![]() It might be instructive for you to do some experiments with an LED, a resistor, a battery, and a multimeter.Īn LED is a diode. If the 5v logic output must drive a significant load, that 3.3v logic source must sink its current too. If you lower them, the poor 3.3v logic source must work even harder. It won't be fast, because their values are high. This circuit has wide latitude in choosing these values. That forces the transistor "off".Īnswer to your question (4):Depends on what you're driving. There can be no voltage between base and emitter. That "logic 0 input" is working hard to pull transistor's emitter down to 0v.Īnswer to your question (3): Run Kirchoff's voltage law around the base-emitter loop of the schematic (right). The 3.3v must supply current through the 2.2k resistor, since the transistor base is being dragged down by the logic source providing "logic 0 input".Īnswer to your question (2): Yes, a clash. The collector resistor is left on its own to pull the output up to +5v.Īnswer to your question (1): Base is at 0.65v. ![]() Since it is off, no collector current flows. The base-emitter junction of the transistor has no voltage drop as before, so no base current can flow through the 2.2K resistor. In this case base and emitter are kept at the same voltage. On the right show the case where the logic input is in its "high state", up at 3.3v. ![]() Simulate this circuit – Schematic created using CircuitLab That current must flow through the transistor, and the input driver must sink this too. Not shown is the current that might flow from the 5v logic output. Many saturated transistor circuits allow collector current to be ten or twenty times larger than base current, but not in this case - no current gain is required of the transistor (actually less than one). It is so saturated that collector voltage actually falls below the base voltage, and provides a logic low at the output. The current ensures that the transistor is heavily saturated. Considerable base current flows (about 1.2 mA). The transistor's base has no choice but to follow, and is about 0.65v higher. The transistor's emitter is pulled to zero volts by that logic source. The logic source that provides the logic low (on the left) will have to work at it, because it must sink current from both the 3.3v supply, and from the 5v supply. The case for a logic low input (0v) is on the left, the case for logic high input (+3.3v) is on the right. I have re-drawn your schematic, because I like to see more "positive voltage" supplies at the top, "less-positive" voltage supplies lower down, and ground at the bottom. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |