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I am trying to learn about transistor biasing, and can’t seem to wrap my head around it.

First of all I’ll explain what I’m trying to accomplish. I’m trying to build a linear 0-100 V power supply which will be capable of supplying loads with a varying impedance. I would like to be able to control the voltage to a motor or soldering iron which usually have quite a low impedance. I also intend to use this power supply for anodizing titanium, which has a relatively high impedance (~3-5 kΩ).

Ideally I would like to be able to control the power supply with a potentiometer, but I’ve blown up almost every single one I have now.

enter image description here

I’ve attached a photo of the circuit I currently have. Along with it you can see the spice analysis and transistor arrangement with a 0.7 Ω resistor as the load. The test is a DC sweep at V1 from 0-100 V with the potentiometer fully open.

While testing my circuit I found that the transistors aren’t fully saturated which would be why I’ve damaged so many. Also I do know that using a motor with this low an impedance would use a huge amount of current and requires multiple transistors in parallel to provide it.

Could someone please help me to understand how to correctly bias a transistor with a varying load impedance?

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    \$\begingroup\$ That's not how you design a power supply. Where's the feedback? Why is Q3 upside-down? A linear supply for a low-impedance load is usually not a good idea. \$\endgroup\$
    – Hearth
    Commented Mar 7, 2023 at 1:54
  • \$\begingroup\$ This is the first circuit I’ve tried to design and I didn’t go to school for this so I’m working from the basic knowledge I’ve found on Google. Originally I was trying to use a single npn configuration which the potentiometer didn’t have the current available to power the motor especially with the low gain since it wasn’t fully saturated. At least that’s what I’ve come to understand so far. \$\endgroup\$
    – Infernoman
    Commented Mar 7, 2023 at 1:59
  • \$\begingroup\$ Also from what I understand this could be considered a negative feedback common collector. Instead of the load being placed on the collector it is placed on the emmitter. Ref. - allaboutcircuits.com/textbook/semiconductors/chpt-4/feedback \$\endgroup\$
    – Infernoman
    Commented Mar 7, 2023 at 2:05
  • \$\begingroup\$ So I think the biggest issue I’m having is trying to build this with varying impedance. Currently the configuration I have “works” just not very well. But high a high impedance load in place of the 0.7 ohm resistor it greatly reduces the voltage and that would be why it doesn’t work. If I were to place a load resistor in parallel with the load. Would this fix the issue with the high impedance load not conducting enough to turn the transistors on? \$\endgroup\$
    – Infernoman
    Commented Mar 7, 2023 at 3:21
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    \$\begingroup\$ The emitter resistor doesn't really function as feedback in a circuit like this. What you need is something to sense the output voltage and compare it to some kind of stable voltage reference (a TL431 is a good choice for that); your conventional linear regulator circuit consists of an op amp driving a current gain stage, with feedback as described above. \$\endgroup\$
    – Hearth
    Commented Mar 7, 2023 at 3:52

1 Answer 1

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The fun arrangement is slightly different:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ 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? \$\endgroup\$
    – Infernoman
    Commented Mar 7, 2023 at 8:03
  • \$\begingroup\$ @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. \$\endgroup\$
    – greybeard
    Commented Mar 7, 2023 at 8:20
  • \$\begingroup\$ 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 \$\endgroup\$
    – Infernoman
    Commented Mar 7, 2023 at 8:27
  • \$\begingroup\$ 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. \$\endgroup\$
    – greybeard
    Commented Mar 7, 2023 at 9:12
  • \$\begingroup\$ 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? \$\endgroup\$
    – Infernoman
    Commented Mar 7, 2023 at 9:41

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