Op-Amp Help - Page 1
Forum section: electronics / general

author: 08mechan
author_rank: Newbie
timestamp: August 09, 2011, 06:58:50 am
content: I am actually working on a project which is a high rate discharger, for battery capacity test.and i have to design a control circuit for current regulation, so that the battery is being discharge at a constant current.i was provided another 20 Amps discharger and i reverse engineered that and the final circuit was this (Attached Image).but the problem is that i am not able to understand its working. (i'm not much good at analogue circuits)so plz anyone help me in this regard.Thanks in advance.

author: qno
author_rank: Frequent Contributor
timestamp: August 09, 2011, 09:19:44 am
content: Check outEEVblog #102 - DIY Constant Current Dummy Load for Power Supply and Battery Testing

author: amspire
author_rank: Super Contributor
timestamp: August 09, 2011, 09:25:16 am
content: First, there are a few obvious errors in the schematic.Q1 needs to have the collector and emitter swapped, R5 is definitely not 8.2M ohms - it could be 8.2 ohms, R9.R10 and R11 will not be 47 ohms - 0.047 ohms would work, and most importantly, the junction of R9, R10 and R11 connects to the other end of R8.D1, D2 and D3 will probably be 1n4004 diodes, and my guess is that Q1 will be providing a positive supply of 5.6v or 6.8 V, and zener D9 will be supplying a similar negative voltage. If zener D10 is around about 6.2V or 7.5 V, then negative temp coefficient of the Q1 base emitter voltage will cancel; out the positive temp coefficient of D10 giving a pretty stable reference voltage.The use of the OP777 is a big clue - this is a low offset voltage op-amp, so the current regulator is working with very low voltages. R9, R10 and R11 just balance the current between the power transistors. Not sure what D1, D2 and D3 are doing - perhaps they are there to protect the op-amp in the case of a power transistor failing.The current shunt R8 is sure to be a very low resistance - it could be 0.001 ohms or smaller.I think the MJ3000 is a darlington, so the op-amp only has to output a few millamps to get RCA power transistors to sink the 20A.With the schematic corrected, the circuit will make more sense.Simply, the op-amp will increase or decrease the battery discharge current until pin 3 goes to 0 volts.If I make up some numbers. let say the junction of R4 and R5 is 82 mV. Then when pin3 is at 0 volts, the current through R6 and RV4 will be about .082/105K = 0.8 uA.At maximum current, RV3 will be 10 K, and it will have the same current through it : 0.8uA, so that means the voltage on R8 would have to be about -8mv. In these made up numbers, this would mean that R8 would be .008v/20A = 0.0004 ohms.If RV3 is set to 2K (20%) then to keep pin3 at 0V, the opamp will force the current through R8 to make the R8 voltage 2K ohms x -0.8uA = 1.6mV. -1.6 mV across R8 means a current of 4A.RV4 is there to calibrate the 20A when RV3 is at maximum.Hope that helps

author: johnmx
author_rank: Frequent Contributor
timestamp: August 09, 2011, 04:13:19 pm
content: Check the circuit again and draw it from the start. The one you drew is definitely not the real one.