# Tag Info

4

I am 99% sure it is a temperature sensor. App Note AN12.14 from Microchip can tell you a little bit about it. (This is a pretty common cheap way to measure temperature on a circuit board)

0

The answer given by @OKHS seems to be correct with the given assumptions. Just for the sake of completeness, please note that when there is a resistance $\mathrm{R_s}$ seen at the course of an NMOS transistor, the output impedance is (at low frequencies) $$\mathrm{R_{out}=R_s+r_{ds}+g_m r_{ds} R_s}$$ The derivation is ...

0

Go here https://www.youtube.com/watch?v=iNYeww1Sjzk You can use two opto isolators to switch equally hard between High and low

1

Take a look at how the MOSFET conducts. Essentially, as you add the gate voltage, you are accumulating more and more holes to aid in the conduction. Less voltage means less holes, but never 0. This is important in that there will always be SOME conduction, but the construction of the MOSFET plays a huge role in how much is allowed through. Subsequently, ...

2

Hints If any one of these steps makes you scratch your head then ask What is the voltage at the emitter (broadly speaking) Knowing Ve, what is current through 10k resistor Knowing hFe or beta, what is current thru collector What is voltage at collector These are fairly straightforward but take your time.

3

You have in and out the wrong way round. It's meant to be a colpitts oscillator.

1

Yes - there is another (intuitive) approach based on feedback theory. As we know - the resistor Rs (I am using a large symbol because it is an ohmic resistive part) provides negative current-controlled voltage feedback. According to feedback theory, the output resistance in this case is always increased by the factor (1-loop gain LG). For the given circuit ...

0

You're basically running 100x more current in Q2 than Q1. You can do this with just diodes, and BJTs with collector=base act as diodes. So, remove RB2, short RB1. You'll get somewhat higher current used from the supply. Now, your scope measurements are a little difficult, and there is basically a large offset that needs to be removed. So, exchange the ...

3

They might be MJW16060A NPN high-voltage transistors, the packaging matches. The 9926 and 9915 are date codes most likely.

1

Yes, Silicon FETs can operate 'easily' down to around 77 K (-196 C), although the characteristics will (obviously) change -- some (e.g. on-resistance) for the better. Threshold voltages will be significantly (~400 mV) higher. Diffusion processes don't stop at cold temperatures -- but the doping (which makes it n- pr p-type) freezes out which basically means ...

-1

At 0 Kelvin: everything stops so no, a MOSFET will not work. Above 0 Kelvin: it will probably work. Also see the datasheet of your favourite MOSFET, the manufacturer will list the minimum (guaranteed) operating temperature. That does not mean the MOSFET will not work, it's just not guaranteed (and not tested) to work.

2

Well if you take one example, say a IRF740 power MOSFET. Look at the datasheet it will give you a maximum operating temperature, and a storage temperature. The low end of the storage temperature is -65deg. Of course this would be a starting point. As soon as you start operation and switch a load through the FET, current will flow and inevitably cause the ...

4

In the base of a transistor there is obviously more then a simple resistance. The so called voltage drop of 0.7V is due to the diode like action within the Base to Emitter connection. In a partial approximation the Base to Emitter connection can be viewed as a small 0.7V battery and a resistor in series. This small battery opposes the voltage that is ...

1

I think, to answer your question it is - as a first step - important to know the difference between linear and non-linear resistances. For example, the base-emitter resistance of a BJT is strongly non-linear. In such a case, we always must discriminate between (a) the STATIC resistance Rbe (as a ratio of DC values only) and (b) the DYNAMIC (differential) ...

5

This is because of what you might call 'differential impedance': the change in current per unit change in voltage. A resistor's differential impedance is linear: for every extra volt across it, a fixed number of additional amps flow through it, determined by that resistor's resistance value. Semiconductors don't exhibit linear differential impedance; a unit ...

0

Cannot find as DExxxx, but these may be equivalent parts from other manufacturers: Dual N-Channel 2.5V Specified PowerTrenchÒ MOSFET https://www.fairchildsemi.com/datasheets/FD/FDS9926A.pdf N-CHANNEL ENHANCEMENT-MODE. POWER MOSFET http://www.siliconstandard.com/Documents/SSM9915H,J.pdf If the descriptions don't seem to match what you have then verify that ...

0

If you've got a switched 12 volt source, then you only need 277 ohms in series with the squib to make the simulation work:

1

In general (there are exceptions for fast changing signals where the lumped element model breaks down) components only care about the relative voltages between their terminals. Given the same gate source voltage and the same load you should see the same drain source voltage in both cases. The problem is that generating that gate source voltage just got much ...

3

Assuming you are querying if there is any appreciable difference in the MOSFET's characteristics when comparing sinking current or sourcing current to a load: simulate this circuit – Schematic created using CircuitLab As long as you drive the gate of the MOSFET with an equal Vgs (risetime and source capability if PWM'ing) then no, there is no ...

0

Where are you connecting the audio jack? Depending on that, you need to place a potentiometer (variable resistor) and a resistor and connect it to the Base (Pin 1) of the TIP31. Try this. It works for me.

1

Your assumptions seem correct. However, just make sure: 1) The voltage of the power supply is the same on the input and output of the optocoupler, it could be different and could cause trouble. (I'm assuming it is based on your drawing) 2) The resistors you use, limit the current accordingly. 3) You can go for the first option, but you need to place a ...

0

maybe hfe is too low to turn leds on, try using a tip142, it has a hfe of 1000, so leds may turn on even with low volume. hfe on bd241 is 10, tip31 is around 30, tip142 would be too high but if the led light, you just need a trimpot between p2 jack and tip142 to regulate gain. remember to put a good heatsink on tip142.

1

Consider something like a 74HC74 flip-flop. For this purpose you can ground the clock (CP) input and D input or use it for reset purposes. The FF powers up in an indeterminate state. It can be reset to Q low by bringing RD low momentarily, or a low-to-high transition on CP with D held low. Bringing SD low will cause Q to go high and it will stay high ...

1

I think this is the circuit you are asking about because applying an external short circuit is the only sure way to force the c-e voltage to 0 V. simulate this circuit – Schematic created using CircuitLab In this circuit, the collector current is not necessarily 0. If the base voltage, supplied by V1, goes negative by more than a few 100 mV, ...

5

What you're looking for is called an SR latch. It's not a transistor - though it can be constructed from discrete transistors - and it's available in IC form, for example the CD4043/CD4044. The latch has two input pins, (S)et and (R)eset. A high level on S turns the latch on, and a high level on R turns the latch off again. More generally, if you're ...

1

Zero volts across collector and emitter can only mean zero current into the collector. It's basic ohms law. There may indeed be base emitter current flowing but that current flows out thru the emitter (NPN) and not the collector. If collector emitter voltage is held at zero volts and the base voltage is significant then the collector base junction is ...

1

It looks like you are smoothing the output rather aggressively with that 47 microfarad capacitor. The IS1U621 is a high frequency remote control receiver, designed to receive pulses at a frequency around 38kHz. That capacitor is effectively a short circuit at that frequency. Try removing it, or better yet, connecting it's negative end to the negative supply ...

1

First of all, you said that for a device that is twice the size of the original device, then the current is twice the amount for the same voltage. That is simply a definition of your choosing. (For example, for many things, twice as big would be defined as twice the mass.) Given that as the definition: $i_1 = f(v)$ for device1 if device2 is twice the ...

1

R5 controls the current gain through the transistor path, which allows you to control the current through the 8 ohm resistor at the end via bjt gate current equations. If you were say, going to QUT and taking enb245. It would have been useful because the peizo speaker you have connected as that 8ohm resistor has a 250mW rating, so you can use R5 to limit ...

2

Consider a case ,say S1 is closed and S2 is open .Transistors T2 and T4 are OFF and virtually absent from the circuit . The base of T3 will be pulled to 3.3 V so the transistor will be in hard saturation. Thus, the output at its collector will be Vce(saturation) i.e. 0.3 V above ground and the transistor will conduct the maximum current .This output is the ...

3

If you want to make an all-NPN H-bridge, the schematic and associated logic is as follows: Do not use these power transistors straight from a uC though. You will not be able to saturate them and they will likely burn. Lower power transistors will work though, beta in saturation is about 10. Also this circuit doesn't show any flyback diodes, so it's just ...

1

For typical uses (very generally 10 - 100 V; 1-10 A, < 10 MHz), the physics of silicon mean that the capacitance of the FET structures (and parasitics associated with them) have values which have a more significant circuit effect than the inductance (generally associated with bonding wires and the package structure). However, at high frequencies ...

0

The parasitic inductance is the sum of the bonding wire and the PCB traces .So its really a function of package type and board layout.Most of the time the parasitic capacitance is more significant .When you compare the energy stored in C to L you find that C is much worse.This is why for powermos its generally more rewarding to implement ZVS rather than ZCS ...

1

The structure of the MOSFET accounts for variety of capacitances which include junction capacitances,sidewall capacitances which come into picture at high frequencies to limit frequency response. Talking about the inductances,to operate, the MOSFET must be connected to the external circuit, most of the time using wire bonding These connections exhibit a ...

6

Googling SOT23 marking K1P the first hit links to the MMBT2222 datasheet which contains the following text ( on page 2 in a section called"Marking Information" ). K1P = Product Type Marking Code YM = Date Code Marking and a further table decoding the "ym" date code. The table only goes up to W (=2009) but it's reasonable to guess that Z4 means April ...

1

You are getting a very low value of Beta because your transistor is working in 'saturation region. Check out that Vce = 83.85mV which is much less than the rated 300mV of 2N2222. Therefore it is obvious that Beta will be less than the expected one (∼50). It is changing with the resistances. If you consider that Vce ∼ 0.2V, Ic ∼ (Vcc - Vce)/(Rc + Re) ...

1

The transistor is in saturation, not the forward-active region. You can tell because $V_{\text{CE}} \approx 84\text{mV}$; when $V_{\text{CE}} < 0.2\text{V}$ or so, the transistor is in saturation. $\beta$ is high in the forward-active region but is lower when in saturation. To get the transistor in the forward-active region, you need to lower the ...

4

One major parameter which decide biasing in BJT transistors is Bias Stability. As β (hFE) widely varies from transistor to transistor. An stable biasing will provide minimum alteration in the Q-point on wide changes in β. Mathematically stability factor is denoted by, S= delta Ic / delta Icb. S depends on the circuit configuration and the bias resistors. ...

3

Q1 is a P-channel MOSFET used to connect/disconnect the PV from the circuitry on the downstream side of its [Q1's] drain. R3 functions to pull Q1's gate up to its source, turning Q1 OFF when T1 is turned off by Cde being low, and T1 is used to pull Q1's gate down to ground, turning Q1 ON, when T1 is turned ON by Cde being driven high.

1

These are fabricated transistors on a silicon wafer and are not "known" by their indivual characteristics because they are not testable or provable as individual entities. You might find some information from a manufacturer about these things but generally I think you may be searching for a needle in a haystack.

2

You have basically answered your own question: they set the default voltage to a value so yhat you don't get a floating high or low value. This is especially important in TTL which needs to have defined inputs to have defined output(s). If there is nothing driving an input, or what would typically drive that input is in a high impedance (tri-state) ...

1

This won't work. FETs don't conduct current the same way as BJTs (NPNs, PNPs) do. They are majority carrier devices, which basically means that there isn't a current flowing across a PN junction. Even if you could, GaAs transistors are not engineered to be efficient light emitters, and wouldn't emit in the visible region. You wouldn't see anything. PN ...

3

Found a great animation here: http://www.falstad.com/circuit/ (then choose from the menu: circuits/transistors/multivibrators/astable multivib

0

It is basically an RC circuit. The capacitor is loaded through the resistor and the transistor works as a key. (Saturation and Cut-off modes) When the capacitor is fully loaded, the voltage is enough to sature the transistor and turn on the LEDs. When it is discharged it cant sature the transistor and it enters in cut off mode, so the LEDs are off. The ON ...

1

Collectively the network of Q2, Q3, Q4, Q5, R4, R6, R7, R8, R22, C4 and D2 are what you would call a gate-drive circuit. The purpose of such a circuit is evident in its name; in this case it is to switch the PFET Q1 by controlling the charging and discharging of its gate-source junction. Both Q2/Q3 and Q4/Q5 are wired as a Sziklai Pair in order to achieve a ...

3

I'm not sure, but it seems they are used to switch off Q1 quickly. Follow my reasoning and see if it makes sense to you. First of all you should consult the datasheet for the controller chip BQ2031. It describes the chip operations and tells that its MOD pin is the PWM output that allows to control the charging cycle through (ultimately) Q1. At page 10 ...

6

As @FakeMoustache hinted in a comment to your question, the explanation lies in the behavior of a reverse-biased PN junction, because that's what Q1's collector-base junction is in your circuit. From a macroscopic point of view any reverse-biased PN junction acts like a parallel-plate capacitor whose capacitance (called transition capacitance $C_T$) ...

0

Are you sure that you provided all information on this circuit? As far as I can see, this device only produces some leakage current. There is no chance for a current flow. Either Q3 will do something, nor Q1. The only possible current might flow in Q2 form the Emitter to the base but there's still no way to continue. Even a timing analysis might not ...

2

Both the question and configuration seems to be ..broken, somehow, but making a few assumptions/changes, i'll try to illustrate how i would go about solving a similar task. I'm guessing that my assumed configuration is not the intended one, but i can't see any obvious alternative. So: Assuming all transistors are NPN (and changing parameters for the ...

1

The only reason to further isolate mcu and relay coil is if you have yet a different isolated PSU that drives the coil, otherwise I see no explanation. Items are sold on ebay and people wire all kind of stuff together, therefore I presume the sellers introduced this feature to prevent MCU boards to burn. Of course the singnal from MCU needs to be buffered ...

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