1
\$\begingroup\$

I think I have designed a current regulator. This is the circuit diagram:

schematic

simulate this circuit – Schematic created using CircuitLab

The operation is very simple. Let's say the transistor turns on at a voltage Vo.

For 0<V1<Vo if we increase V1 a little bit according to Ohm's law the current in the overall current will be increased but the current going through the transistor will be increased as well (a little bit) and according to KCL the current through RL will remain costant.

The bigger the value of RL, the better it works. Also the more heavily doped the transistor the better it works.

However if V1>Vo the transistor becomes almost a short circuit and current through RL goes to 0. That's why I need transistors with on voltage somewhat bigger than 0.7 (2.5V would be okay) in order for this current regulator to have a practical voltage range of operations. Where can I find these transistors?

Where can I find transistors with on voltage > 0.7V?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ The two legs are in parallel so the current in one is independent of the current in the other. The current in one leg is the same, regardless of what is going on in the other leg, if it is even there at all. \$\endgroup\$ – DKNguyen Jul 19 '20 at 5:07
  • 3
    \$\begingroup\$ There's no regulation, current or voltage. Your source voltage is always directly applied to the load. So the load always experiences the current determined by the source voltage divided by the load resistance. Assuming your source voltage approximates an ideal source, anyway. \$\endgroup\$ – jonk Jul 19 '20 at 5:07
3
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's circuit redrawn in conventional layout. (Current flow from top to bottom of schematic.)

I think I have designed a current regulator.

If your voltage source is a true constant voltage source then you haven't. The current through RL will be V1/RL no matter what Q1 is doing.

You may be confusing your invention with a shunt regulator.

enter image description here

Figure 2. A shunt regulator block diagram. Source: Electronics Notes

The shunt regulator or shunt voltage regulator is a form of voltage regulator where the regulating element shunts the current to ground.

The shunt regulator operates by maintaining a constant voltage across its terminals and it takes up the surplus current to maintain the voltage across the load.

Source: Electronics Notes

The classic Zener-diode + resistor is one form of shunt regulator but note that it relies on the series resistance. Without that your circuit does nothing but waste energy and, probably, burn out transistors.


Your question title refers to a "NOT gate" and you have tagged the post as "digital logic" and "constant-current". Is is none of these.

  • It's not "logic" because it's an analog circuit. While the circuit may remind you of a simple BJT NOT gate it is not working as digital logic in your design and it is missing the pull-up / load resistor.
  • It's not "constant-current" because it doesn't provide any constant-current regulation. As discussed already the load current depends purely on V1 and Q1 doesn't change it.
\$\endgroup\$
0
\$\begingroup\$

This is an interesting idea, but I'm afraid it will need some refining, even beyond transistor Vbe (Vo).

Let's start with Vbe, though. Unfortunately here, Vbe is a property of the semiconductor materials used, which means we can't set it arbitrarily. The 0.7V value you quoted wasn't some arbitrary standard we chose; it's what you'll find on any silicon BJT. Germanium transistors - the other popular variety of which I'm aware - have Vbe even lower. If someone knows of significantly higher Vbe's, please let me know in the comments so I can add them here for completeness.

There's a more fundamental problem, however. Here in your idealized circuit, RL is in parallel with your transistor, and V1 connects to both sides of RL. This means that your transistor will have no effect, whatsoever, on what RL experiences. To see this, just apply Ohm's law for RL. The current IL = V1/RL, no matter what's going on in other parallel branches.

Your transistor, however, will indeed draw more current as V1 increases. Assuming a beta of 100, your current into the BJT collector will be 100*((V1-0.7)/(1kOhm)). This will heat up the transistor, which causes it to draw more current (since it's a BJT). I'm not sure if the conditions are right here for thermal runaway, but I'd definitely want to think about it pretty hard before I wired it up to a car battery. And yes, as you noted, it will hit a point where the transistor is drawing more current than it's capable of, and the air will be filled with the lovely aroma of magic blue smoke.

But hey, that's the smell of learning! I've practically burned enough transistors that I could have recycled them into my diploma. :)

Note: There is a circuit, conceptually similar to yours, that could work.

schematic

simulate this circuit – Schematic created using CircuitLab

The Zener Diode will drain any voltage above its set voltage, keeping the voltage across RL from exceeding that set point. Of course, by Ohm's law, this means that the current through RL is also limited. It's not the most energy-efficient way to accomplish this, but variations on this theme are common where price and/or size constraints are more important.

The Zener diode doesn't burn, since R1 is in place to limit the total possible current through V1 to I1 = V1/R1. That would hold even if RL or the voltage regulator failed short.

There are also ways to do this with voltage regulators, but as your question wasn't really about alternative ways of accomplishing this feat, I'll stop here for now.

\$\endgroup\$
10
  • 1
    \$\begingroup\$ Your voltage regulator has its output and ground pins shorted together. That's not likely to work well. \$\endgroup\$ – The Photon Jul 19 '20 at 5:43
  • \$\begingroup\$ it works better if you put the R1 between out and GND and put the RL where R1 is. \$\endgroup\$ – Jasen Jul 19 '20 at 5:55
  • \$\begingroup\$ Thanks; it's been while since I set up a circuit like this. I'll remove it for now to avoid confusion. \$\endgroup\$ – Wayfaring Stranger Jul 19 '20 at 6:05
  • \$\begingroup\$ @ The Photon - Oh, duh, you're absolutely right! I was treating it like a Zener diode. Thanks for pointing that out. \$\endgroup\$ – Wayfaring Stranger Jul 19 '20 at 6:11
  • 1
    \$\begingroup\$ @Helena, that's not correct. It's a shunt regulator but it regulates voltage to the load provided the input voltage is high enough. \$\endgroup\$ – Transistor Jul 19 '20 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.