DC-DC buck converter

Thread Starter

electric77

Joined Mar 18, 2016
9
Hi
Iam doing a project that requires to step down 36 V to three loads that need about 5V and 2A, 2A and 5 A.
if i connect my loads in parallel, to my buck converter that has the following secifications:
input voltage: DC 5-40V; Output voltage: DC1.2-36V; Out current: 12A(max.); Output power: 100W
will i damage my loads

loads are arduino or raspberry pi (5V 2A)
ultrasonic sensor (5V 2A)
and servo motor ( 5V 5A)

question is, is my buck converter always producing 12 A?
if yes then best option would be resistor to reduce current?
and also how do i protect circuit from feedback current due to servo (any diode?)
plz plz plz
help
 

Sensacell

Joined Jun 19, 2012
3,419
Your loads draw current- the DC to DC converter only produces current demanded by the loads.
No no load, no current.

No problem as long as the total load current is less than the maximum rated capacity of your DC to DC converter.

"feedback current" is an undefined and unlikely problem.
 

ronv

Joined Nov 12, 2008
3,770
Hi
Iam doing a project that requires to step down 36 V to three loads that need about 5V and 2A, 2A and 5 A.
if i connect my loads in parallel, to my buck converter that has the following secifications:
input voltage: DC 5-40V; Output voltage: DC1.2-36V; Out current: 12A(max.); Output power: 100W
will i damage my loads

loads are arduino or raspberry pi (5V 2A)
ultrasonic sensor (5V 2A)
and servo motor ( 5V 5A)

question is, is my buck converter always producing 12 A?
if yes then best option would be resistor to reduce current?
and also how do i protect circuit from feedback current due to servo (any diode?)
plz plz plz
help
You may need to be careful that you don't reverse the motor before it has stopped.
 

JJacobs

Joined Feb 7, 2015
1
Basics...
Your buck regulator regulates the Voltage to the loads. It will attempt to hold the voltage at the output to 5V regardless of the load on it. That is the function of a voltage regulator.

The current, up to the supplies maximum, is drawn by the load.

The current rating of the supply is simply the maximum current the load can draw before that voltage starts to drop. You should seldom if ever be at the maximum.

According to Ohm's Law: I = E/R, since E (voltage) is fixed by the voltage regulator, I (current) varies according to R, which is what the load draws.
 
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