analysis, the DC sources are omitted from diagrams in order to make the analysis simpler (superposition principle). You just have to take it for granted that there's a DC source in series with the AC source shown on the emitter side so that the actual instantaneous current is ...

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I have tried it too, it's due to Q4 connected to Q3, instead connect the emitter of Q3 to the base of Q4. Download the Falstad Java simulator and import this, it will make a simulated circuit showing how it works. (http://www.falstad.com/circuit/) Falstad Code: \$ 1 5.0E-6 78.57719942274176 85 5.0 50 R 368 128 368 80 0 0 40.0 9.0 0.0 0.0 0.5 r 368 128 368 ...

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A more common set up for a current source is as follows: simulate this circuit – Schematic created using CircuitLab The voltage across R1 will be reasonably constant at $2×U_D-U_{BE} = 2× 0.7 - 0.7 = 0.7\text{V}$. Program the current by calculating $R_1 = \dfrac{U_D}{I}$ Pick a current through R2 that is approximately 10 times bigger than ...

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It's not going to be an accurate constant current source because the voltage-current slope of the BJT is not that great but if you can live with this then that's OK: - Ideally, once outside the saturation region, the lines would be horizontal to the base line of Vce - horizontal means that no-matter how you change the resistance of the load (which ...

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No, you need to keep the base voltage constant. Try Figure 5 at http://en.wikipedia.org/wiki/Current_source, which uses an LED for this purpose. You could switch right at the load, if you need to

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