Timeline for Transistor biasing arrangement
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2023 at 20:16 | vote | accept | Infernoman | ||
| Mar 7, 2023 at 10:03 | comment | added | greybeard | Investigate the basics of linear power supplies - if nothing else, it helps you fix what you think to need. There is one place for SCRs in contemporary power supplies: as a crowbar device. Note that the configuration of the transistors is fundamentally different: in a "pseudo-SCR", each collector is connected to the base of the complementary transistor, while in a (complementary) Darlington there is one "input" transistor the base of which is used and an "output" one with collector and emitter used to interface with the rest of the world/circuit. | |
| Mar 7, 2023 at 9:41 | comment | added | Infernoman | Thanks for clarifying that for me. At least I’m slowly starting to understand things. While doing some searching I found a feedback based motor controller for universal motors which is very similar. - nutsvolts.com/magazine/article/scr_principles_and_circuits figure 16 is the one I’m talking about. With a bridge rectifier and 2 transistors configured as an scr. Would this cover all of my bases? | |
| Mar 7, 2023 at 9:12 | comment | added | greybeard | The overall circuit is a potentiometer dividing the supply voltage, followed by an emitter follower "sourcing" a load: the goal is not to keep transistor losses low - no way to reduce losses with "linear" circuitry -, but to have the output voltage close to the input voltage. There is no way to (forward)bias an Si-BJT's BE diode to 5 V without destroying it - you may be mixing this up with (enhancement-mode) (MOS)FETs. | |
| Mar 7, 2023 at 8:27 | comment | added | Infernoman | I could be misunderstanding how the circuit works as well. So forgive me for my ignorance but the vbe is the base voltage minus emitter voltage correct? And to switch the transistor fully on to minimize heat losses you need a voltage of +5v for an npn transistor. But both q1 and q3 range from 0.6-1.4v | |
| Mar 7, 2023 at 8:20 | comment | added | greybeard | @Infernoman: I'm afraid I don't get what you're after: Which transistor/\$V_{BE}\$ are you asking about? I picture the transistors used in continuous rather than in switch mode. | |
| Mar 7, 2023 at 8:03 | comment | added | Infernoman | This definitely seems like it will work from what I can see using circuit lab. It also works for both high and low impedance loads. The vbe for a low impedance load is ~0.6 volts and ~1.4v on q1 and q3 and 0.6v and 0.6v for a high impedance load. Is there any way to raise the vbe to switch the transistor fully on? | |
| S Mar 7, 2023 at 6:29 | history | answered | greybeard | CC BY-SA 4.0 | |
| S Mar 7, 2023 at 6:29 | history | made wiki | Post Made Community Wiki by greybeard |