Integrated Regulated power supply - Page 1
Forum section: electronics / beginners

author: techguru
author_rank: Regular Contributor
timestamp: June 09, 2018, 06:53:08 pm
content: Hi to all,I found this circuit in internet, please explain the variable current limiting function in LM723 IC?Thank you all..

author: jeroen79
author_rank: Frequent Contributor
timestamp: June 10, 2018, 12:32:37 am
content: http://www.ti.com/lit/ds/symlink/lm723.pdfThe built in current limiter consists of a single transistor that with it's collector connected to the output transistor's base.Parallel to this transistor's base and emitter is the current sense resistor.When the voltage drop over that resistor exceeds 0.6V the base will start to draw current, turning the transistor on.This transistor will then draw current away from the output transistor's base, reducing the output current.Interestingly, in this circuit they put the current sense resistor on the low side, rather than on the high side.

author: techguru
author_rank: Regular Contributor
timestamp: June 10, 2018, 04:22:18 am
content: I think there is a error in Schematic. Check this . I have attached the circuit, with red color to mark the error.

author: techguru
author_rank: Regular Contributor
timestamp: June 10, 2018, 04:51:29 am
content: How does this make Current limiting function?I have calculated node voltages at each node. now my current sense transistor will experience 6.35v or 5.6V in its base and 0V at emitter at no load condition . This will damage my current sense transistor, Is it correct??Please clear my doubt...

author: DC1MC
author_rank: Super Contributor
timestamp: June 10, 2018, 04:58:44 am
content: NOTE: This message has been deleted by the forum moderator Simon for being against the forum rules and/or at the discretion of the moderator as being in the best interests of the forum community and the nature of the thread.If you believe this to be in error, please contact the moderator involved.An optional additional explanation is:

author: techguru
author_rank: Regular Contributor
timestamp: June 10, 2018, 05:03:13 am
content: If it is, Then guide me in understanding the complete circuit.

author: DC1MC
author_rank: Super Contributor
timestamp: June 10, 2018, 05:13:22 am
content: I want to, but I can't, calculating 135mA and voltages on that schematic showed that you don't have the basics and you need much more learning before starting with this kind of schematics, sorry.

author: techguru
author_rank: Regular Contributor
timestamp: June 10, 2018, 05:18:35 am
content: Here is the clear calculation sir.. Please help me...

author: cellularmitosis
author_rank: Unknown
timestamp: June 10, 2018, 07:30:38 am
content: Quote from: DC1MC on June 10, 2018, 05:13:22 amTake another look, he didn't write 135mA. Why don't you make fun of his name some more, accuse him of trolling again, and continue teasing him with additional unhelpful comments?techguru, it looks like your mistake was that R6 is 68R, not 68k.

author: cellularmitosis
author_rank: Unknown
timestamp: June 10, 2018, 07:47:42 am
content: This is clever -- the 723 is floating on top of the current shunt. When current flows through the shunt, the 723 lifts itself up, which also lifts the base of the current limit transistor.It looks like when the 500R pot is turned all the way down, there's 0.09V at the transistor base (assuming 7V Vref). That leaves 0.51V to turn on the transistor, which will happen when 3.4A flows over the shunt.When the 500R pot is turned all the way up, there's about 0.75V at the transistor base, so the output is totally shut off.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: June 10, 2018, 07:49:14 am
content: The problem with this arrangement is that if P2 has a bad spot on its track the output will go to max and likely damage the circuit under test. Better to have the pot in the +ve input (pin 5) and a fixed feedback ratio. Only thing is, to get low voltages you then need an aux supply for the 723.

author: xavier60
author_rank: Super Contributor
timestamp: June 10, 2018, 09:41:51 am
content: Quote from: techguru on June 10, 2018, 05:18:35 amKeep in mind that you should be calculating divider voltages With Respect To ground, pin 7 of the LM723.

author: Simon
author_rank: Unknown
timestamp: June 10, 2018, 09:52:49 am
content: Quote from: DC1MC on June 10, 2018, 05:13:22 amThen why reply to the thread? if you think something is amiss then report it. I encourage you to spend the next 3 days thinking about the extreme value in keeping your big mouth SHUT if you have no intention of helping and are considering stirring things up. Well what you do with the next 3 days is up to you but you certainly won't be spending them on here doing very much.....

author: techguru
author_rank: Regular Contributor
timestamp: June 14, 2018, 06:31:23 am
content: Thanks a lot cellularmitosis...

author: techguru
author_rank: Regular Contributor
timestamp: June 14, 2018, 06:44:52 am
content: hi to all..1)why we have Q1(pnp) and Q2(npn)?According to me it is Q2 which is going to act as external series pass element.2)But why do we have Q1 there and Internal series pass element collecter connected to base of PNP transistor what is the logic present there?According to me collector of the series pass transistor should be connected to maximum positive voltage(not exceding 40V). what will happen now?3)why emitter of internal series pass transistor connected to 220 ohm ant that is connected 68 ohm. A voltage divider is formed and it is connected to output?Emitter of internal pass transistor should be connected to external series pass transistor but why it is being connected to voltage divider and to output?still i am not clear in this schematic? Kindly help me to understand?

author: xavier60
author_rank: Super Contributor
timestamp: June 14, 2018, 01:51:06 pm
content: Quote from: techguru on June 14, 2018, 06:44:52 amR8 and R7 have nothing to do with each other. R7 is there only to bleed off charge from the Base of Q2 to help it turn off more quickly when needed to lessen voltage overshoot at the output when load is suddenly removed. R4 is for the same reason.At up to about 3mA of output load, the internal transistor supplies all of the output current via R8. The transistor's Collector draws this current from the unregulated rail via R4.Above 3mA, the voltage drop across R4 exceeds 0.6V causing some of the current to flow from Emitter to the Base of Q1 causing it to partly turn on causing current to flow from its Emitter to Collector. This current is also supplied to the output via R7.At some higher load current, the voltage drop across R7 will exceed 0.6V causing Q2 to partly turn on and to supply current to the output.The way that Q1 drives Q2 results in very high current gain, very approximately 1000. For example with 5A of load current, the internal transistor only need pass about 10mA.R8 limits the amount current that the internal transistor can supply, likely to protect it from overload in case something goes wrong.

author: techguru
author_rank: Regular Contributor
timestamp: June 14, 2018, 05:28:09 pm
content: Dear all,Thanks to cellularmitosis and xavier60 for their kind help.we have professor who does'nt spend time with us. all the good hearts in this Forum are thanked for this Service.again thanks a lot..xavier60 Sir,I will be happy if u share the information how do you acquired this knowledge?Because i didnt thought of 3mA concept.but the relationship reveals Vbe=Vt ln(Ic/Is),if we substitute that we will get 660mV.Which will help to turn on the transistor.. Thanks a lot Sir.ByStudent..