Basic current limit - uh... why won't it work? - Page 1
Forum section: electronics / projects

author: c4757p
author_rank: Super Contributor
timestamp: January 22, 2014, 09:00:28 pm
content: The local regulator in my PSU kit was very easy to blow up while probing around, so I figured I'd add a simple current limit to it.R4 senses current, R5 blocks it from Q4's B-E path, Q4 stops the current through D1 when the limit (about 18mA) is reached, and R6 provides a discharge path for C1 (time constant = (10k+3k3)(47uF) = 625ms).In real life it doesn't work. If I short the output, the bench PSU's (not the DUT's) current limit kicks in at its setpoint of 400mA, with about 200mA flowing through the load (God knows where the other 200mA is), and the magic smoke comes out of Q3 in about three seconds, leaving it shorted C-E. R4 is also crispy, though still measures 39 ohms...What am I missing?All I could think of was Q4's VCE(sat), not fully shutting down Q1, but that has a maximum of 0.3V (and I've tried replacing it in case it was bad).

author: megajocke
author_rank: Regular Contributor
timestamp: January 22, 2014, 09:48:53 pm
content: I'd suspect Q3 gets destroyed as soon as the short is applied and that 47 uF cap discharges into its BE-junction.The simulator is probably not understanding what time step is appropriate across the load resistance step and underestimating the peak current for that reason. Try forcing a minimum timestep of 1 us or so or use something with a defined switchover speed as a load.

author: rrpilot
author_rank: Newbie
timestamp: January 22, 2014, 10:21:13 pm
content: This seems over complicated for what you're trying to accomplish. Have you taken a look at the circuits suggested inhttp://en.wikipedia.org/wiki/Current_limiting?Something to keep in mind with this sort of circuit is the power dissipation under a short condition.Rough numbers, you have almost 37 V across the pass transistor with 18 mA running through (if the current limiting is working), that's over 600 mW. Are you using a little TO92 or SOT23? Probably need something bigger to withstand the heat for any length of time.

author: c4757p
author_rank: Super Contributor
timestamp: January 22, 2014, 10:39:25 pm
content: Quote from: rrpilot on January 22, 2014, 10:21:13 pmIt's one transistor and a current sense resistor... All the rest is the regulator itself, and I can't make it any simpler unless I really don't like line rejection.QuoteIt's a TO-92. Will be a 1W rated PZT2907A, but they're expensive and I'm sick of killing them, so I'm using PN2907A instead right now.In fact, I might change it to a PN2907A just because they are cheaper, even though it does look a little bodgey to have one PTH transistor on an SMD board.Quote from: megajocke on January 22, 2014, 09:48:53 pmYou're probably right. I completely missed that C1 will dump its charge through the transistor.I'll try adding a series base resistor.It's always something simple, isn't it?Now, on to the next problem: wanna help me find something to blame it on? Too much coffee, not enough coffee, flux fumes... I don't want to admit that I made that mistakeEdit: I don't think that's it, actually, because I was able to very briefly short it out without damage, and still the limit did not work.I can see damage accumulating with each charge dump, but then wouldn't the current limit work until the damage has been done?

author: sync
author_rank: Frequent Contributor
timestamp: January 22, 2014, 10:59:29 pm
content: While C1 is discharging it will provide a base current for Q3 -> high collector current. C1 is bad.

author: c4757p
author_rank: Super Contributor
timestamp: January 22, 2014, 11:16:45 pm
content: I may restructure it to actually pull Q3's base directly up, but that will require another transistor. I was trying to keep it simple.

author: sync
author_rank: Frequent Contributor
timestamp: January 22, 2014, 11:46:01 pm
content: What's about a classic HP reference supply regulator. It provides a regulated supply voltage and a reference voltage. Adding a current limit should be easy.Screen shot is from a HP6113A manual. But HP uses this kind of circuit in many power supplys.

author: C
author_rank: Super Contributor
timestamp: January 23, 2014, 12:09:54 am
content: R6 is not good newsC1 will discharge with out R6The zener D1 will discharge C1 as will Q3A shunt across a zener hides the zener's current changeso by adding R6 you made the very sharp current change of D1 in to mush.Q1 now has to handle more current due to R6If you need more change on C1 in the discharge direction just increase your operating point a little higher on the curve of D1With out R6 how large does C1 need to be?A split R7 would let you control the ratio of D1 charge/discharge of C1 vs Q3 discharge of C1.Edit: think I did not say above good, so no changes above just add below.A voltage regulator with a current limit is an or like function.For a Voltage regulator you want a fast change around it set point. R6 changes what was fast in to slow.All these transistors and none are working at making the voltage regulation better of even as good as before with just Q3.The function that R6 could provide is a faster change in the off direction but this would be better if the fast off only happened when over in voltage OR current. A constant fast off means more work to stay on all the time.Killing the current to D1 will drop the output but this change will slow as you hit the more vertical part of the D1 voltage curve.In the on direction you could use an AND. Voltage & Current are less than set point add More output.In the off direction Voltage or Current is at limit, No more output.C

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 12:17:08 am
content: R6 was to "help" C1 fully discharge if Q1 didn't quite shut off all the way. I added it on a hunch and will of course remove it if not absolutely necessary, even just to keep the part count down.R6 shouldn't affect the needed size of C1, it will stay the same.I'm thinking a resistor between C1 and Q3's base will probably fix it. I'll follow up later when I get a chance to try.

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 02:54:45 pm
content: OK, I really like this one in theory. I'll see how it works in practice in a few minutes.Differential pair regulation should improve the output impedance, though I'm not going to use another matched pair here. I'm nottooconcerned with output impedance as the load will be roughly constant.The Zener diode no longer needs a current source (and no, a resistor to V- isn't good enough, it needs constant current for line rejection) - by dropping the Zener voltage I was able to run it off its own output. R9 supplies just enough current to bootstrap it, but is swamped by R1 when the regulator starts up.R1 also pulls the reference voltage to ground if there is a short circuit, giving a sort of foldback protection.Still only four transistors.

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 04:01:30 pm
content: Works an absolute treat, including the foldbackEdit - plus a 3.3nF capacitor across R3.Edit 2 - I just realized how similar it is to that HP schematic, which I hadn't even noticed until just now...

author: sync
author_rank: Frequent Contributor
timestamp: January 23, 2014, 04:35:00 pm
content: Congratulation! What is the purpose of R6 and C1? Slow start?

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 04:38:54 pm
content: LPF. The line rejection is pretty good, but I wanted the lineripplerejection to be a bit better. This regulator comes right after the main rectifier/capacitor.Of course, this boostrapped, self-biased version hassignificantlybetter rejection, so it might not be necessary. But it also filters out the Zener breakdown noise, which is actually somewhat significant, at least with the diode I used for testing (some 1N5... part, I didn't have the BZX diode in 15V)And, the final version:

author: C
author_rank: Super Contributor
timestamp: January 23, 2014, 07:15:16 pm
content: Look at the action of R1.When the output is low R1 is trying to make the ref voltage lower.When the output is High R1 is trying to make the ref voltage Higher.Does it make sense to take the results of a zener diode which is not great to start with worse?To say this a different wayif the output power is less than ref voltage make ref voltage less.if the output power is greater than ref voltage make ref voltage greater.So R1 is not good for the voltage side of things, that leaves the current side for it's reason for being.Granted This idea is often used but if this is being done for the current side of things should it not be done more directly to the current side?Why?In both versions you have a current sense resistor that changes values not just with the load but also with V1 due to it's connection to the UN-Reg side.For the current reg side to work really well you need a current reference. This current reference needs to change with the effects of V1 so that when it is compared to the load current sensor V1 gets canceled out.In the second version you could split R9 and have a reference like this.Voltage & current = power. So while you want to control each separately, when combined it is power. So in both versions Q3 is a power control. Could part of the reason or need for R1 be the fact that Q4 is modifying the power control with a signal the changes both by both Load current & V1 voltage.If you want fold back current limiting then you really want an action like R1 that effects the current limit directly to keep it from effecting the voltage part.So while the circuit may work,R1 makes voltage regulation worse.With the way Q4 is currently connected it effects Power regulation all the time when it should have it's large effect happening due to a difference of reference current to output current.C

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 07:20:29 pm
content: I understand why it "should" regulate worse, and I was wondering about that myself, but actual data disagrees. It regulates very, very well (though I don't have recorded numbers to prove that... yet.) Much better than the original (simple constant-current driving the Zener directly)I'm not sure what you mean about the "current side". The current limit, or the current through the Zener?I don't really care about doing the foldback "correctly", it's really just a nice bonus. The regulator works fine without it. It wasn't even intentional at first.

author: megajocke
author_rank: Regular Contributor
timestamp: January 23, 2014, 08:32:02 pm
content: An alternative way to make sure the circuit starts up is to put a resistor in parallell with the output transistor. But maybe the worsening of line regulation caused by your current startup resistor is negligible compared to other effects in the circuit.To form a bridge (3 resistors + zener) in that way you have done is really quite a neat and commonly used technique to get good performance out of a zener diode based regulator. You take the most stable voltage you have in your design from the zener, amplify it using a noninverting operational amplifier to get a higher voltage that isn't much worse in terms of stability and then use the difference in voltage (which will also be very stable unless the operational amplifier performance is bad or it is set to a gain of close to 1) to establish the zener current.This way the stability of the zener current will be almost as good as the reference voltage itself, worsened only by nonidealities of the operational amplifier and the three resistors in the bridge. (feedback divider + current setting resistor + zener)Wouldn't it also be possible to use the same zener as the main reference element for the output voltage and for the bias regulator? That's the way most HP designs seem to do it, using a 6.2 V temperature compensated zener and a bias voltage of 12.4 V. Your design would need a higher bias voltage of course, but it shouldn't cause any problems to choose a higher amplification factor.A simplified schematic of a (positive) reference is attached. Note that there is no way for the power supply to easily influence the output. The "vref" output gets even more line rejection than the output because the variations are further attenuated by the resistor/zener divider. The load regulation of the vref output is worse however.

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 08:38:18 pm
content: Quote from: megajocke on January 23, 2014, 08:32:02 pmYes!Thank you for cutting out a block of my system for me......I was using a main reference Zener powered by another specifically because each stage wasn't that good. This one works amazingly well and caneasilybe the main reference.Though the thought didn't even occur to me...I don't even really have to change anything at this point - just take out the 6.9V reference and move the voltage pot into a voltage divider down to the -32.4V rail, which is now stable enough tobethe reference.

author: megajocke
author_rank: Regular Contributor
timestamp: January 23, 2014, 09:15:58 pm
content: I was thinking mote along the lines of using the same zener voltage in this circuit as you had used for the main reference, and feeding the pot directly from the zener in this circuit.

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 09:24:06 pm
content: It doesn't really make much of a difference, I think.

author: megajocke
author_rank: Regular Contributor
timestamp: January 23, 2014, 10:23:36 pm
content: I guess it would depend on how good you can make the reference amplifierBut wouldn't you save a resistor by using the voltage over the zener? Also, it would make the reference amplifier performance less critical. But maybe it's good enough already?

author: c4757p
author_rank: Super Contributor
timestamp: January 23, 2014, 10:33:18 pm
content: You make a good point. The reference amplifier won't bethatgood, and it will save a resistor.I'm just pleased at how the schematic is cleaning up.I already had a large page (11x17in / 280x430mm) that was somewhat full, and was looking for space for annotations... This helps a good bit. Only saves a few parts, but integrates blocks such that the schematic is neater.

author: C
author_rank: Super Contributor
timestamp: January 24, 2014, 03:12:21 pm
content: Quote from: c4757p on January 23, 2014, 07:20:29 pmZener current, You know this, I just did not say it well. if I have correct data sheet Fig 3Current-voltage characteristicThe effect of R1 is making a large change of the operating point on the curve with what happens on the output. With out R1 the operating point changes up/down with changes in V1 some. For a zener you pay more money to make this change small. When you add R1, in a way you are saying the curve shape is more important. A LED's curve could work just as well in some cases.For good voltage regulation you want the change on D1 as small as you can make it with a huge change of V1. Now R1 will help do this and you are saying that the output is better then input source by a 3k3 / 100k ratio but then do nothing to keep the dynamics out that this causes from getting to D1. C1 is an after effect filter.Add a resistor between this junction and D1 & put a filter cap on the junction. This filter cap would do some or all of what C1 is trying to do while keeping the change on the curve smaller.If the above changes break the requlation then the rest of the circuit is counting on and using that less voltage regulation and curve to function.At the extreme and if you can stand the a very slow startup, you could totally remove the 100k to V1 and have it start with just leakage currents. I see some paths that will allow this to happen.Current limit and current sideQ4 & R8 is the total of the current side i see.Note that R8 will change based on V1. You are doing nothing to counter this change so if you are using Q4 as a switch then a V1 change will change when the switch happens. A better current sense would compare R8 to a ref current created by having a known load at the same output voltage. This would let you cancel the V1 change in R8 as the ref current could also be effected by V1. This may be more then you need or want here.Make sense this time?C

author: c4757p
author_rank: Super Contributor
timestamp: January 24, 2014, 03:34:29 pm
content: C, I have a tough time following you sometimes!But I think I may have addressed a couple of your comments. Here's what I have now. (The arrangement is different - I moved things around so that in the documentation, I'd be able to easily replace the amplifier with an amplifier symbol without moving anything - but it's essentially the same circuit.) R9 no longer sources current to the Zener, instead R40 slightly bypasses the output transistor. So once settled, the Zener current should be fully under the control of the amplifier. R40 just makes Q18's leakage current look worse.I still have the large filter capacitor, just to filter out some of the breakdown noise before it's amplified onto the output. Though the BZX series diodes don't seem to be nearly as noisy as this god-awful 1N5245... The other end of the reference output (you can see in the screenshot where it goes somewhere else) is also similarly filtered.I tried removing the bootstrap resistor but the startup waswaytoo long. Around five seconds. Unacceptable for a power supply - I like to be able to just flick them on and off as I test a circuit. The bottleneck now is the filter cap, which is acceptable to me.I don't see how the current limit will be affected by V1. The limit sense transistor's B-E junction is directly across the sense resistor, which has nothing else across it. It should be seeing the current-induced Ohmic drop directly.Sorry for the sudden, possibly confusing change of reference numbers...

author: C
author_rank: Super Contributor
timestamp: January 24, 2014, 06:09:46 pm
content: Quote from: c4757p on January 24, 2014, 03:34:29 pmThis is not a problem as far as I an concerned so don't worry about asking for clarity or a different way of saying same thing.My quick look at your latest, only easy change that might help is a cap between R4 & R5. would prevent some of the quick changes from effecting zener.QuoteTo see this think of your output being fixed in voltage and current. You then have Q18's collector at a fixed level. As V1 changes, the change has to go some place and in latest that is R40 (thanks you made that easy). So R40 is changing due to V1 and you have R39 & Q18 in parallel to R40 so they also have to change with a V1 change.QuoteI am afraid to even try to tell you about some equipment that I have worked on that curred the bootstrap problem in what most would say an unworkable way.The Crazy circuit to replace bootstrap resistor used a small amount of leakage to start turning on a transistor. The transistor suppled a positive feedback so it quickly went to full on with the only thing holding it back a cap. It stayed in a positive feedback mode until close to operating point when a part of the positive feedback circuit became a negative feedback circuit that exactly balanced the forward feedback. In physical terms that circuit acted like you trying to run across a room as fast as you can and out the door but you hit wall and stuck to it.So you had one cap charging from the positive rail with a second charging from the negative rail with crazy circuit connected to them.With the bootstrap resistor connected to one cap of the two caps. So at power up you had full power supply voltage across bootstrap resistor. When finished the crazy circuit had the bootstrap resistor power end at the same voltage as the operating point making it an effective open circuit.Quotenot a problem, At one point I worked with systems that used house power panels for power safety. Not talking just one house panel here, Blew a breaker one day, it was one of the many small ones just 660V 800A. A lot of electronics to get to that level of power.C