agilent E3620A issue - Page 1
Forum section: electronics / projects

author: electros6
author_rank: Regular Contributor
timestamp: June 08, 2015, 03:03:55 pm
content: Hi guys,I tried to of agilent E3620A and it works fine. I have one question. as shown in the attachment ,even after omit the components Q6,R29,CR10 and short as shown the power supply works fine, then what is the need for that componentts

author: bson
author_rank: Unknown
timestamp: June 08, 2015, 05:33:18 pm
content: Looks like a current limiter for the output.But, it's hard to tell where the feedback is coming from due to the circuit being chopped off to the right so it's just a WAG.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 08, 2015, 10:06:46 pm
content: CR10 is there to protect the output transistor from excessively negative voltage (lower than - 6 V).Q6 gives some extra current gain, so that the OPs don't have to deliver so much current. However 10 mA of static current are not so bad in this case. But during transients more current (way more than 10 mA) helps to turn off the output transistor faster. Delay in turning of Q6 limits the speed the output current can shoot up.

author: free_electron
author_rank: Super Contributor
timestamp: June 08, 2015, 10:08:38 pm
content: that is a current gain stagr. the opmps don;t have enough drive power to control the main transistor directly. especially under impulse load this improves the response.

author: mikerj
author_rank: Super Contributor
timestamp: June 08, 2015, 10:48:27 pm
content: Quote from: electros6 on June 08, 2015, 03:03:55 pmHow much testing have you done? Transient performance at high loads? Power on/off overshoots?

author: dom0
author_rank: Super Contributor
timestamp: June 09, 2015, 08:07:05 am
content: It will basically work, just not as good. Stability may be reduced if you short out Q6(*). Transient response will be worse. The diode is mainly a protection measure for transient events at on/off. R29 complements the diode OR gate in front of the ops and is of course replaced by R28 if you short out Q6.(*) especially with large power transistors or MOSFETs as the SPE.

author: macboy
author_rank: Super Contributor
timestamp: June 09, 2015, 02:55:12 pm
content: You should be more concerned about the lack of a pull-up resistor on the output of U18B (LM393 comparator) on output #1. Examine what happens without it, specifically, what is the effect on the unregulated voltage between TP1/TP2 as you adjust the output voltage? Compare this to the voltage between TP3/TP4 as you adjust output #2 voltage. If you really want to experiment with and improve this supply, leave Q6 in there, and fix this comparator!

author: dom0
author_rank: Super Contributor
timestamp: June 09, 2015, 05:03:25 pm
content: There areno comparators in the linear controller there, and U10 is not an LM393. There is also nothing to fix in this circuit.

author: macboy
author_rank: Super Contributor
timestamp: June 09, 2015, 05:26:22 pm
content: Quote from: dom0 on June 09, 2015, 05:03:25 pmI said nothing of U10.U18 is a LM393 comparator controlling the switched-tap pre-regulator for supply 1. A design flaw prevents it from ever turning off Q9, and the result is 40 VDC at the input of the pass transistor at all output voltages. This causes excessive power dissipation when using low output voltages at high currents.

author: dom0
author_rank: Super Contributor
timestamp: June 09, 2015, 05:30:59 pm
content: Okay, I assumed you read the number 10 (with crossed-through zero) as 18, since the piece of schematic you are talking about wasn't in this thread before and actually isn't the revision OP posted (you can see it at the left of OPs snippet that it's not the same schematic as yours).Edit: OPs snippet is of channel two, which as a remarkably complicated TRIAC control circuit entirely different than that of channel one. Channel two has apparently three levels on the filter caps ; I guess the thermal design of the case was not sufficient for another channel with just two levels.

author: dom0
author_rank: Super Contributor
timestamp: June 09, 2015, 05:47:13 pm
content: Still it doesn't make sense what you say.The reference at U18B- is negative.The other operand is the output voltage, which is negative, too(!), since the positive side of it is the regulators GND (A1).So, if U18B+ is lower than U18B-, which means, that the output voltage is *above* a threshold set by the reference, it's output swings to it's negative rail, sinking current from Q9 BE junction limited by 10 k?, so Q9 is definitely in full saturation here and thus you have about 11.x V on U3D1 and R6.If U18B+ is higher than ref (so the output voltage is *lower* than the threshold) the open-collector (emitter at -12 V!) does not conduct, and no base current from Q9 is drawn, thus it does not conduct and the Opto-TRIAC does not turn on, which in turn doesn't turn the power TRIAC on.Kudos to HP for drawing this part of the circuit partly bottom up, I guess we have to be happy they didn't write some labels in the mirror :-)e: typo(s)

author: macboy
author_rank: Super Contributor
timestamp: June 09, 2015, 10:13:30 pm
content: Quote from: dom0 on June 09, 2015, 05:47:13 pmLogically this is true, but I have six of these supplies and I assure you that in all of them it doesn't work that way. There is just enough leakage through the base of that transistor to to cause just enough collector current to light up the optocoupler just enough to activate it's output transistor and activate that triac. The pull up transistor is absolutely necessary, or the pre-regulator does not work. This is easily confirmed by just measuring the test points i noted earlier. Even with that fix the threshold is a ridiculous 7 V or so, so that needs adjustment also.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 10, 2015, 05:07:03 pm
content: Leakage and the current through R62 may be enough, so that Q6 might conduct enough to case trouble. A base to emitter resistor (e.g. 1 K) at Q6 should solve the possible problem.