Going over the application notes, I see that VCC provides many option but with minimum of 8V (default). I need 5V to supply my encoder and the line driver.
Should I just use a voltage divider to VCC and get my 5V out of it? Sorry if I am answering my own question, but I want to make sure. Should I also filter the power with capacitors? Any suggestions will be appreciated.
Accepted Solutions
Hello @PCama.1​ ,
You are right, the VCC buck regulator is not able to generate a 5 V supply voltage.
Voltage divider is not the right way to obtain a supply voltage because the output of the divider strongly depends on the consumption of the connected circuitry.
One of the easiest way to obtain a regulated voltage is though a Zener diode and a BJT as shown below:
The key points are:
- R1 must be tuned to provide the right bias current to the Zener
- Rout guarantees a minimum load current making the circuit stable
However, I would recommend a cheap linear regulator as the L78L or the LM217L.
They provide a cost-effective and compact solution for a reliable supply voltage generation.
One important point: the current consumption of the 5 V will contribute to the overall buck output current, so take care not exceeding regulator limit.
Hello @PCama.1​ ,
You are right, the VCC buck regulator is not able to generate a 5 V supply voltage.
Voltage divider is not the right way to obtain a supply voltage because the output of the divider strongly depends on the consumption of the connected circuitry.
One of the easiest way to obtain a regulated voltage is though a Zener diode and a BJT as shown below:
The key points are:
- R1 must be tuned to provide the right bias current to the Zener
- Rout guarantees a minimum load current making the circuit stable
However, I would recommend a cheap linear regulator as the L78L or the LM217L.
They provide a cost-effective and compact solution for a reliable supply voltage generation.
One important point: the current consumption of the 5 V will contribute to the overall buck output current, so take care not exceeding regulator limit.
