# Boost Converter: Power Losses

#### declan2693

Joined Feb 2, 2018
37
Hi,

I am trying to figure out the total losses of boost converter.

So my input is 2.9V and 100mA so that (2.9*100mA) = 0.29 Watts

My output is 5V and with 50 ohm resistor, 25/50 = 0.5W.

What am I missing as obviously there should be power losses not the other way around.

Circuit image -> boost.PNG

#### Attachments

Last edited by a moderator:

#### artmaster547

Joined Jan 6, 2016
410
Hi,

I am trying to figure out the total losses of boost converter.

So my input is 2.9V and 100mA so that (2.9*100mA) = 0.29 Watts

My output is 5V and with 50 ohm resistor, 25/50 = 0.5W.

What am I missing as obviously there should be power losses not the other way around.

Circuit image -> boost.PNG
hmm this is weird does your converter deliver current to your load? Is there any way you could measure this the output power should definitely be equal or mostly likely less than the input

#### declan2693

Joined Feb 2, 2018
37
hmm this is weird does your converter deliver current to your load? Is there any way you could measure this the output power should definitely be equal or mostly likely less than the input
As you can see from the image of the circuit, the amp meter reading from the load is -0.375ma to +0.080ma but the curcuit does work as intended.

#### declan2693

Joined Feb 2, 2018
37

#### ericgibbs

Joined Jan 29, 2010
9,116
My output is 5V and with 50 ohm resistor, 25/50 = 0.5W.
Hi d,

E

#### declan2693

Joined Feb 2, 2018
37
Hi d,

E
What is the problem?
P = V^2/R = 5^2/50 = 0.5W

#### ericgibbs

Joined Jan 29, 2010
9,116
hi D,
Woops, misread it, Sorry about that. E

#### dante_clericuzzio

Joined Mar 28, 2016
213
Hi,

I am trying to figure out the total losses of boost converter.

So my input is 2.9V and 100mA so that (2.9*100mA) = 0.29 Watts

My output is 5V and with 50 ohm resistor, 25/50 = 0.5W.

What am I missing as obviously there should be power losses not the other way around.

Circuit image -> boost.PNG
I think boost converter IC does increase the current at certain output check the datasheet how much current it draws

#### kubeek

Joined Sep 20, 2005
5,650
As you can see from the image of the circuit, the amp meter reading from the load is -0.375ma to +0.080ma but the curcuit does work as intended.
The current is not continuous, therefore you cannot use some simple average or whatever to calculate the input power. You need to mulitply current times voltage for each point in time and then average that, there should be some way to do that in your simulator.