Timeline for How do I use this voltage regulator with a digital pinout?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2012 at 19:28 | vote | accept | Matt | ||
| Jul 31, 2012 at 3:58 | answer | added | The Photon | timeline score: 5 | |
| S Jul 30, 2012 at 19:47 | history | suggested | m.Alin | CC BY-SA 3.0 |
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| Jul 30, 2012 at 19:32 | review | Suggested edits | |||
| S Jul 30, 2012 at 19:47 | |||||
| Jul 30, 2012 at 17:53 | history | edited | stevenvh |
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| Mar 18, 2012 at 20:51 | history | tweeted | twitter.com/#!/StackElectronix/status/181482952297943041 | ||
| Mar 18, 2012 at 11:11 | answer | added | Russell McMahon♦ | timeline score: 1 | |
| Mar 18, 2012 at 11:06 | comment | added | Matt | And where would I learn such datasheet-reading skills? | |
| Mar 18, 2012 at 10:58 | comment | added | Russell McMahon♦ | You say that you cannot understand the datasheet. The DS has very clear an versy specific examples of how to use the part. If you cannot understand simple circuits such as are in the xx317 datasheet then you need to learn to do so as it is fundamental to learning the art. | |
| Mar 18, 2012 at 10:39 | vote | accept | Matt | ||
| Aug 1, 2012 at 19:28 | |||||
| Mar 18, 2012 at 10:35 | comment | added | shimofuri | @Matt if you have an analog signal for input, you can simplify things by building an amplifier instead to control the motor which can be accomplished by a single high capacity transistor. In your case, you need to build and inverting amplifier. | |
| Mar 18, 2012 at 10:24 | comment | added | Matt | Sorry for making this confusing. You're correct - the signal described here is not strictly digital, I should really have called it an analog signal between 0 and 5V. The application is a robot, and I'm trying to control the speed of the motor by changing the voltage of an analog pin coming from the computer. | |
| Mar 18, 2012 at 10:20 | comment | added | shimofuri | I think we have some confusion here... my understanding of digital signal is that it can only have two states: a HIGH and LOW (the zen philosophers will argue for a third one typically called HIGH IMPEDANCE). We cannot use the actual voltage of the digital signal to directly program the regulator. HIGH is not equal to 5V, nor is LOW equal to 2.5V. Can you please provide more details on your application: Why do you want to reduce the voltage of the regulator? | |
| Mar 18, 2012 at 10:06 | comment | added | PetPaulsen | @Matt - The voltage regulator requires a minimum load current. Did you provide a load, or did you measure the output pins without a load? Usually the minimum load current is given by the voltage divider. | |
| Mar 18, 2012 at 9:58 | history | edited | Matt | CC BY-SA 3.0 |
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| Mar 18, 2012 at 9:49 | comment | added | Matt | Vnorm=7V if signal is low, Vred=2V (or the lowest value the regulator can manage) if signal is high. The input voltage, Vin=7V. I think I can vary the voltage of the digital pinout, so if I set it to 2.5V then V should be reduced to around 5V. None of this has to be exact, I just need some way of decreasing the voltage from the pinout. | |
| Mar 18, 2012 at 9:38 | comment | added | shimofuri | What exactly do you want to do? Please provide values for normal voltage (Vnorm = ?) that you are expecting if digital signal is LOW, reduced voltage (Vred = ?) if digital signal is HIGH, and the input voltage (Vin =?) for the regulator. I will modify my answer to help you achieve what you want. | |
| Mar 18, 2012 at 9:34 | answer | added | shimofuri | timeline score: 3 | |
| Mar 18, 2012 at 9:07 | history | asked | Matt | CC BY-SA 3.0 |