Timeline for How to calculate this current limit circuit?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2018 at 1:57 | answer | added | lukeluck | timeline score: 1 | |
| Jun 27, 2018 at 1:26 | history | edited | lukeluck | CC BY-SA 4.0 |
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| Jun 26, 2018 at 6:26 | comment | added | jonk | @CristobolPolychronopolis Hopefully, that helps a bit. | |
| Jun 26, 2018 at 6:26 | answer | added | jonk | timeline score: 3 | |
| Jun 26, 2018 at 5:47 | comment | added | jonk | @CristobolPolychronopolis G36 used a very straight-forward approach to solving the problem and provides a useful formula that predicts the sign of the change in load current vs input voltage, too. It's a fine approach and simple to explain. But since G36 doesn't appear to be answering directly, I'll walk you though G36's approach. | |
| Jun 26, 2018 at 3:57 | comment | added | lukeluck | @G36 how can you figure out IL = ((R2 + R4) Vbe - R2 Vin)/(R1 R4)? | |
| Jun 25, 2018 at 16:06 | comment | added | G36 | Notice that now widout the load current Vbe voltage will reach 0.6V if Vin > 1+R4/R2 *0.6 = 7.2V and IL = ((R2 + R4) Vbe - R2 Vin)/(R1 R4) Try to look at this circuit electronics.stackexchange.com/questions/299389/… | |
| Jun 25, 2018 at 12:41 | comment | added | Cristobol Polychronopolis | Using second pic--R2, under no-current conditions, has a voltage of about 0.41V from a 5V supply. You can go from there and figure how much voltage you need across R1 to turn on the transistor. If you run it from a higher supply, R2 will have a higher no-current voltage, and Q1 may be on all the time, cutting off the output by default. From a lower supply voltage, more current will be permitted. See it? | |
| Jun 25, 2018 at 10:25 | history | edited | lukeluck |
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| Jun 25, 2018 at 7:31 | history | edited | lukeluck | CC BY-SA 4.0 |
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| Jun 25, 2018 at 7:03 | history | edited | lukeluck | CC BY-SA 4.0 |
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| Jun 25, 2018 at 6:34 | history | asked | lukeluck | CC BY-SA 4.0 |