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66

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 ...


22

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 ...


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 ...


17

If you know the component is a BJT transistor, you can use a diode tester and test the pins pairwise. If you find two diodes with a common anode, and no conduction between the two cathodes, then you have a NPN transistor. If you find two diodes with a common cathode and no conduction between the two anodes, then you have a PNP transistor. Other options, if ...


15

Because both transistors are biased into their linear regions all the time, this circuit is not a "hard" clipper such as two inverse-parallel diodes, a transistor that goes into saturation or cutoff, an overdriven opamp, etc. This circuit has a softer clipping action, more like compression than clipping. This greatly reduces the odd harmonics in ...


13

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 ...


13

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. ...


11

"Reinforced diode". Diode connected BJT aka "active diode" is simply a transistor which collector is connected to the base. Thus the collector-emitter part of the transistor is connected in parallel to its base-emitter junction so we can think of this combination as of a "reinforced diode". The current through this "composed diode" is beta times bigger than ...


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:


10

A diode connected BJT has much better ideality factor than a regular diode and is used where close to ideal behaviour is required, such as in silicon temperature sensors. Many of these sensors operate by pulsing two different levels of current through a diode, but to be accurate the ideality factor must be close to 1 (i.e. it is as close as possible to the ...


10

It's not oscillating What you are seeing is ripple from C1 causing excessive variation of the current through D1 and varying its voltage. That voltage then directly affects the output voltage. You could make the filtering of the voltage to the zener more effective by splitting R1 into two resistors of 1k each and putting C2 at that point to ground. A filter ...


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 the ...


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 ...


8

Usually in very low leakage applications, e.g. high impedance input clamps to power rails, or log amplifiers, where leakage would disrupt the current to voltage relationship your hoping to make use of Downsides are usually higher capacitance and much lower reverse voltage breakdown, but if you have small signals they can beat some of the better diodes you ...


8

Even if you set the pin to high impedance, voltage on it must not exceed the maximum ratings in the datasheet, which is probably close to 3.3V. There could be an exception if the pin is labeled "5V tolerant". So if the base of your PNP is above that voltage, you'll need an extra transistor to control it.


8

The method dscribed by @user2233709 works to find the base and to determine if the device is PNP or NPN. Which is C (collector) and which is E (Emitter) and a very rough indication of Beta can be determined as follows. To differentiate between collector and emitter assume one of to possible terminals is E (Emitter), connect a 1M - 10M resistor from B (Base) ...


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

Zach, this circuit is pretty easy to understand if you understand the BJT first. (You will understand diodes, if you understand the BJT, so that's a given.) Everyone struggles with these things at some point, so it's fine you don't apprehend this well right now. Take it one step at a time. There is plenty of information on diodes here (and elsewhere.) You ...


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 ...


7

Generally speaking you should keep output voltage within Vdd or within some limit that is usually less than 6V. Otherwise the protection diode (when present, typically on non-5V tolerant pins) will conduct which will keep the PNP transistor from turning off, and could cause damage. I suggest using a dual pre-biased transistor to save PCB space, for example ...


7

Out of curiosity, I simulated this circuit using LTspice XVII. Using some guestimates for the bias voltage (I went for 4.5 V, half of the pot) and a 100 mV input signal at 1 kHz, I see the following (blue) output: Green: VSIG; 100 mV, 1 kHz sine wave. Blue: VOUT; output waveform. The schematic is a copy of yours with appropriate NXP transistor models. Part ...


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

From Vintage-Radio. Transistor polarity There are two different types of transistor - NPN and PNP. The circuits for the two are similar, however the polarity of the power supply for PNP types is the opposite way around to that for NPN types. With NPN types the emitter is negative, whereas with PNP types it is positive. To anyone familiar with valve ...


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