Don't worry about it, the important points of the argument still stand. *note: This is a simplification, the required voltage depends on the resistor of choice and the current required by the relay. And even if you were to put 12 V on the input, with zero base resistance, you'd still have only 11.3 V across the relay, compared to 11.7~11.9 in the left circuit. The transistor will still amplify the current, so the microcontroller doesn't have to supply as much current as the relay requires, but the voltage becomes a limiting factor. But the microcontroller probably can't output 11 volts it can only do 5 V and 0 V (or 3.3 V and 0 V, or 1.8 V and 0 V, depending on its supply voltage). So the voltage at the base needs to be 10.7 V relative to ground, and your input needs to be more than about 11 volts. For decades the quasi-complementary output stage made sense but because PNP and NPN power transistors are now equally available and have more closely matched performance characteristics, modern audio power amplifiers often use equivalent topologies for both pairs: either 2 Darlingtons or 2 Sziklai pairs. Remember, for the relay to be turned on, we need it to have at least 10 volts across it, which means that node B is at least 10 V relative to ground. In the right circuit, we need at least 0.7 volts between the base and node B. Then the transistor will saturate, the voltage at node A will be around 0.1~0.3 V, and the relay switches on.* In the left circuit, if we want to switch the relay on, all we need is to put at least 0.7 volts between the base and ground, which we can do easily by putting essentially any voltage > 1 volt on the input node. Let's assume that to actuate the relay, we need to have a minimum of 10 V across it (a reasonable assumption for a 12 V relay). Simulate this circuit – Schematic created using CircuitLab I'm sure you can imagine flipping them over to the PNP equivalent if you want to. In that case, the second stage is simpler done using an NPN.Here's two circuits for you to look at. Depending on what is added to the silicon, it will be either N-type or P-type. A bipolar junction transistor is made up of three pieces of silicon. The practical result is the direction of current flow. With a PNP in the second stage, none of this is needed and you can just slam it right there, without any gate or emitter resistors.Īs suggested by, the same arguments hold when the first stage is PNP. NPN and PNP refer to the arrangement of the pieces that make up the transister. AC coupling) but for DC signals you would need to bias the second NPN in a way that its emitter never goes near GND. For AC signals, you can do the level shift with a capacitor (i.e. If you place another NPN after it, you need to level shift the voltage because the second NPN wants a low input voltage near its emitter (e.g. The reason is because of the different input voltage requirements of a PNP or NPN second stage, respectively: As stated before, the output voltage of the (e.g.) NPN common emitter first stage is rather high. Therefore, a PNP is often a better match for a follow up stage, because you can use very small (or no) emitter resistors with it. In particular, it cannot reach very close to Ground (assuming the input is somewhere midway between the supplies). The PNP case works because the voltage polarity across the series resistor matches up with the polarity required to make the PNP become active. If you use an NPN Common Emitter first stage, its output (collector) cannot drop below \$V_ - V_D\$, where \$V_D\approx0.6V\$ is a diode drop. Lets look at the situation for both PNP and NPN: simulate this circuit.
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