| Electronics > Beginners |
| LM217 voltage regulator question |
| << < (2/2) |
| Wimberleytech:
--- Quote from: Chris Wilson on May 19, 2019, 11:36:18 pm ---Thanks for the very fast replies! Will an HEF4013 flip flop (which it is powering) draw that much all the time or do i need to redesign the divider? It *SEEMS* to work OK on the breadboard... EDIT, hmm it seems the HEF4913 only draws Ua levels maybe I need to change this :( IDD supply current all valid input combinations; IO = 0 A 5 V - 1.0 - 1.0 - 30 - 30 uA 10 V - 2.0 - 2.0 - 60 - 60 uA 15 V - 4.0 - 4.0 - 120 - 120 uA --- End quote --- Add a dummy load to make up the difference. |
| magic:
Funny, mine also says 10mA for the LM317 but still recommends 240Ω without specifying it's for LM217 only. I always assumed 240Ω is enough, good to know. |
| Chris Wilson:
--- Quote from: Wimberleytech on May 19, 2019, 11:41:15 pm --- Add a dummy load to make up the difference. --- End quote --- 240 ohm 1/2 watt resistor OK? Although using R1 at 240 Ohms and adjusting R2 to around 900 Ohms should give circa 6V. Thank you again. |
| Wimberleytech:
--- Quote from: Chris Wilson on May 20, 2019, 12:00:14 am --- --- Quote from: Wimberleytech on May 19, 2019, 11:41:15 pm --- Add a dummy load to make up the difference. --- End quote --- 240 ohm 1/2 watt resistor OK? Although using R1 at 240 Ohms and adjusting R2 to around 900 Ohms should give circa 6V. Thank you again. --- End quote --- 1.252/240 = 6.5mW so 1/2 watt is overkill but it is fine. |
| Zero999:
--- Quote from: Chris Wilson on May 19, 2019, 10:12:40 pm ---I was looking for around 6 to 7V out from 12V in and I had a 5k ten turn pot to hand. using a cadence resistor box I found R1 at 1k gave a nice adjustment range but the spec sheets cite R1 as usually being a lot lower resistance, circa 240 Ohms. Is there anything bad in using 1k for R1 in the divider? Thanks. --- End quote --- As mentioned above, the resistance needs to be lower because the LM217 has a minimum load current requirement of 5mA. The reason for this is it's actually an op-amp and voltage reference connected to a big Darlington pair, along with some protection circuitry. The negative rail of the op-amp is also connected to the output terminal of the regulator, so in order for the op-amp to work properly, the current flowing through the output needs to be high enough. At higher currents, the op-amp still draws around 5mA, it's just the output transistor turns on to allow more current through the output terminal. |
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