| Electronics > Projects, Designs, and Technical Stuff |
| Adjustable Voltage Supply: Schematic Concerns |
| << < (5/6) > >> |
| Jay_Diddy_B:
Hi, Take a look at this: Not finished, but an idea for you to work on. Regards, Jay_Diddy_B 1074 neg.asc (3.41 kB - downloaded 57 times.) |
| LoveLaika:
Thanks for your reply as well as your example circuit. Sorry for asking like this, but I've been going back and forth with some others about my project, and I've been rethinking my original circuit. I was hoping that I could get some feedback on my thoughts. With the LM2673, I can use the current limiting feature (I_adj) with a pot to limit the switch current, so as my load current increases, the output voltage will drop if it exceeds that limit. I verified that in SPICE, so in that sense, I can more or less current limit my regulators, plus or minus 25%. This is done automatically within the regulator, so that takes care of one problem. I'd like to add an LED indicator of sorts so I can 'indicate' when a limit has been reached, so I still need a current sense feature, but now, it doesn't have to be folded back into the feedback loop. The key thing was my sense resistor at 1 ohm. I have to shrink it down to something like 1 milli-Ohm in order to have a negligible voltage drop, and this way, I can place it alongside NEG_OUT and POS_OUT without worrying about feedback and voltage drop. Does this sound more or less alright? Following this, it seems that the only thing I have to decide now is picking a current sense op-amp. |
| Jay_Diddy_B:
LoveLaika, In the circuit I gave you. The differential amplifier U3 measures the output voltage directly. The voltage from U3 output to the GND pin on U1 is the same as the voltage across the load. The 1 \$\Omega\$ current sense resistor is to illustrate that this method works even if there are large current sense resistors. You can, and should, use a smaller value resistor. 1 milliohm sense resistor have other problems when you are trying to measure small currents. The I-adj pin on the LM2673, will give you current limiting. Most bench power supplies are constant current. If you implement the current limiting with the I-adj pin the output current will vary with output voltage. You can still use a current sense amplifier. Jay_Diddy_B |
| LoveLaika:
Thanks for your reply. With your warning though, about current limiting with the I-adj pin, isn't that the same idea? If the load draws too much current, then the output voltage will decrease (or increase in the case of the negative output voltage). This occurs without the current feedback amplifier loop depending on how I_LIM/I_ADJ is set, so wouldn't that make the diff-amp redundant (unless you're using it for something else)? |
| Jay_Diddy_B:
Hi, Even if you use the I-lim pin on the chip for limiting, you will probably need a sense resistor for metering. Jay_Diddy_B |
| Navigation |
| Message Index |
| Next page |
| Previous page |