# Tag Info

## New answers tagged transistors

1

You did not specify how large your input signal is and how large are the "spikes and high voltages". If your input signal is below say 0.5 volt peak, then you can use back-to-back diodes connected from your input to ground. The diodes will not conduct for the signal and thus appear as high impedance devices but will conduct for anything above about 0.7 ...

1

Have you considered a flame diode or triode? These are vacuum tube equivalents that operate inside the flame of an alcohol lamp instead of in a vacuum. This guy has built both, and the site has pictures of his devices and descriptions of the materials and techniqued he used. You will need a source of high voltage DC (he uses 200VDC) and you will have to ...

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NTE161 Silicon NPN transistor have fT 600MHz and have low Output Capacitance of about Cob:1.7pF. Emitter–Base Breakdown Voltage is about 3V

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PN3569 seems to be the only one with equal or greater than 600MHz fT and equal or greater max Ic on Mouser with the same package. The max E-B voltage is 5V instead of 6V like the 2N3904, if that matters to you.

2

You can use the thumbwheel for the transistors as well. Keep the emmiters and collectors connected but have the base switch. The only problem with this is the leakage current for all of the unconnected transistors. I do not think this will be a big problem, but your gain will be reduced slightly.

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at Lg ~< 10 nm Still some solutions exist like: http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6523122

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The log circuit (or diode circuit) is for the triangle to sine converter- distorting the triangle waves so they like more sine-y. They have nothing to do with making it oscillate. This is the core building block at work is a slightly more complex version of this: The comparator has a +/- output and some hysteresis. When it reaches the positive trigger ...

0

The actual design is NPN where the middle "P" is the base. The base-collector is reverse biased. Mental picture time: if you forward bias the base-emitter, just a little, some electrons will flow from emitter to base. However, since the base junction is Quite Thin, several of those electrons will "overshoot" the target... ending up in the Collector. Which is ...

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Yes it is true that for "linear" amplification purposes it is required that Vce>Vbe. As a consequence, the base-collector pn junction will be reverse biased with Vc>Vb (I assume, it was this condition you have read somewhere). Only under this condition the carriers (electrons in case of npn) which are set free at the emitter - if Vbe>(0.6...0.7)V - are able ...

2

It's not clear what you read, but the conditions for forward active operation of an npn transistor are $$V_{be} > V_f$$ $$V_{ce} \ge V_{ce}(\mathrm{sat})$$ $V_f$ is typically around 0.6 or 0.7 V and $V_{ce}(\mathrm{sat})$ is typically around 0.2 V. Some texts will say that $V_{ce} > V_{be}$ is required for forward active operation, and this ...

1

Your logic looks like a sum of products (ANDs feeding into an OR). In that case, you can replace both the ANDs and the OR with NAND gates. Also, you can replace (~a ~c) with ~(a + c), which saves a gate. Putting it together, that gives 1 NOR, 1 NOT, and 2 NAND, for a total of 14 transistors.

2

"Common Collector" means that the Collector is a common point for both input and output signals (ie. Ground, or a fixed voltage relative to it). In a true Common Collector circuit the value of RC is zero Ohms. Larger values will still work, but with reduced output voltage swing. In your circuit the Emitter can only go up to slightly less than 9V, because RC ...

0

You could add a piezo buzzer in series with your LED, put the high pitch one in series with the HEIGH LED and the a lower pitch in series with your LOW LED. They are quite cheap, small and work can work at low voltages so it would be perfect with to fit your circuit.

1

I'll just mention that an old favorite transistor among hobbyists is the 2N2222, an NPN which is good to 800mA, and has an hFE of better than 100. Also, if you run into a situation where you need a higher current AND high gain, consider a darlington transistor. I'll agree with Spehro, a MOSFET has a lot of advantages as a switch. But you asked about a ...

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Google "electronic buzzers" and choose one with high pitch and one with low pitch. Here is a page from Maplin that gives some ideas: - Here's a tri-tone one (on maplin page also): - There are plenty on the web and they work from DC so should easily interface to your circuit. Here's a very small one that works from 1.5 volts to 5V: -

3

You'd be better to use a MOSFET for this kind of application. R1 keeps the Arduino output current low during switching (the MOSFET gate looks a bit like a short during the brief switching interval), and R2 makes sure that if the drive gets disconnected the MOSFET switches 'off' and does not hang around half-off and half-on, burning itself up. ...

0

I use microcap (which uses standard transistor spice files) and there's plenty of times I'm simulating in the low hundreds of MHz for analogue and digital signals. If I don't pick the correct transistor it doesn't work (and neither would it in the real world). I've pretty much designed radio front ends at 400MHz and got reasonable results when implementing ...

1

A transistor may be used to amplify current, voltage, or both. In the indicated circuit, it is being used to amplify both. When using the emitter as an "output", the output voltage will vary almost 1:1 with the input voltage. In some circuits, that 1:1 voltage behavior is very desirable, since the accuracy of the unity gain won't be significantly affected ...

2

LED would indeed be brighter if you put it after because on the before transistor there will be hfe*Ib ammount of current, and it will have (hfe+1)Ib after the transistor. You will probably not notice this because hfe 100 or more in most cases and 1% extra current will not cause visibly more light. If you have a transistor with hfe of say 5, then you will ...

1

OK first a little on notation. Your Vgs(off) is also called the Threshold voltage (Vt) and the pinch-of voltage (Vp), I think they all mean the same thing. The voltage where the device is off. So I cracked open "Art of electronics" (2nd ed.), They talk about two regions of operation. A linear region where the the drain current is proportional to the gate ...

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Human eyes have a log response, so losing a bit of brightness is a lot less undesirable than some brightness where it doesn't belong. The "right" way to do this, in my opinion, is to insert some dead time so that the drivers have a chance to settle out. IOW, turn off the one driver, wait and wait and wait (that's 3 microseconds perhaps) and then turn the ...

7

I was thinking that the current had to go though the transistor to be increased. This highlights a dangerous misconception. Current is the flow of charge. Charge, like energy, is never created nor destroyed. Thus, you will never find a device where the total current flowing into the device is not equal to the total current flowing out. In more formal ...

4

Replace the transistor with a simple switch, so you just have a battery, an LED, a switch, and a current-limiting resistor. Note that when you close the switch, current flows on both "sides" of the switch. "Why does the switch control current "before" it as well as "after" it?" Because it has to be: current in a simple closed loop has to be the same ...

7

Perhaps it's a little easier if you think of the circuit as being drawn something like this: simulate this circuit – Schematic created using CircuitLab Your finger acts as a resistor (a largely-unpredictable variable resistor), connecting the positive terminal of V2 to the base of the transistor. That allows a little current to flow through the ...

14

You are looking at a loop of current, so after and before don't have much meaning. Other than the tiny base current, the current in the red and blue paths are the same. Your concept of before and after in electronics isn't applicable. You need to understand some basics before you can make sense of this, but that would be too much to try to teach in a ...

20

The collector of the amplifier (the pin labeled "C") is actually AFTER the transistor (using the asker's reference frame, which is somewhat misleading.). The Base (B) is the input, and the active transistor creates a situation where the current on the collector is many times that of the base. So, the Base current is the input, and the Collector current is ...

3

A diode (or a transistor connected as a diode as you have) does not behave like a resistor. Rather it has a voltage drop that doesn't change much with current, and has a slight negative temperature coefficient. If you use the recommended 47K resistor (passing about 180uA through the diode), then the voltage drop should be around 600mV and change about ...

3

Any bipolar transistor with collector connected to base can be used to measure temperature. No, it's not really a temperature-variable resistor. It's more like a temperature-variable voltage source assuming some minimum current is passed thru it. With C and B shorted, the transistor looks electrically much like a diode. To use it as a temperature ...

0

A typical D latch will have a single data input along with an enable; internal logic will combine those so as to yield "write-zero" and "write-one" behaviors. An SRAM cell will have one or two bidirectional data wires and one or two enable signals (if two, one will control each data wire; if one, it will control both. To read a memory cell, drive an enable ...

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An SRAM cell is functionally equivalent to a D latch (but not to a D-type master-slave flip-flop). As you say, the implementation details are different: An SRAM cell is typically built so that it is written and read using the same pair of wires ("bit lines") — the difference is whether or not the bit lines are being driven by the control circuitry. On ...

2

First off, you dont need a series resistor with the fan. If its rated for 1.68W at 12V, then as you calculated it, the current through it at 12V will be 140mA. The usual rule of thumb for saturating (turning it fully on/off) a transistor is that the base current should be 1/10 of the collector current. That means that the base current would have to be ...

0

${{5\text V−0.7\text V}\over220\Omega} \approx 19.5\text{mA}$ You cannot source that much sustained current from many outputs without damaging the device. Either increase the value of the 220$\Omega$ resistors or switch to something like a ULN2803A instead. As a bonus, the ULN' includes the clamp diodes.

3

The power resistors will be the easiest. You can get them with all sorts of resistances. So 10W is 10V @ 1 amp (10 ohms) or 30V @ 0.33A, (~100 ohms). I'm not sure how you are mounting them but I like the TO-220 pack that has a screw hole. I've also used transistors as heaters. Sometimes by themselves, but also in series with a power resistor. For a ...

1

Short answer: No Explanation: Essentially what @Passerby said. While it sounds like a nice idea in theory, the variability in supply voltage, control voltage, max brightness/current, etc. all mean that these devices would have to be made in relatively small volume. Small volume -> high price, so that product line likely wouldn't survive very long since the ...

3

You don't mention a specific type of transistor so I am going to explain generally only the physical principles. You can think of a transistor as two connected pn junctions. You have holes flowing from the p-type region to the n-type region. Also you have electrons flow from the n-type to the p-type. When holes reach the n-type region they will disappear ...

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After a lot of research and experiments with a software simulator, I could be able to redesign my circuit like this- It works perfectly in the simulator.

3

Depending on which mode you are in, you have two different equations for Id. Also Vds in this case, is the same as Vd because Vs = 0V. You write Vd in terms of Id. $$V_{ds} = V_{dd}-R_d*I_d$$ The thing with these types of circuits is that, its a guessing game. You have to assume on mode of operation, and use the equations for that mode. The math will ...

2

Assume that Rd is insignificant and Vdd is high enough to saturate the FET, then calculate what Id would be. Now calculate what Vds would be with that current flowing from Vdd through Rd. Which region is the FET in now?

2

Because of how transistors work. If you consider the NPN bipolar transistors for example (the ones on your diagram), they only amplify the input signal if the base is positively charged with respect to the emitter: that's when the isolation layer (called depletion layer) that spontaneously form between the base and the emitter is no longer enough to stop ...

0

"Why should an extra voltage VBB be needed when there is already an voltage VCC available to produce Ic current?" If this would be correct, you would describe the properties of a resistive two-pole. But the transistor has three nodes. And it is a device that can be used for amplification purposes. Hence, you need a third terminal for connecting the signal ...

0

Read some more and build some circuits. There are several transistor models. At it's simplest a transistor has only one polarity. It only controls current flow in one direction. So for an npn you need to push current into the base to get current to flow from C to E. Pulling current out of the base, (base negative with respect to the emitter) does ...

0

Bipolar transistor are "current control" device. The $I_{c}$ are "controlled" by $I_{b}$, so you must have a voltage source (some times a current source may ok) to bias the base-emitter to supply $I_{b}$. Please read the book carefully. I think it will show you better than me.

5

The base to emitter voltage is pretty stable at 0.6V, you are not adding any feedback with the transistor. You actually built a emitter follower or common collector. The main property of this circuit is that the emitter is pretty much at the same voltage as the base voltage. If you are hinting at the non ideal behavior of the transistor, even the slight ...

1

A little while ago I was trying to think of a way to replicate the basic functionality of a LM3909 (now obsolete). I wanted a short flash, followed by a long-ish delay (about 5-10 seconds) and I ended up with a circuit which looked something like this: simulate this circuit – Schematic created using CircuitLab I was also trying to waste as ...

0

Brhans is right you can't put NPN transistors there you either need a pullup resistor or PNP there but i fail to see the point since inside max u already have a NPN transistor driving that line. I think you will have better luck if you explain what you are trying to accomplish, im guessing some weird dimming or something similar witch wont work anyway. Have ...

3

If you short the output terminals then Q3 will be dissipating 250mA * 11.3V = 2.8W. BSR14 is an SOT-23 transistor with an $R_{\theta JA}$ of 357°C/W (typical layout) so the junction would head for >1000°C, without pausing at the absolute maximum temperature of 150°C. That would tend to be terminal. If you replace Q3 with a much larger transistor (with ...

1

When light is present, the phototransistor will conduct better than in dark conditions. This results in a lower voltage on the base of the transistor (the phototransistor and resistor form a so called voltage divider). When the voltage on the base of the transistor is lower than about 0.7V (transistor base-emitter voltage) + 1.8V (LED forward voltage) = ...

3

A direct connection is not a good idea since there are too many unknowns. The safest thing is to have the microcontroller activate something that acts like a switch. A relay definitely does this. However, an opto-isolator will most likely work too, and will be smaller, cheaper, and take less power. You should be able to drive it from a microcontroller ...

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There is an older version of the WS2811, that's the WS2801. This chip also has a constant voltage option, so you can control a FET with it that PWM's all your LEDs. Have a look at page 10 of the datasheeet! Edit: You would need some LEDs in series (say 3) with a resistor, you can then connect 10 of these sets of "3 leds and a resistor" in parallel, to ...

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You can charge the battery using an Arduino as the controller. Sense the voltage of the battery with the ADC, making sure you scale the twelve volts down with a voltage divider as the Arduino will only handle an input of 5V on the inputs. From this you can work out the duty cycle of the signal to send to the battery to charge. Once the battery is fully ...

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