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0

I think there is a typo in the equation for Cobo in the Infineon application note. I think it should read: Cobo = Ccb + (Cce * Ceb)(Cce + Ceb) Please correct me if I am wrong.


1

When I look at the datasheet for a 2N3906 manufactured by Fairchild (https://www.fairchildsemi.com/datasheets/2N/2N3906.pdf), I see on page 3 that it has an Emitter-Base breakdown Voltage of -5.0V. So if your SW12 switch is closed then you're applying 12V to the base of your 2N3906. Its emitter is tied to ground through RaT1-2, so you're effectively applying ...


2

Page 2 of this datasheet tells the story. Under absolute maximum ratings it states that the maximum reverse voltage of base and emitter is 5V. You have exceeded this by using 12V and although the base-emitter region is probably still intact (due to the 100k resistors in your design) you can't expect leakage current to be insignificant. Don't expect ...


1

This is expected. Edit: It's not leakage- you're breaking down the E-B junction in reverse by exceeding the breakdown voltage (typically rated at 5V with actual breakdown 6~9V). A p-channel MOSFET might give you what you want since the gate is insulated. There are also (somewhat rare) 'symmetrical' transistors that have equal E-B and C-E breakdown voltages ...


0

First of all, keep in mind that LEDs are basically current driven, not voltage driven. Most commercially available LED strips are voltage driven to the outside, because the simply add some dropper resistors. And here we touch the reason why the first option is often preferred: Efficiency. Not taking power needed for the controls into account, if you run ...


2

I'm afraid your transistor will not particularly like -110V on its base. Nor is your transistor still a switching device, it's better not to have an emitter resistance if you want a sharp and fast response, since it'll want to be an emitter follower rather than a switcher. You should make a few small adaptations, though your initial thought is a reasonable ...


0

If you are indeed using a 2N3904 with the circuit shown, you will probably blow out the transistor. The transistor can handle 6V of emitter-base voltage, when you have a -110V signal in, your EB voltage will be far more than that. However, if you had a transistor that could handle these voltages, then the circuit should work the way you want it to. I ...


2

First let's clear up some of your misunderstandings. This will take some time. Doping does not change the net charge. For every phosphorus atom introduced, a free electron is introduced. But also, an additional (compared with silicon) proton is introduced. After doping, the silicon is still electrically neutral. Doping does change charge carrier density. ...


-2

The emitter resistance connected means the transistor will go to saturation , but the base resistance and collector resistance will remain same .Batter you draw a circuit and calculate base current ,then you will good result.


1

To invert your signal, use another transistor. simulate this circuit – Schematic created using CircuitLab Any of the BJTs will work. I'm personally partial to the 2N3904, but that's habit more than anything. Of the MOSFETs, I'd recommend first the IRF520, then the RFP70N06, and the IRF510 last, although any of them will work. If you use the ...


2

It's accepted engineering practice - in switching circuits - to force transistors' betas to 10, so with that in mind, if Q2's collector current is 300 mA, then its base current should be set at 30 mA. Q1's collector current would then be 30mA plus whatever came from R1, say 1mA, for a total of 31 mA. The ATMega would then have to force 3.1 mA into Q1's ...


2

I am having trouble understanding the three operation modes of a transistor. When we talk about the modes of operation of a transistor, we're usually talking about cut-off, forward-active, and saturated operation. The rest of your question seems to be about the different fundamental amplifier configurations, rather than the operating modes, so that's ...


0

Your T9F is a PLVA665A Zener Diode. Specs for it can easily be found through google. Such as here. That is a 6.5V Zener, rated for 250mA. Code T9F found in this book - SMD-PHILIPS Marking Codes Small Signal Transistors and Diodes. Other copies of the book can be found as PDF through google. I can't find anything at all about the 2U90. Could you post ...


0

I think you need to understand what those numbers mean. 10 A is the maximum rated current. 30 VDC is the maximum recommended operating voltage, the voltage that that the relay is intended to isolate when the contacts are open. 10-28 VDC is the coil operating voltage range, the voltage required for the relay to toggle its state.


4

1) Driving 35 amps through contacts rated for 10 amps is a great way to kill the contacts. 2) The contacts will fail either by welding shut or by "burning" so that they can no longer make contact. Which is more likely depends somewhat on your load. If your load is purely resistive or has a turn-on surge (like incandescent lamps or capacitors) then welding ...


1

Power MOSFETs normally have relatively high Vth, which requires voltage higher than MCU will output. Also the gate capacitance will normally require high current (sometimes even 5A!) for like tens of nanoseconds each PWM cycle, which is way beyond what MCU can do. And of course the high side MOSFET's source voltage rises to Vbus, so to drive it's gate you ...


0

The controller you show is a full H-bridge controller. Here's the rub: The motors will be running on a voltage higher than the microcontroller's voltage. The two extra transistors are to transition from the microcontroller voltage to the motor voltage, allowing the MOSFETs to fully open and close through the voltage on their gates. Imagine this: ...


0

If I recall it right, Jeri Ellsworth made a few transistors, depletion-mode NMOS be exact, and recorded videos logging the process of making it. You can check YouTube for how she made it and her tests.


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Sure it can be done, if you have ninja skillz: Homebrew NMOS Transistor Step by Step - So Easy Even Jeri Can Do It Making Microchips at Home - Cooking with Jeri Part 1 I wanted to post more links but I don't have enough reputation points to allow that. Just search for the followup videos to the 2 links I have provided.


0

A very similar question has been asked before. My answer there included this link to the website of a someone who has actually made thin film transistors out of zinc oxide. That site also includes links to otehr people doing similar work - at home and in the garage. So, it is possible to make transistors at home, but from what I see it is more of a learning ...


3

Well, yes and no. You can make a transistor by hand in your kitchen. It will involve some nasty chemicals, but it is doable. A friend of mine made a LED a couple of years ago. Not a transistor, sure, but the same process was involved. (She was studying semiconductors at that time, so she had help. The process itself was quite similar to producing ...


0

Your question is wrong, because size is one of the key attributes that defines a "modern" BJT or MOSFET. Given that you really mean "is it possible to make a BJT or MOSFET using obsolete techniques in your garage" I would say it would be so difficult as to be almost impossible both from a technical and financial standpoint. Using just used, refurbished ...


0

It's possible, but not so easy and the result is not going to be anywhere near the quality of components that you can buy. The main issue is that in order to actually form the semiconductor junctions inside of the transistor, you need to carefully change the properties of a piece of semiconductor in a controlled manner. This requires rather specialized ...


0

For the quiescent point, \$V_{GS} = V_D\$ since there is no current flowing into the gate of the MOSFET, and there is a capacitor blocking any current flowing into \$R_C\$. And since you have an expression for \$I_D\$ you can write \$V_D = 15V - I_D*R_D\$; from there you should be able to solve this.


0

First of all, thank you for all your replies! I researched a bit more and decided that a transistor and everything similar were going to be to dangerous. So I decided to use an octocoupler, more specifically a K827P, so that the Raspberry Pi is totally isolated from the 24V. As seen in the diagram the microcomputer is protected, and a resistor is added ...


2

1) Why are R4 and R1 required at all? Why would we limit current flowing into the transistors and charging the capacitors? Transistors operate over a range of collector currents - you can look this up[ for different types (see https://www.sparkfun.com/datasheets/Components/2N3904.pdf) . The minimum values used for R1 and R4 will determine the maximum ...


1

The two capacitors provide a negative coupling between the two transistors: when one goes into conducting, this keeps the other out of conducting. 1) Without R1 and R4 there would not be a negative going voltage on the collectors when a transistor gets into conducting: it would just be the power supply, no change. And a capcitor does no 'conduct' a 'no ...


2

The reason for the sawtooth shape at the collectors of the transistors is caused by the B-E junction of the transistors and C1 / C2. You can insert resistors in series with the base of each transistor - I'd start with 10k and see what happens. Be sure that the resistors are in series directly at the base of each transistor. That should give you a much ...


0

Your post/tags are confusing. If you want to use an NPN BJT transistor you should remove the MOSFET tag. If you want to use a MOSFET you should correct the terminology in the post. However, either way the effect is the same. As Vbe or Vgs goes up and the transistor starts turning on more and more, it will draw more current into the collector/drain. The ...


2

If you look at the data sheet https://www.fairchildsemi.com/datasheets/BU/BUZ11.pdf you'll see the thermal resistance to ambient, which for a non-heatsunk TO-220 (such as you are using) is in the vicinity of 75 deg/W, if the transistor is dissipating 2 watts the chip will get to ~175 C (2 x 75 + 25 ambient), which is more than enough to do damage. If you ...


3

Mirroring what Brian Drummond said above, it may not be switching on 'very hard'. What this means is that the MOSFET is not entering the 'active' region of operation. In the 'active' region, you can generally expect the MOSFET to have an \$Rds\$ that matches the specified \$Rds_{on}\$ (in the case of the BUZ11, 0.040 ohms). What is instead happening if ...


-1

You have to be around 10V higher than V(gth) (Usually; check graphs of Vgs first) for maximum switching speed, otherwise you will spend a massive amount of time trying to 'turn on' the FET, or alternatively the next signal might overlap the previous, meaning you garble the I/O data. On top of that, you have to make sure you have enough current to 'pump' the ...


0

An N-Channel MOSFET will start to saturate at V_gth, which isn't necessarily that high. A cursory glance on digikey turns up: http://aosmd.com/res/data_sheets/AO3404A.pdf The idea of an N-channel is that you need the gate to be a little above source to start conducting. In a P-channel, you'd need to drive from source to drain, the full swing. note - I'm ...


0

Use a buck switching regulator (output 3v3) connected to your headlight wiring. No need for a switch. Don't use a linear regulator for this because it might dissipate too much power and burn (load dependant). Buck regulators are near enough 90% efficient at transferring power so it will burn very little power. One word of warning is that with any regulator ...


2

Could be leaky, could be pickup of AC hum from the mains or RF from AM radio etc. as the transistor will act as a detector. To tell the difference, put a 1uF ceramic (low leakage) capacitor from base to emitter and see if the issue disappears. In any case, you can reduce the sensitivity simply by connecting a relatively high-value resistor from emitter to ...


5

Both circuits invert your input signal. The second one will not work properly. If you want a non-inverting level shifter you can use this circuit: simulate this circuit – Schematic created using CircuitLab When the input is close to 3.3V, the transistor is off and R2 pulls the output up to 5V. When the input is close to ground, the ...


2

Based on the updated information in the question, I believe Option A is working as I would expect. It both inverts the signal and level shifts it. I guess the examples you found didn't emphasize the inverting nature of this circuit. Although I would suggest using 10k in series with the base. You want the collector current to be 10x or 20x the base current. ...


0

Try connecting a slow monostable 555 oscillator to a simple, shift register frequency divider. This way you can shift the pulses in and they will stay on until the shift register is cleared. Learn more about shift registers here: http://www.allaboutcircuits.com/vol_4/chpt_12/1.html Shift Registers as frequency dividers: (Not nearly as hard as it looks BTW) ...


1

Also check out Seeedstudio's Open Parts Library: http://www.seeedstudio.com/wiki/Open_parts_library


0

Transistors: 2N3904 2N4906 BC546 BC547 BC548 MSPA44 MSPA92 MSPA46 2N2222 Diodes: 1N4007 1N4148 BA159 These are available from multiple distributors, i list the brand-names, you can also search just for the number eg. 3904 or 2222,


3

Firstly: current flows into the base, through the emitter. secondly, current flows through the collector, and out of the emitter. The total current through the emitter is that through the base plus that through the collector. You will need a datasheet to determine the exact voltage drop. Also bare in mind, however, that no two transistors are identical. ...


0

Tunnelling isn't to do with electrons passing across the bandgap. That's electron excitation. Let's take Silicon (because that's one material I can remember the figures for): bandgap is 1.1ev. The bandgap is the difference in energy levels between the bottom of the conduction band and the top of the valence band. So to move the electron from valence band ...


0

One application where you can only use one type is when you implement the BJT as a "parasitic" PNP in a standard CMOS process. In a typical CMOS process NPNs are impossible to implement but a useful PNP can be made in which the collector is implemented in the substrate, the base is an n-well, and the collector is a p diffusion within that n-well. For more ...


3

There have been times when transistors with specific characteristics were feasible in NPN, or PNP, but not vice-versa. Two examples : The 2N3055 power transistor was NPN, at a time when PNP transistors of the same power were simply not available (115W I think, later pushed to 150W). That led to the "output triple" - a combination of 3 transistors that ...


-1

for me a simple opto isolator will do. pic seg output(+)-- 1k---opto pin 1(diode anode) * gnd --------------- to opto pin 2(diode catode) * (+9v to +15v) ---1k-- to opto pin 3 * opto pin 4 to segment * repeat the above for all the 7 segments (A to G) + dot * common display catode ---- to npn collector (bc457) * pic mpx out --5k-- ...


1

Looks like a typo- here is the curve from the similar part number FJPF13007 And your datasheet: You may find it useful to refer to the datasheet of the original manufacturer- it was Motorola, spun off some years ago as ON Semi.


3

From page 3 of the datasheet, it appears the input resistance of the codec is between 15k-30k (nominal, depending on the gain setting). This is of the same order of magnitude as your JFET amplifier's output impedance, so the signal reduction is consistent with what you have observed. Regarding the clipping, the datasheet states that the input signal must be ...


0

You need to capacitively couple the pot wiper to the input to the CODEC with something like a 1uF capacitor (C2 in the below schematic) so that the proper bias (between AGND and AVDD can be maintained. The pot pulls the DC level close to AGND (the CODEC input probably looks like two ~20K resistors, one to AGND and one to AVDD).


0

check this link http://www.instructables.com/id/Digital-Logic-Gates-Just-Using-Transistors/ NOT gate OR gate AND gate


0

Looking at the datasheet and the maximum voltage ratings on page 6 the datasheet says this: Here the analogue supply voltage (AVDD) is specified between -0.3V and +3.63V. In the same table the voltage range for the analogue inputs should be between AGND -0.3V and AVDD +0.3V Which means that you have a signal headroom at a maximum of 3.66V which is much ...



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