What is this LM358 doing  ? - Page 1
Forum section: electronics / beginners

author: lordvader88
author_rank: Frequent Contributor
timestamp: March 03, 2017, 12:57:08 am
content: I understand what the LM317 is doing, and it outputs 15V, and I more or less get U570A half of LM358But what is the U470 LM358 doing to control 8V below?pin5 is 8V, its split on a voltage divider, and a LM358 can sink current I learned recently. So are the diodes on the outputs really sinking that much current??15V 8V regulators picture below

author: lordvader88
author_rank: Frequent Contributor
timestamp: March 03, 2017, 01:09:22 am
content: how are those transistor making 8V

author: Paul Moir
author_rank: Frequent Contributor
timestamp: March 03, 2017, 01:22:03 am
content: The bottom one is doing voltage regulation while the top one is doing current regulation. The diodes are to pick which of the two is in charge at the moment. As described, the bottom op amp is "fighting" to keep its inputs balanced at 8v (top input from the 10v reference divided down to 8, the bottom from the regulator's output.) The top one is doing nothing at this point, but if its bottom input gets over it's top input, it'll take control by going lower and forcing the output voltage below 8v.Make sense?

author: wraper
author_rank: Unknown
timestamp: March 03, 2017, 01:22:28 am
content: Where that 10V ref is coming from? it can be used to disable the regulator output. on top of that higher value resistors can be used in divider as a bonus.

author: David Hess
author_rank: Super Contributor
timestamp: March 03, 2017, 02:25:50 am
content: Ah, this is one of my favorite circuits.The LM358 is using the LM317 as a protected power pass element to produce a precision +15 volt output from the +10 volt reference. Accuracy is only limited by the voltage divider and LM358's input errors.

author: MrAl
author_rank: Super Contributor
timestamp: March 03, 2017, 09:02:13 am
content: Hello,U570A, the LM358, is acting as an error amplifier and integrator. It's one of the simplest parts of a control stage where we see the error as:Verr=Vref-Voutand the output of the op amp is the time integral of Verr which is known to produce low steady state output error. That is highly desirable in a power supply.An advantage of doing it this way is that we get rid of the temperature dependence of the LM317, which although may look good on the data sheet, has one serious flaw: the internal reference diode heats up as the IC itself heats up meaning the output would change more without this op amp addition.The drawback is somewhat slower response because the LM358 slew rate is very limited. A faster op amp would help here but a possible oscillation with various different kinds of loads has to be checked for carefully.

author: David Hess
author_rank: Super Contributor
timestamp: March 03, 2017, 04:05:31 pm
content: Quote from: MrAl on March 03, 2017, 09:02:13 amThis does not diminish your point about the temperature dependence but the LM317 actually uses a bandgap reference.Even if the reference was perfect, the differential error amplifier stage in the LM317 is also affected and using a separate error amplifier in the form of an external operational amplifier separates it from the heat produced by the LM317 pass transistor. This is why operational amplifiers in precision applications cannot drive heavy loads while maintaining precision; often this shows up as poor settling time.QuoteThe slow LM358 does not limit performance in this case as Tektronix felt it necessary to add C575 to slow it even further. Using a faster operational amplifier would yield no benefit and even in an ideal circuit, would require external compensation to slow it down anyway.I wonder what Tektronix instrument this is. I looks to me like they intended the +15 volt output to slowly follow and then track the +10 volt reference for proper power sequencing. The more common configuration for this circuit is much faster and higher performance even with a slow LM358.

author: lordvader88
author_rank: Frequent Contributor
timestamp: March 07, 2017, 04:28:37 am
content: This is from my tektronix2430A that lays in pieces for 2months , and I haven't powered it up in ages, just busy, lazy and trying to learn a bunch of stuff along the way. And draw/map out the circuits on the PSU itself(with a marker, and on pictures)I haven't powered it up to measure around in weeks.U470A and B, of that LM358, how are they controlling Q465, Q479 in order to get the voltages written there.The unregulated +8.63V on collector Q479 comes from the primary transformer. Where does its 8.73V base voltage come from? That would mean about say a Vce of 6.3V for Q465. And something a bit more than 9.4V on base of Q465, round it to 10V say. Then meaning 5V is dropped across R565, at 1mA, which splits into the base, and CR466.Looking at the approximate or average U470A,B output voltages, and those diodes, practically no reverse current flows through CR465.And some flows into CR466 into pin 7. And that fine tuning and feedback splits/controls the base current into base Q465 and then controls the base of Q479 ? Is that sort of what its doing, with the C485?So fine tuning Vce of those transistors is what really makes the 8V ?Also U470pin 3 is directly connected right back to the right side of C595, and so 1 resistor away from pin6Here's the manual link, this is on page 410. The manual explains it on another level, on p131, I need to re-read as I learn more.http://w140.com/2430-SM.pdfI've done some basic op-amp calculations, but not of the calculus type (but I've done lots of calculus)

author: David Hess
author_rank: Super Contributor
timestamp: March 08, 2017, 04:24:52 am
content: Quote from: lordvader88 on March 07, 2017, 04:28:37 amThis is a series pass regulator with foldback current limiting.U470B is the voltage error amplifier which compares the +10 volt reference divided down to +8 volts to the output and drives the pass transistors through CR466. U470A monitors the output current through shunt R473 and limits the current to about 2 amps.The two diodes are used so that either the voltage or current control loop can drive the pass transistors as needed.