[sdiy] switching dc power sources

Sarah Thompson plodger at gmail.com
Wed Oct 28 12:29:47 CET 2015


Use a LDO that has a feedback input -- normally this is fed from a resistor divider from the output voltage. If you mess with the feedback voltage, e.g., by biassing it slightly via a resistor and diode fed from a logic level output from your microcontroller, you can persuade the LDO to generate two different voltages under software control. 

This also works for switching converters if you need better efficiency.

Sarah

Sent from my iPad

> On Oct 27, 2015, at 8:02 AM, Nils Pipenbrinck <n.pipenbrinck at hilbert-space.de> wrote:
> 
> Not strictly synth related but...
> 
> 
> In one of my projects I want a low impedance power supply that can
> either source 3V or 1.8V. The load is kind of weak, it's a CPLD IO bank
> that I'm going to power, so 20mA would fit my bill, 50mA would be better.
> 
> The voltages itself are available as well, with required current source
> capabilities.
> 
> At first I thought: Just generate the voltages from my microcontroller
> with a resistor divider and run that through an opamp voltage follower
> to generate the current, but no: My load requires decoupling capacitors
> and opamps don't like capacitive loads. Those who can cost a fortune.
> 
> Then I thought: Screw it, I'll just take a ordinary adjustable 3
> terminal LDO and switch one of the voltage defining resistors. That does
> not work either because LDOs can only source current, not sink any. If I
> switch from 3V to 1.8V the output voltage will just stay high until I've
> consumed enough energy to drop it down.
> 
> 
> So, any idea how to buffer a reference voltage in a way that is fine
> with capacitive loads? I searched long and hard on the internet but
> haven't found anything cheap with low part count.
> 
> Probably also good to know:
> 
> - I have an additional voltage-source of 5V available.
> - input/reference voltage won't be modulated. It's either 3V or 1.8V.
> - I don't need super fast switching. It is okay for the output voltage
> to take a quater second to reach it's final voltage.
> 
> 
> Best,
>  Nils
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