# Tag Info

1

Do you have $20? You can solder them for$20. Or for free, if you already have a heat gun. Why do you think you'll over heat the package? I solder large power transistors, including TO-247 size power transistors and FETs to much more substantial copper than anything in your photo. Like a solid copper heatsink that weighs several pounds. I've yet to have ...

1

This is a horrible question, because it's not related to how we'd normally calculate the operating point of this circuit. It's so annoying that rather than try to teach you anything, I'm just going to give you (part of) the answer. I'm doing this because the question isn't well-posed to teach you anything anyway. You can get $I_C$ very simply as I_C = ...

0

If the boost converter is non-isolating i.e. relies on 0V being common to its input and output then using a MOSFET this way is probably not a good idea. Any current through the MOSFET (when it is on) will cause the 0V on the drain side to be noisy compared to the 0V on the source side. This might mean that some analogue functionality is upset or at least ...

2

R2 cannot be 0 ohm; I'm guessing this is a typo. When the Fet turns on it has an on resistance of .024 ohms, so you would essentially short to ground. The part you have will be fine, use the "gate threshold voltage" in the data sheet to choose parts that will switch at your logic level.

3

Your circuit will work at 5V too. It may flash a little faster and the LEDs may not be quite as bright but both those issues can be adjusted by tweaking the values of the 75K and 820 ohm resistors accordingly. I mention this because of the plentiful supply of cheap 5V USB device chargers around these days. One of these may be a good alternative to powering ...

3

As already stated, a modern 'smart' battery charger won´t make a good power supply. A smoothing capacitor across the output may work, but maybe you should consider some other options, especially for the future: You could: 1) Buy a couple of PP9 rechargable batteries to use with your projects - and so make proper/full use of your charger. 2) Buy a ...

-1

Using Battery Charger is not good way, since the charger only source 10% -20% of current requirement to charge the battery. for example if the battery is of 3V,100mAh then using constant current charging method which normally this kind of battery charger use source only 10mA of current to charge the battery. Also battery charger considers no-load ...

3

After your edit to mention that it works correctly on a 9V battery, I expect the problem is that your 9v charger has some "smarts" and only charges when it "thinks" it sees a 9V battery attached. The charger could be checking a few things to see if it should start the charge cycle. Existing voltage - the charger checks to see if there is a partially ...

0

I think the error is caused by the microphone - T1 combination. If no sound, T1 floats, right? If there is sound, the microphone will float, so T1 wont. Isn't it the problem, that sound is very high frequent and never static and your capacitor is a bit big for high frequences? I haven't made any calculations, just a thought..

0

"High-side switching" is the term you are looking for. Typically, this involves P-channel MOSFETS, although it can be done with N-channel MOSFETs as well. If you use a PMOS, then you can switch it easily from your 5V signal. However, an NMOS will require a boost circuit, as the gate voltage needs to exceed your supply voltage. Hence, a PMOS circuit (like ...

1

NPN or Logic Level N-Channel mosfet as a driver for a PNP transistor/P-Channel Mosfet, which controls the High side of the leds. simulate this circuit – Schematic created using CircuitLab

1

You can see that, in the first circuit, the drain-to-gate connection allows you to write an equation involving both the gate-source voltage as well as drain current through M1 (the second equation you wrote): this gives you space to choose proper values for Vgs, V1 and R1 in order to obtain the Id you want (remember that a current mirror serves the purpose ...

1

The connection is basic to how a current mirror works. The two gates are connected together so that both FETs are driven to the same level. This assumes well-matched FETs, as would be the case with them being next to each other on the same IC. The way the current mirror works is that the desired current is forced D-S thru M1. With its gate connected to ...

1

The amplitude you have observed (~0.8V, probably closer to 0.7 V) is given by your p-n junction between base and emitter. A p-n junction has a maximum voltage around that value because the voltage vs. direct current curve is a steep exponential - see the first quadrant of the figure below. To get amplitudes greater than 0.7 V across a p-n junction means ...

1

Where are you measuring this amplitude drop? If you connected the 555 output directly to the base of the transistor, and the transistor's emitter is grounded, this drop is expected, and you are stressing the base-emitter junction of the transistor. The base-emitter junction is effectively a diode and will hold the base voltage to about 0.7 volts above the ...

3

Yes, you can do that, you can even buy transistors with more closely defined hFE. For example, the 2SC1815Y is guaranteed to have an hFE within a 2:1 range (120~240). It might vary another 2:1 from -25°C to 100°C, so that's a 4:1 range total, plus perhaps 50% for Vce variation. In SMT, the BCX70x is even more tightly specified (< 2:1 at 2mA). It's ...

2

The J505 is still available I believe here; Link to RS. It's a JFET device with gate connected to source like this: - And has a VI response like this: - According to the data sheet it typically draws 1 mA and works across a range of voltages from about 1.2 volts: - All the way up to 100V

5

No, don't try to use the constancy of HFE to run your transistor, it varies with temperature, voltage, device, too many things. As you have 3.3v available, use emitter degeneration like this. This is a standard technique for biassing transistors and creating current sources. simulate this circuit – Schematic created using CircuitLab The two ...

0

As an aside, most electrical engineers would not use a bipolar junction transistor at all. If you use an N-channel MOSFET to switch in ground, all of these issues go away. You can place the resistor before or after the LED, it doesn't matter.

0

As a push-pull amplifier your circuit has two major errors. The input to Q2 is referenced to a voltage 5V lower than the output. I suspect your intention was that V3 would create a 'virtual earth' at +5V, in which case the input should also have been referenced to +5V. For simulation it may be easier to put the earthing point at V3's positive terminal, ...

0

Because R4 is connected to the base of Q2 -- when Q1 is off, it forms a load on Q1's collector (R1), so the voltage there won't rise above about 6 V. This doesn't really affect the performance of the NOT gate -- its threshold is about 0.8 V, so any input higher than this will be a '1'. Given that the transistors have a beta (gain) of well over 50, you ...

1

This might help: - So, look up what the transition frequency is and imagine a line falling at 6dB/octave crossing the point marked $f_T$. The horizontal/flat line (at lower frequencies) is the "DC" hFE and, where the 6dB/octave line crosses the horizontal line, that is the place where hFE starts to decrease with frequency. Basically this means that you ...

0

To me, the problem (question) is clear: You must discriminate between DC current gain B=Ic/Ib and the differential current gain "beta"=d(Ic)/d(Ib)=ic/ib. The DC current Ic is physically determined by the base-emitter voltage Vbe according to Shockleys exponential equation Ic=Ic,o[exp(Vbe/Vth) - 1] The corresponding DC base current Ib is just a small ...

2

The reason you haven't seen anything similar is that it's mostly pointless — and a waste of power. Ignoring the signal variations on V2 for a moment, the action of Q2, Q3 and Q4 is to maintain a constant voltage drop across — and therefore a constant current through — R1. Now, when the signal voltage rises, there is a need for load ...

0

Well looks like I solved it on my own. I used progressively increasing sized (W/L ratio) inverters of even numbers which acted as a buffer in between the VCO and divider. The full swing at the last stage was achieved because of the higher W/L ratio of the inverter at the last stage.

2

You are misunderstanding the purpose of R1. You seem to think it has something to do with supplying power to the probes in order to make the light turn on. Actually, R1's purpose is to make the light go off when the probes are not touching each other. When the probes are touched together, current flows from the battery (+) through the emitter to the base, ...

2

Assuming battery + is on the transistor/R1 connection, and battery - to the lamp/Probe 2 connection: By disconnecting the wire going to the 100K you are only removing the part of the circuit that tries to ensure the transistor stays off when it should. The 100K makes the circuit a bit less sensitive so that only a somewhat lower resistance short (between ...

1

An amplifier never creates power, it always limits it. The amount it is limited is the "gate door opening and closing". We call it an amplifier because the amount of electrons allowed through the door is much greater than the number of electrons controlling the door, so it is the control signal that is amplified, or appears to be amplified. For instance the ...

0

There is nothing unusual about this NPN transistor. Since you are reading a normal junction voltage on what you say is the base, it's unlikely (but not impossible) it's completely fried. Please make sure that the B C E order of pins is respected in your wiring (it's not E B C, which would give the symptoms you describe). Also note that the tab is common ...

1

Options: - Your transistor (a decent choice) is broken You haven't wired it up like your schematic Your power supply is not connected (check volts between 5V and emitter) Your meter is intermittantly faulty.

0

Your diode is correct. If it weren't, you would basically have a short circuit. The Battery Depending on the battery, you may have more than enough voltage to run the motor, but not enough voltage when the motor is at load. Be sure that you can run the motor on the battery alone! If you can't, then the rest won't help. Switching Device Your circuit ...

3

I do know the motor has a high current draw. I tried my device with a micro mini motor and even attached directly to the battery it wasn't powerful enough to turn the wheel it was geared to. [My emphasis above.] I suspect this is your problem. Adding a transistor does not allow any more current to go to the motor, than is possible by directly ...

2

As Bart suggested, I'm assuming by backwards, you mean the emitter and collector are switched. Yes current can flow in both directions. An NPN transistor backwards is also an NPN. There will still be a reverse beta, however, the backwards NPN transistor won't work as well as a correctly oriented one will. It's not recommended. You're thinking of diodes ...

7

A NPN transistor basically is a stack of three differently doped areas of a semiconductor. The first is N-doped, the middle P-doped and the last N-doped. So yes, on the first sight, you can swap collector and emitter, and the transistor might still work. But there's more magic: A real transistor is optimized to fulfill its specs when connected correctly, ...

1

Anything that increases thermal mass will help with cooling, including but not limited to thermally bonding the package to another metal mass with solder or thermal paste.

0

If you have added a .model statement into standard.bjt for the mmbt3904lt1g then in your schematic you should be able to right-click on the transistor and select "Pick New Transistor" and select the mmbt3904lt1g from the list.

1

If the model definition is a whole subcircuit (starts with .SUBCKT), you need to use the X prefix: Ctrl + Right click on the component, and set the prefix value to "X". Also check that the Value is set according to the name of the SUBCKT. Also check that you have an ".include" directive somewhere on your schematic, set with the correct path of the file. ...

2

The circuit makes no sense to me. You should have 5V connected to the PIR's Vcc, the transistor is never going to conduct since its base is always above ground, there's no current limit on the LED and it emitter is above the collector. Don't you want an NPN transistor with its base connected to the output, and the LED and a resistor in series between Vcc and ...

1

simulate this circuit – Schematic created using CircuitLab

0

I got it. Whilst I was looking around for the datasheet of the PIR, I wanted to look at the one of the transistor one more. I noticed that one of the drawings I found online gave me the C and E pins all the way around!! Once I simply rotated the transistor of 180 degrees (so what was connected to the collector is now connected to E and vice-versa), it ...

0

This schematic is fine, provided that the load doesn't exceed the nominal collector current of the transistor. Your microcontroller should be safe. And yes, you can measure the current consumed by your load if this info is missing. Connect it in series with an amperemeter to the nominal voltage (is it really 37 volts or did you just reuse someone else's ...

4

My question: "Under what conditions can extra current come into the base?" Whenever you exceed the specs of the transistor. If you break the transistor by connecting it to a higher voltage, current or power than it can handle, it can be damaged in unexpected ways. As a non-isolating device, it could in theory fail closed, connecting the collector to the ...

0

first thing find any number on the transistor this number is the key of transistors with the help of datasheet or the internet, ie you can find the type of transistor and the characteristics of it secend if you don't find any number on the transistor you can make this circuit to find the kind of the transistor, but befor that you shoult know the Transistor ...

0

Note that because of the marvel of modern, inexpensive microcontrollers, and clever open-source software, there are small, inexpensive pieces of "test gear" which will identify and measure a wide variety of transistors, NPN, PNP, BJT, FET, etc. and thyristors SCR, Triac, etc, resistors, capacitors, inductors, etc. There are perhaps two dozen slightly ...

-1

https://www.youtube.com/watch?v=2vk7JS-0ir0 use this vedio u can get full vision about testing the npn and pnp transistors

2

Either one will work. There are two differences. With the PNP transistors, as you noted, you get a one-Vbe drop. This may or may not be significant, depending on your supply voltage and the type of LED display (blue or white LED displays will obviously be worse than red, green or yellow displays). The NPN transistors avoid some of the drop (it will be ...

8

This is pretty easy. I used to do this routinely in high school when salvaging parts from unknown discarded boards. As Spehro said in a comment, sometimes you can find a part number. In that case, you can find the datasheet and get the parameters outright. However, all too often there is no manufacturer's part number, or its just a short code, or its a ...

1

Honestly it is better to type up the first line of numbers/letters you see on the front face of the transistor, add the word "datasheet" and search it up on google (and try to go for the PDF results). You will able to determine if it is PNP, NPN, some other transistor like a mosfet, or a miscellaneous chip. You can also know all their characteristics. If ...

2

From 555 datasheet (TI) During the timing cycle when the output is high, the further application of a trigger pulse will not effect the circuit so long as the trigger input is returned high at least 10 μs before the end of the timing interval. However the circuit can be reset during this time by the application of a negative pulse to the reset terminal ...

0

In SPICE modelling you use dependant sources to model ideal circuits, these can be anything from an ideal transistor as mentioned by others, to whole blocks that might eventually be implemented with op-amps etc. In fact you can also use dependant sources to model the non-idealities of a "real" circuit/device to gain insight, to act as a placeholder until ...

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