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| Need help with bi directional constant current source (±100mA) |
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| OM222O:
--- Quote from: duak on November 12, 2018, 05:04:44 am ---OM, here's an awful sketch of a current booster. Ri handles the opamp's output current up to the point at which its voltage drop turns on one or the other transistors (~0.6 V / 100R = 6 mA). D1 or D2 turn on when the inductor voltage tries to go beyond the power supply rail voltages. If these weren't here, the inductor current would try to force its way thru the opamp. D3 and D4 are zener diodes that limit the power supply rail voltages when the energy in the inductor has to be dumped. Since this is an actuator in a mechanical system, there may be an energy source may be external to the inductor (such as an arm) in addition to the energy stored in the magnetic field. If there's a lot of energy, you might need some big zener diodes, or a more capable voltage limiter. Do you know how to calculate the worst case power dissipation in the transistors? It's not always obvious, but with a simple inductive load for the case of just holding, you can have maximum current with essentially zero voltage across it. The transistors would then have to drop the rail voltage and so dissipate 100 mA * 24 V = 2.4 W. If the actuator is moving, there will also be a back EMF impressed on the output voltage and further increase the power dissipation. Tons of fun, eh? Good luck! --- End quote --- thank you very much. I just had one final question: shouldn't there be current limiting resistors in series with all diodes? D1 through D4. the entire transient will take about 28ms so I'm not sure if the diodes can dissipate that power as it's just a short burst (actuator won't go from -100 to +100 rapidly, I just need the response time to be as quick as possible to react to the environment, not to rapidly move all the travel range) something like alex recommended seems like a better choice but is still missing resistors for D3 and D4? I'm also not sure why he is using 4 transistors ... which one is dissipating the major amount of power? |
| Alex Nikitin:
--- Quote from: OM222O on November 12, 2018, 12:15:48 pm --- thank you very much. I just had one final question: shouldn't there be current limiting resistors in series with all diodes? D1 through D4. the entire transient will take about 28ms so I'm not sure if the diodes can dissipate that power as it's just a short burst (actuator won't go from -100 to +100 rapidly, I just need the response time to be as quick as possible to react to the environment, not to rapidly move all the travel range) something like alex recommended seems like a better choice but is still missing resistors for D3 and D4? --- End quote --- The energy in the coil is not that great, so any mid-sized diode should be OK. --- Quote from: OM222O on November 12, 2018, 12:15:48 pm ---I'm also not sure why he is using 4 transistors ... which one is dissipating the major amount of power? --- End quote --- The output pair obviously dissipates more than the previous stage. The opamp and the first pair of transistors (Q1 and Q4) make two voltage controlled current sources - one for each polarity, and the output pair (Q2 and Q3) together with resistors and diodes work as two current mirrors with a combined output. The ratio is 10:1 , so for 10mA through Q1 or Q4 the output will be 100mA through Q3 or Q2 respectively. This circuit is convenient as you essentially limited only by the size and voltage handling of transistors. The circuit as is is working in Class B so is not very linear in the area of low output currents. If that matters, just add a resistor between bases of Q2 and Q3, providing about 0.5mA current (so for +/-40V output stage supply it should be ~150K), however that may create a small offset current due to an asymmetry of two current mirrors. It could be trimmed down if required though. Cheers Alex |
| OM222O:
Thank you very much for the detailed explanation. as I mentioned before size is a concern on my PCB so I will probably go with the circuit that duak provided. if it proved to be unsuitable after testing, I will give your circuit a shot as well. |
| Alex Nikitin:
--- Quote from: OM222O on November 12, 2018, 02:28:44 pm ---Thank you very much for the detailed explanation. as I mentioned before size is a concern on my PCB so I will probably go with the circuit that duak provided. if it proved to be unsuitable after testing, I will give your circuit a shot as well. --- End quote --- You should not forget that in either circuit output devices may dissipate considerable power so should be sized accordingly and placed on a heatsink (100mAx40V = 4W for example). Cheers Alex |
| duak:
OM, you will not want to put resistors in series with the clamp diodes (D1 & D2) as they are trying to bypass currents that could damage the transistors or opamps. Putting resistors in series with the zener diodes will allow the rail voltages to rise further. Are you able to quantify how much mechanical energy may have to be dumped? You could implement a more capable energy dumper by augmenting the zener diodes with some transistors and resistors. If you have a number of drivers & actuators, only one energy dumper for each rail is needed. Best o' luck, |
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