Hot answers tagged

65

This is a crude current source. The two diodes are to create a voltage drop of approximately 1.4v, which is then applied across the transistor VBE and R21. This gives approximately 0.7v on R21, or 10mA through it. Those two diodes could be replaced by a resistor to give the same voltage drop, but then if the supply voltage varied, the current would vary ...


20

Any amplifier that can be made with an NPN BJT can also be made with a PNP. Whether its inverting or not really depends on how the output is interpreted. More on that later. To convert this NPN common-emitter amplifier to a PNP common-emitter amplifier, just mirror the entire thing except for the supply voltages top-for-bottom: Equivalently, you can just ...


20

What is the role of the pnp transistor network? It's a differential voltage to current converter followed by a load (R34 and R35). The voltage between P+ and P- sets a voltage across R31. This (minus 0.7 volts) sets a voltage across R33 and that causes a current to flow out of the collector (largely irrespective of what load the collector has). Given the ...


18

Both diodes together produce a volt drop of about 1.4 volts, hence the base is at a voltage of (5 - 1.4) volts. This means the emitter is 0.7 volts higher (due to the base-emitter region being a forward-biased diode) at 4.3 volts. This forces 0.7 volts across R21 and means that a current of 0.7/68 amps (about 10 mA) flows through the emitter and then to the ...


14

You are trying to create a high-side switch. simulate this circuit – Schematic created using CircuitLab Figure 1. (a) NPN open collector switch. (b) A failed attempt to make a PNP version. Does any one can tell me is that the above configuration is correct and circuit can be use safely? I can tell you that the above configuration is not correct ...


12

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 ...


12

Figure 1. In this example Vss is greater than the 5 V supply of the micro-controller. The protection diodes keep the transistor always on. Figure 1 shows the internal schematic of a 5 V powered GPIO in "output" mode. A pair of transistor switches pulls the output high or low. (Only one can be turned on at a time.) Note the internal protection diodes. The ...


11

The basic idea is something like this: simulate this circuit – Schematic created using CircuitLab There are problems with this, though. For example, \$Q_3\$ might oscillate. Some added base resistance is a common fix. But there are other approaches. In this case, I don't think there's much likelihood, though. Just mentioning it, in case it matters. ...


10

You don't even need the transistor. A couple of resistors will do. If you pick 10 k\$\Omega\$ for R1 and 20 k\$\Omega\$ for R2 an input voltage of 5 V will be scaled down to 3.3 V out. In general: \$ V_{OUT} = \dfrac{R_2}{R_1 + R_2} V_{IN} \$ edit Apparently the 3.3 volt is enough for the Arduino Rx Olin rightly points this out. It may not be ...


10

In a very simple form, relative to VCC, think of the transistor as either coming before or after the device. If the transistor is connected between VCC and the device, it is sourcing current. If the transistor is connected between the device and ground, it is sinking current. (Image from CircuitsToday.com) Some articles that describe things in more ...


10

JYelton is right, and probably this is what whoever said "NPN transistors are sinking and PNP are sourcing devices" had in mind. But, that's not the only way to use a transistor. For example: simulate this circuit – Schematic created using CircuitLab This configuration is called common collector or emitter follower. Now the NPN is sourcing, and the ...


10

Change your load-driving PNP for a P-channel MOSFET. Also, you want a pull-up resistor on the gate of that MOSFET to ensure it switches itself off when the NPN switches off. The NPN should also really have a pull-down on its base to keep it off when its input is HiZ or disconnected. This is a circuit I use all over the place:


9

As far as I understood, you are trying to make some kind of a sound level detector, which will let you detect if there is a sound with a certain volume or not. You can do this with minor changes to the schematic you have. But before that, you should understand the circuit. Let's break that circuit down. First of all the part with the microphone. R1 is for ...


9

Overview Just to get the schematic into the editor supported here and to draw it in a slightly better form for analysis: simulate this circuit – Schematic created using CircuitLab I've dumped the switch (implied.) I've stopped busing power around and just used a ground and \$V_\text{CC}\$ indicator, instead. This helps to focus a little better on ...


8

Remove R5 and you will have what you describe. The configuration of Q5 is called common collector or emitter follower. Essentially, the voltage at the emitter is the voltage at the base minus 0.6V, but the emitter current can be much more than the base current, because the gain of the transistor will draw more current from the collector. Thus, it's a current ...


8

suspect ... not sourcing current, they are sinking Don't guess or go by heresay, READ THE DATASHEET. That will tell you outright how much current particular port pins can source. Most likely, most port pins can source some current. It doesn't take much current to be amplified by a transistor to drive a LED, so even for a rather weak current source ...


8

That's actually pretty easy. You have a schematic editor you could use here. I'll use it now. I've paired them, side by side, to match up with your pictures in your question: simulate this circuit – Schematic created using CircuitLab As you probably already know, \$R_1\$ is used to sink the needed base recombination current for \$Q_1\$, when it's ...


8

Another approach : a Schottky diode from Q1 base to collector; or possibly on both transistors (with the appropriate orientation). As each transistor starts to enter saturation, Vc falls below Vb and the Schottky diode becomes forward biased. This drains further base current preventing the transistor entering full saturation, which is a major cause of ...


7

If you have the exact NPN equivalent you only need to mount the circuit as it is drawn, so that the base, collector and emitter terminals of the NPN are connected where base, collector and emitter of the PNP should go, and then reverse the power supply polarity. This trick works in most cases, but it depends on the circuit we are speaking about. The exact ...


7

It doesn't matter what kind of transistor you use if the anode of the LED is tied to ground and you don't have a negative supply. "LED" stands for light emitting diode. To light a LED, you put forward current thru it, meaning in the direction so that the inherent diode is forward biased and conducts normally. The anode is then more positive than the ...


7

Yes you can - its the same device. Convention - print PNP voltages as negative, NPN types positive


7

In your schematic, the PNPs need to be on the high side, and NPNs on the low side, all in the appropriate direction (+ → -). But you are overthinking it; using the MCU's standard push-pull outputs (which you've almost reconstructed in your schematic) with opposite levels will light the LED assuming the output voltage is high enough for the LEDs, and you can ...


7

This can be made to work as an SCR type circuit. There are couple of things to consider though. One item, as pointed out by David Tweed, is the need to put bleeder resistors across the BE junctions of both transistors to keep leakage from instantly turning on the coupled PNP-NPN pair. Another item is that a fast turn on rate of the supply voltage can ...


7

You're using the PNP as an emitter follower so no wonder it adds a voltage to the bottom side of the waveform. What you need to do is use an NPN or an NMOS transistor. Then the polarity of the signal from opamp U1 (used as a comparator) is "upside down" but that is easy to fix: just swap the + and - inputs of the opamp. In the case that you use an NPN, do ...


7

There is a standard way to do this with a P-channel MOS transistor (PMOS). simulate this circuit – Schematic created using CircuitLab This is the standard way of doing it. You may or may not need to add C1. If the load has capacitance, then sometimes when you turn on M1, the 5V rail will suddenly dip, and that can cause problems for anything powered ...


7

This is a constant current source. The two diodes will drop a fairly constant voltage with respect to supply voltage and current. As a result, the voltage across R21 in series with the base-emitter voltage will have to equal about 1.4V. As the base-emitter junction is in essence a diode, it too will drop about 0.7 V (or at least a fairly constant voltage ...


6

NPN and PNP transistors are different. Electrons are more mobile than Holes Which means that PNP is not as good as NPN. For Si BJTs the PNP types are behind when it comes to breakdown voltage and really high power. For general purpose devices like BC337 / BC327 things for all intent and purpose are the same but if you wanted to do a off line SMPS it wouldnt ...


6

In most applications it doesn't matter. The first circuit will turn off the PNP a little faster, which would only matter in high speed switching, like if the load was being PWM controlled. Here is yet another topology to consider: This only works if V+ is about a Volt or more higher than the highest VCTRL can be, but can be useful when that is true. I'll ...


6

In the datasheet you will see the packages thermal chatacteristics: You can see that for the SOT-23 package, for every watt the junction temperature rises by 357°C above ambient. The maximum dissipation (without a heat sink) is also given as 350mW (for an ambient of 25°C) We can see why the 350mW is given: 357 * 0.35 = 124.95°C - the maximum ...


6

This circuit doesn't protect anything. The transistor only adds to the stress on the power supply by turning on, and providing one more path for current to flow. If the power supply is an ideal voltage source, then the transistor does not reduce the current which flows to the shorted output. The only way this circuit reduces the current is by adding to the ...


Only top voted, non community-wiki answers of a minimum length are eligible