CV-CC Bench Power Supply----How's it work ? - Page 1
Forum section: electronics / beginners

author: huu_tri0101
author_rank: Contributor
timestamp: March 17, 2013, 04:13:00 pm
content: First: How can it control the current, the voltage by the transistor ? what's the function of the transistor ?Second: I have the bench power, which has the problem that when I set it at 5.5V then it jump around 5.51V- 5.5V. It does that on every voltage that I set it (just jumping +/-0.01V). How can I fix it ? It just a regulated power supply 0-18V:0-5A. It come with 2 transistor 2n2055, some op-amps...Thank you for reading.

author: Len
author_rank: Frequent Contributor
timestamp: March 17, 2013, 06:39:46 pm
content: Quote from: huu_tri0101 on March 17, 2013, 04:13:00 pmIf you're looking for basic info about how a CV-CC power supply works, Dave made a good video:http://www.eevblog.com/2011/11/28/eevblog-221-lab-power-supply-design-part-1/

author: huu_tri0101
author_rank: Contributor
timestamp: March 18, 2013, 09:00:09 pm
content: Please help me

author: tld
author_rank: Regular Contributor
timestamp: March 19, 2013, 09:13:14 am
content: Quote from: huu_tri0101 on March 18, 2013, 09:00:09 pmHe did.Watch the video(s), it/they will explain a lot.tld

author: eevblogfan
author_rank: Frequent Contributor
timestamp: March 19, 2013, 10:21:20 am
content: helloplease enter the following link , I't will help you a-lot !http://powersupplyblog.tm.agilent.com/2012/07/how-does-power-supply-regulate-its.htmlCheers

author: BravoV
author_rank: Super Contributor
timestamp: March 19, 2013, 10:34:12 am
content: Not sure, but I had the impression that the OP need a help in modifying the PSU to have a CC mode feature and also fixing the voltage regulation.Looking at the question, he doesn't even understand the basic thing like transistor, those Agilent's literatures or advanced topic pointed might be too overwhelming. Also there is a possibility in language barrier as well.Again, just a wild guess.Tôi có chính xác?

author: rdl
author_rank: Super Contributor
timestamp: March 19, 2013, 12:55:30 pm
content: QuoteThat's only 10 mV, it may be normal - just a bit of jitter in the least significant digit, or it may be as good as the power supply can regulate.

author: olsenn
author_rank: Frequent Contributor
timestamp: March 19, 2013, 01:11:51 pm
content: Most power supplies don't offer true constant current capabilities, they only limit the maximum current that may be sourced (this is useful to protect your circuits from turning to smoke when wired incorrectly).A transistor is (in the case of a BJT) a current controlled conductance source (not a current controlled current source!) or (in the case of a FET) a voltage controlled conductance source. It's actually a psuedo conductance source, since the maximum amount of collector-emitter current is independant of the collector-emitter voltage... it is purely dependant on the base-emitter junction

author: huu_tri0101
author_rank: Contributor
timestamp: March 19, 2013, 04:07:32 pm
content: Quote from: olsenn on March 19, 2013, 01:11:51 pmBJT has 3 states:+Cut off+Saturation+Forward-activeIs that BJT uses "Saturation" in this case ?

author: huu_tri0101
author_rank: Contributor
timestamp: March 19, 2013, 04:15:31 pm
content: Quote from: rdl on March 19, 2013, 12:55:30 pmI have 2 same power supply. One has that problem, one doesn't. I try to switch the sensor board (using ICL7107: change Voltage & Current to 7 segments indicator) -> so the problem is not from the sensor board.

author: hammil
author_rank: Regular Contributor
timestamp: March 20, 2013, 11:22:32 am
content: The following shows a very simple constant voltage supply:http://falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+5.0+50%0Az+496+400+496+320+1+0.805904783+12.0%0Ar+496+320+496+240+0+1000.0%0Ad+352+384+352+336+1+0.805904783%0Ad+352+336+352+240+1+0.805904783%0Ad+384+288+384+240+1+0.805904783%0Ad+384+384+384+288+1+0.805904783%0AT+272+288+320+336+0+4.0+0.05+0.0010693020078094506+6.829016051985159E-17+0.999%0Aw+320+288+384+288+0%0Aw+320+336+352+336+0%0Av+272+336+272+288+0+1+50.0+324.0+0.0+0.0+0.5%0Aw+352+384+384+384+0%0Aw+352+240+384+240+0%0Ac+432+240+432+384+0+0.0010+14.061342394204042%0Aw+384+384+432+384+0%0Aw+384+240+432+240+0%0Aw+432+240+496+240+0%0Ag+432+384+432+416+0%0Ag+496+400+496+416+0%0A174+544+320+544+384+0+1000000.0+0.0050+Resistance%0Ag+544+384+544+416+0%0Aw+544+320+496+320+0%0Aa+592+368+704+368+1+15.0+-15.0+1000000.0%0At+704+368+736+368+0+1+-1.4832552488673265+0.6950886176585325+100.0%0Aw+592+352+560+352+0%0Aw+592+384+592+416+0%0Aw+592+416+736+416+0%0Aw+736+416+736+384+0%0Aw+496+240+736+240+0%0Aw+736+240+736+352+0%0Ar+736+416+736+480+0+100.0%0Ag+736+480+736+496+0%0Ao+27+64+0+35+20.0+0.8+0+-1%0Ao+29+64+0+291+20.0+0.8+1+-1%0AThe mains voltage is transformed down to a lower voltage, which is then rectified and filtered with a high-value capacitor.A stable, low-current voltage is achieved using a zener diode, which is then adjusted with a potentiometer and fed into a simple follower.The high current capability is achieved using the NPN transistor, whose voltage at the emitter will increase as the base voltage increases. However, most of the current is transferred through the collector, which is connected to our high-current but noisy supply. (it's a bit more complicated than that really, but the explanation suffices here)Just ask if you'd like more explanationConstant-current source will be coming shortly.

author: hammil
author_rank: Regular Contributor
timestamp: March 20, 2013, 11:38:08 am
content: Here's a constant-current source:http://falstad.com/circuit/#%24+1+5.0E-6+10.20027730826997+50+5.0+50%0Az+496+400+496+320+1+0.805904783+1.0%0Ar+496+320+496+240+0+1000.0%0Ad+352+384+352+336+1+0.805904783%0Ad+352+336+352+240+1+0.805904783%0Ad+384+288+384+240+1+0.805904783%0Ad+384+384+384+288+1+0.805904783%0AT+272+288+320+336+0+4.0+0.05+0.5145176466675584+8.928802764292767E-17+0.999%0Aw+320+288+384+288+0%0Aw+320+336+352+336+0%0Av+272+336+272+288+0+1+50.0+324.0+0.0+0.0+0.5%0Aw+352+384+384+384+0%0Aw+352+240+384+240+0%0Ac+432+240+432+384+0+0.0010+14.17732142193744%0Aw+384+384+432+384+0%0Aw+384+240+432+240+0%0Aw+432+240+496+240+0%0Ag+432+384+432+416+0%0Ag+496+400+496+416+0%0A174+544+320+544+384+0+1000000.0+0.18320000000000003+Resistance%0Ag+544+384+544+416+0%0Aa+592+368+704+368+1+15.0+-15.0+1000000.0%0At+704+368+736+368+0+1+-8.288432706641135+0.6872434436690291+100.0%0Aw+592+352+560+352+0%0Aw+496+240+736+240+0%0Aw+736+240+736+352+0%0Ar+496+320+544+320+0+9000000.0%0Aw+592+384+592+448+0%0Ar+592+448+592+496+0+1.0%0Ag+592+496+592+512+0%0Aw+592+448+736+448+0%0A174+736+384+736+416+0+100.0+0.5891000000000001+Resistance%0Aw+752+400+752+448+0%0Aw+752+448+736+448+0%0Ao+31+64+0+33+0.078125+0.4+0+-1%0AThe supply system is identical to that of the CV source, but the control system is different.Firstly, our reference voltage is very low, 0-100mV. The reason for this will become apparent soon.Secondly, we have a 1 ohm fixed resistor in series with our variable load. Because we know its value, we know that the current flowing through it is proportional to the voltage across it.The op-amp, rather than comparing the total output voltage, compares the voltage across the 'shunt' resistor to that of the reference voltage, adjusting the output until they are equal. So we can deduce that the current flowing through the shunt (and since they are in series, the current through the load resistor) will be constant.Of course, if your load resistor is too high value for the current you want to push through it (trying to push 1A through a 1K resistor for example, would require 1KV) the system simply saturates at the maximum voltage.

author: huu_tri0101
author_rank: Contributor
timestamp: March 21, 2013, 08:09:17 pm
content: Quote from: hammil on March 20, 2013, 11:38:08 amThank you so much, It helps me a lot.