Puzzling JFET in power supply circuit - Page 1
Forum section: electronics / beginners

author: chipwitch
author_rank: Frequent Contributor
timestamp: April 02, 2013, 04:40:54 am
content: Hi All, I actually have three questions regarding the attached circuit I found in the Onsemi's LM350 datasheet.I went to Mouser's website to order the components. One JFET, 2N3822 (depletion mode, n-type), was not there. A quick search of other sources revealed this thing goes for 7 to 15 dollars. I figured there must be an alternative and, using Mouser's search, began looking for a JFET with similar specs. The Gate/source breakdown voltage is 50v, outside the range of Mouser's inventory. So, I went back to the circuit for a closer examination to see if the 50v. spec was necessary. I simply cannot understand the role of this JFET (2N3822, in the center of the circuit)! I'm hoping someone here will be kind enough to enlighten me. I just want to make sure I'm thinking right about this. The following only focuses on the variable current sub circuit. The second attachment gives more detail on the sub-circuit in question.1) The 2N3822 doesn't really indicate the Source and Drain. Since there is no difference in potential between the Gate and the negative JFET terminal, I assume that leg is the source?2) The two series diodes... Are they for flyback, assuming the pot is wire wound? Why two? I get the 1.4v drop, but why 1.4v?3) Correct me if where I'm wrong. JFET is a depletion mode. Let's call Vout 20v. For simplicity, let's call Vss ground or 0v. So, since the JFET is depletion mode, we basically have a clear path to ground from Vout through the double diode shunt/flyback? What am I missing? SOMETHING must be triggering the gate.4) The pot is 100 ohm. 20v. across it would leave .2A. That's 4W. Is that right?5) How can I choose an alternative to the JFET indicated on the schematic?I really appreciate your help!

author: c4757p
author_rank: Super Contributor
timestamp: April 02, 2013, 05:01:16 am
content: Those JFETs are just acting as current sources. Look them up, find IDSS, and then throw them out and replace with a different current source in that range.Quote from: chipwitch on April 02, 2013, 04:40:54 amMany JFETs are symmetric.QuoteThey're acting as very rough voltage references. 1.4V doesn't matter, it just has to be stable.QuoteWill there be 20V across it with those diodes there?Those circuits are in general rather archaic. Might want to just throw them out.Edit:I actually rather like the second one. Many of the simple LM350/LM317 CC circuits I see have the current going through the pot, this one avoids that. Just use a two-transistor current source, or if you do want to use a JFET, J113 is cheap.

author: chipwitch
author_rank: Frequent Contributor
timestamp: April 02, 2013, 12:22:03 pm
content: Hi c4757p. Thanks for the response. I know this stuff is pretty basic, but I just wasn't sure how to google these questions.Quote from: c4757p on April 02, 2013, 05:01:16 amI don't know why I was thinking the gate-source breakdown was important... something I read and misunderstood I guess. So, exactly what is limiting the current through the diodes (and consequently, the JFET)? Looks to me like 20v (assuming the output and Vss) to ground and infinite current. Intellectually, I know that's wrong, I just don't know why. I'm guessing the JFET is finding a state of equilibrium, neither fully open or closed? It's kind of like "picking yourself up by the bootstraps." The only thing to stop the channel flow is gate current which is provided anytime the gate current is zero? I know the model I use in my head is failing me at this point, I just don't know why.QuoteI did not know that. Cool!QuoteOh, right! If there's 1.4v across the diodes, then there HAS to be the same across the pot! That makes the power requirement for the pot very low (1.4v2/100ohm = 20mW).QuoteReally appreciate the tips! Don't really have anyone to ask this stuff. You're a big help. Thanks again.

author: c4757p
author_rank: Super Contributor
timestamp: April 02, 2013, 04:18:37 pm
content: Quote from: chipwitch on April 02, 2013, 12:22:03 pmWell, because it is! If you're looking for a replacement part you need to consider everything. Good on you for considering that. Of course, you don't have to match the maximum ratings of the chosen part, so much as the maximum conditions of the application. Still, some parts do have ridiculously low maximums, so you still should pay attention. As a current source, it's likely to have a large portion of the power supply across it.QuoteJFETs don't ever "fully" conduct. At VGS= 0, they conduct IDSS(property of the device, "given" in the datasheet - though it can sometimes vary over an entire order of magnitude). As VGSfalls (yes, negative), they conduct less and less, until they finally pinch off. VGSis almost never run positive because then the gate conducts current. There's no "full on", just "full off".

author: chipwitch
author_rank: Frequent Contributor
timestamp: April 03, 2013, 03:33:41 am
content: Argh! For a minute there, after reading your last post, I actually thought I was about to have an eureka moment... then I kept thinking about it and it eluded me.So, referring to the my second attachment in the OP, the channel is open until there is a voltage potential across the gate and one of the two other legs. Since Vss(assume -10v) and the gate are connected, the voltage potential would need to come from above. With 20v applied north of the 100 ohm resistor, we end up with some voltage at the JFET. This causes the gate to begin pinching off the channel, restricting current flow. This particular JFET will have a maximum Idssof 10 mA (I couldn't find the Idssfor the indicated FET so I used the one from the previous attachment, 2N3822). However, that's the stated current at 15v Vgs! How do you determine the maximum current since we have roughly 30v - 1.4v for the diodes?If we have 1.4v dropped across the resistor-diode ckt wouldn't that leave a voltage of 28.6v at the FET? The gate-source cutoff voltage is published at 6v. So, we should have a completely pinched channel, no current flow? Am I thinking correctly?Could a circuit be designed with a source voltage low enough to permit current in excess of Idss, destroying the JFET?I still can't see how this limits the current to the voltage regulator downstream.

author: c4757p
author_rank: Super Contributor
timestamp: April 03, 2013, 03:43:08 am
content: Sorry for a short reply - about to go to bed.The end of the channel that acts as the source is the one with a lower potential. Tying the gate to the source sets Vgs = 0 no matter what happens at the drain, so Ids = Idss. The only way for the drain to influence it is if it were to sink lower than the source and actually become the source. (Ignoring the fact that JFETs are somewhat poor current sources by modern standards and do exhibit a relatively large resistive effect.)More in the morning if nobody else jumps in.

author: c4757p
author_rank: Super Contributor
timestamp: April 03, 2013, 04:01:11 am
content: If you look at the datasheet for a JFET, find the Transfer Characteristics. These are the "curves" the analog old farts like to talk about.You will see that for a constant Vgs, the drain current starts to increase as Vds increases (like a resistor), but then it starts to level off. Idss is just the point it levels off to for Vgs = 0. The current will only vary slightly as drain voltage increases, so it can be considered roughly "constant". As long as you are in this "constant" region, thr current can be approximated to depend only on Vgs. Again, remember that in this circuit, the lower end of the channel always has the lower voltage, so it is always the source. The voltage at the drain will not cause pinch-off.