Hi everyone, I'm looking for a voltage controlled (uno) current source circuit - Page 1
Forum section: electronics / beginners

author: barak198
author_rank: Contributor
timestamp: July 20, 2016, 11:56:35 am
content: if it is possible based on lm317 its purpes is to charge a 6\5 V battety (0.5A MAX)thank you all

author: danadak
author_rank: Super Contributor
timestamp: July 20, 2016, 02:39:36 pm
content: What type of battery are you trying to charge ?litium, Lead Acid, NiCad.......Look at datasheet, it shows some battery charging circuits -http://www.ti.com/lit/ds/symlink/lm317.pdfRegards, Dana.

author: barak198
author_rank: Contributor
timestamp: July 20, 2016, 04:36:31 pm
content: NiMH battery but i'm more intrested in a voltage controlled current source circuitthe thing is that the LM317 has a configuration in which its a current source , the output current is 1.25/Rso i thought that i can control it if i had a voltage controlled potentiomter circuit insted of R, but couldn't find onein any case the controlled current source is mian issuethank you

author: danadak
author_rank: Super Contributor
timestamp: July 21, 2016, 09:24:08 am
content: Here is one solution -https://www.eevblog.com/forum/beginners/best-way-to-digitally-control-lm317-based-constant-current-driver/Here is an electronic load....http://jasper.sikken.nl/electronicload/index.htmlRegards, Dana.

author: barak198
author_rank: Contributor
timestamp: July 23, 2016, 07:18:23 pm
content: thank you for the lm317 current source circuit !I replaced the opamp to lm324and I get 200-300 mA even when the refrence voltage is 0

author: danadak
author_rank: Super Contributor
timestamp: July 24, 2016, 01:52:48 am
content: I simulated the link I gave you, circuits very poor performance. SoI set up the attached circuit in sim, and it seems to function well.Basically the LM317 acts a a fixed .5A current source. Then the OpAmp+ NPN is set up as a V controlled current sink. Basically it robs the LM317current thereby robbing the load of current.I think a better way would be a constant current source using a PNP and OpAmp,no LM317, and just use that as fixed charge source. Note that method requiresa regulated V source for all power used.https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjLmvyp_4rOAhVLHT4KHdbrCp4QjRwIBw&url=http%3A%2F%2Fwww.ecircuitcenter.com%2FCircuits%2Fcurr_src1%2Fcurr_src1.htm&bvm=bv.127984354,d.cWw&psig=AFQjCNEnu_jjTslIBcjDvV_3KDD2copm7Q&ust=1469411681455915Regards, Dana.

author: barak198
author_rank: Contributor
timestamp: July 25, 2016, 09:59:54 am
content: So I finaly managed to get it right ..thanks to a TIP122 I harvested oncecontrolled easly with minimal current drew from the control pin .since I only have 4 small solar cells (4.4v >100mA)and I get a 8.8V source , the voltage remaning on the rest of the circuit is 2.7- 1.8V .note that the control voltage should not get above the collector voltage .I will be back soonthank you allcircuit applet simulationhttp://www.falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+10.20027730826997+64+5+50%0At+496+192+544+192+0+1+0+0+100%0At+576+272+608+272+0+1+0+0+100%0Ad+688+352+688+160+1+0.805904783%0Ar+480+272+528+272+0+7000%0Aw+496+192+464+192+0%0Aw+464+192+368+192+1%0Aw+464+192+464+272+0%0Aw+464+272+480+272+0%0Aw+528+272+544+272+0%0Aw+544+272+544+208+0%0Aw+544+272+576+272+0%0Aw+544+176+544+160+0%0Aw+608+256+608+160+1%0Ar+544+352+608+352+0+70%0Aw+544+160+608+160+0%0Aw+608+160+688+160+0%0Aw+544+352+544+272+0%0Aw+608+352+608+288+0%0Aw+608+352+688+352+0%0Aw+608+352+608+384+2%0Aw+608+160+608+96+2%0A172+48+80+48+48+0+6+0.95+5+0+0+0.5+Voltage%0Aw+48+80+48+128+0%0Ax+405+210+437+213+0+15+base%0Ax+631+115+650+118+0+15+col%0Ax+566+390+594+393+0+15+emit%0Aa+80+176+192+176+1+15+-15+1000000%0Aw+48+128+48+160+0%0Aw+48+160+80+160+0%0At+240+176+304+176+0+1+0+0+100%0Aw+304+160+304+96+0%0Ax+572+238+649+241+0+24+TIP122%0Ar+304+192+368+192+0+500%0Aw+192+176+240+176+1%0Aw+80+192+80+256+0%0Aw+80+256+304+256+0%0Aw+304+256+304+192+0%0Ar+608+384+608+432+0+5%0Ag+608+432+544+432+0%0A162+432+96+608+96+1+6.85+1+0+0%0AR+304+80+304+32+0+0+40+8.8+0+0+0.5%0Aw+304+80+304+96+0%0Aw+304+96+432+96+1%0Ax+371+72+610+75+0+14+bat+simulation+:+6.8+Fwd+Voltage+@+1A%0Ax+73+55+129+58+0+24+var+V%0Aw+368+192+368+288+2%0Aw+416+128+752+128+0%0Aw+768+128+768+368+0%0Aw+752+368+416+368+0%0Aw+400+368+400+128+0%0Ahttp://www.falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+10.20027730826997+64+5+50%0At+496+192+544+192+0+1+0+0+100%0At+576+272+608+272+0+1+0+0+100%0Ad+688+352+688+160+1+0.805904783%0Ar+480+272+528+272+0+7000%0Aw+496+192+464+192+0%0Aw+464+192+368+192+1%0Aw+464+192+464+272+0%0Aw+464+272+480+272+0%0Aw+528+272+544+272+0%0Aw+544+272+544+208+0%0Aw+544+272+576+272+0%0Aw+544+176+544+160+0%0Aw+608+256+608+160+1%0Ar+544+352+608+352+0+70%0Aw+544+160+608+160+0%0Aw+608+160+688+160+0%0Aw+544+352+544+272+0%0Aw+608+352+608+288+0%0Aw+608+352+688+352+0%0Aw+608+352+608+384+2%0Aw+608+160+608+96+2%0A172+48+80+48+48+0+6+0.95+5+0+0+0.5+Voltage%0Aw+48+80+48+128+0%0Ax+405+210+437+213+0+15+base%0Ax+631+115+650+118+0+15+col%0Ax+566+390+594+393+0+15+emit%0Aa+80+176+192+176+1+15+-15+1000000%0Aw+48+128+48+160+0%0Aw+48+160+80+160+0%0At+240+176+304+176+0+1+0+0+100%0Aw+304+160+304+96+0%0Ax+572+238+649+241+0+24+TIP122%0Ar+304+192+368+192+0+500%0Aw+192+176+240+176+1%0Aw+80+192+80+256+0%0Aw+80+256+304+256+0%0Aw+304+256+304+192+0%0Ar+608+384+608+432+0+5%0Ag+608+432+544+432+0%0A162+432+96+608+96+1+6.85+1+0+0%0AR+304+80+304+32+0+0+40+8.8+0+0+0.5%0Aw+304+80+304+96+0%0Aw+304+96+432+96+1%0Ax+371+72+610+75+0+14+bat+simulation+:+6.8+Fwd+Voltage+@+1A%0Ax+73+55+129+58+0+24+var+V%0Aw+368+192+368+288+2%0Aw+416+128+752+128+0%0Aw+768+128+768+368+0%0Aw+752+368+416+368+0%0Aw+400+368+400+128+0%0A

author: danadak
author_rank: Super Contributor
timestamp: July 25, 2016, 10:51:48 am
content: The problem with that approach is that the load, 5 ohms, its valueaffects the current thru it, eg.you do not have a current source. Itmay "measure" close to a current source because the 5 ohms iseffectively in series with the 500 ohm emitter R, so small variationsin the 5 ohms would not impact the current thru it. never-the-lessits a poor current source.Regards, Dana.

author: barak198
author_rank: Contributor
timestamp: July 25, 2016, 11:43:39 am
content: I don't understand the problem ..I control the current that flows through the batterythe resistor purpose is to stabilise the systemsorry for my faults , english isn't my native language