0
\$\begingroup\$

The circuit diagram below shows two simple voltage regulators, one of them is driven by constant current source using a Jfet and the second one also providing a constant current because the voltage is constant.

  • What is the benefit of using a Jfet in a circuit like this? is it related to noise?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
6
  • \$\begingroup\$ Huh, where was this seen? I can't say I've ever seen the top circuit before, and the bottom circuit still leaves much to be desired. Which maybe gives some idea of the effectiveness of these circuits... \$\endgroup\$ Commented Jun 18, 2022 at 6:55
  • \$\begingroup\$ @TimWilliams I can't find it right now, but I remember seeing one using Jfet and another one using LM334. I had this question in back of my mind for a long time... The circuit is just an example, not a complete project. \$\endgroup\$ Commented Jun 18, 2022 at 6:57
  • \$\begingroup\$ The issues are: 1. high dropout reg in front (it's already a regulator, what's the rest for!?) forcing even higher dropout for the final output; 2. JFET is less accurate, more expensive and less available than BJT CCS; 3. op-amp doesn't need it anyway, and what's with the diode? 4. No current limiting (unless CCS * hFE counts... yeech!). 5. No compensation (not that an emitter follower necessarily needs it, but a good precaution). (Granted, the above may be a simplified version, so things like comp can be assumed.) \$\endgroup\$ Commented Jun 18, 2022 at 7:01
  • 1
    \$\begingroup\$ There are some people who like to use unusual parts even when it has no, or negative, value on circuit performance. Audiophile circles are very much this way. An audio amp needs very little performance to sound acceptable, so there is a LOT of freedom to make a poor circuit that still works. Power supplies are often the same way: low bandwidth, who cares about noise or precision. I don't know if what you saw came from such a source, but it's an example where you might see such things. \$\endgroup\$ Commented Jun 18, 2022 at 7:03
  • 2
    \$\begingroup\$ Ah, okay. Well, there are distinctive characteristics that JFETs offer, but they are all more or less irrelevant in this use case. So, basically nothing. \$\endgroup\$ Commented Jun 18, 2022 at 7:12

3 Answers 3

1
\$\begingroup\$

When using a JFET as a 1 mA constant current source, the maximum current that can be drawn into the TIP41 base is 1 mA and, given that the hFE of the transistor is only going to be around 50, the output current into RL is going to be limited to about 50 mA: -

enter image description here

I see or know no practical use of this circuit given the parts and values shown.


In circuit 2 (JFET and R3 replaced by R7), there is a possibility that more than 1 mA could enter the TIP41 base and so there may be a slight increase in the current that can be delivered to load RL under heavy loading (maybe up to 100 mA).


What is the benefit of using a Jfet in a circuit like this? is it related to noise?

I see no benefit and it can't be related to noise (because the closed-loop op-amp control will dictate noise) but, I didn't design the circuits and I don't know what the design aims were.

\$\endgroup\$
2
  • \$\begingroup\$ Does it make any sense if it was for an audio supply? \$\endgroup\$ Commented Jun 18, 2022 at 10:24
  • \$\begingroup\$ A audio power amplifier would not suit this design because of its limited capability to drive current. An audio pre-amp would be better suited to OTS linear regulators @ElectronSurf \$\endgroup\$
    – Andy aka
    Commented Jun 18, 2022 at 10:32
1
\$\begingroup\$

I agree with @TimWilliams that these circuits are a bit strange.

But in general low noise current sources can be made with JFETs, see here. However, in these circuits, this advantage would not materialize, because the noise will be dictated by the integrated regulator voltage.

When using a JFET for current regulation, you usually care less about precision and more about noise and you use it in saturation where it can reject quite a bit of supply voltage variation.

\$\endgroup\$
1
\$\begingroup\$

I suspect that the first circuit is intended to provide a constant current limit for low load resistances. Once the voltage on the load drops below Vref, the op amp is effectively disconnected. Then the current is going to be set by the 1 mA base current times the gain of the pass transistor, which will be more or less constant.

For the second circuit, once the output falls below Vref, the base current to the pass transistor will increase with falling voltage, since the base voltage will also drop and the voltage across the base resistor will increase. This will cause a considerable increase in load current for lower load resistances. In turn, this will increase the power dissipation in the pass transistor, which may or may not be a problem.

The use of the jfet is actually a relatively elegant way to produce a constant current limit, requiring a single added component to provide CCL, but the whole approach is wildly inaccurate. The current limit will depend on the gain of the pass transistor, and that is not well-controlled. Different pass transistors will produce different limit currents.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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