How to calculate power consumption&efficiency of a boost converter?

xw0927

Joined Dec 19, 2010
114

xw0927

Joined Dec 19, 2010
114
anyone has the idea? or clue?? thanks

SgtWookie

Joined Jul 17, 2007
22,221
Could I just simply calculate my power consumption as P=IoutVout
and Power input : P=Iinput*Vinput

and then, efficiency as Power out/Power In?
Yes, that's how it's done.

• xw0927

xw0927

Joined Dec 19, 2010
114
Yes, that's how it's done.
So, to prove it, I just measure it by using multimeter and then calculate it, is it?

Thanks

SgtWookie

Joined Jul 17, 2007
22,221
So, to prove it, I just measure it by using multimeter and then calculate it, is it?
How else are you going to get the information you need to calculate the efficiency?

You should use a number of different "known" loads, like fixed resistors, for the efficiency calculation.

You say the output is set to around 48 volts. Start out with a load of 10mA or so. That would require 4.8k Ohms of resistance, and would require a 1/2 Watt resistor rating. Since 4.8k Ohms is not a standard value of resistance, you might start off by setting the boost converter to output 47 volts and use a 4.7k Ohm resistor, which IS a standard value.

Then you could add another 4.7k Ohm resistor in parallel to the 1st one to get a 20mA load, and another in parallel for a 30mA load, etc.

• xw0927

xw0927

Joined Dec 19, 2010
114
How else are you going to get the information you need to calculate the efficiency?

You should use a number of different "known" loads, like fixed resistors, for the efficiency calculation.

You say the output is set to around 48 volts. Start out with a load of 10mA or so. That would require 4.8k Ohms of resistance, and would require a 1/2 Watt resistor rating. Since 4.8k Ohms is not a standard value of resistance, you might start off by setting the boost converter to output 47 volts and use a 4.7k Ohm resistor, which IS a standard value.

Then you could add another 4.7k Ohm resistor in parallel to the 1st one to get a 20mA load, and another in parallel for a 30mA load, etc.
Erm, basically, I am not very understand with the 0.5watt resistor and 0.25 watt, with my own understand, I assume it to be a resistor with consume 0.5watt when voltage is applied. However, I found it is strange when :

there is a output which is 50V, and it drives a 0.5watt resistor (100K ohms). According to Ohm's law, it the current should be 50V/100k = 500uA. And P = IV, thus P = 0.025watt. How do I explain this phenomenon?

For the method you mentioned above, i explain what I've understood here, and please kindly correct me if i am wrong.

For output voltage 47 and connected to 47K,0.5watt. Then i measure the current, and get Pout. After this, I calculate Pin/Pout. Then, the result of the efficiency is for 0.5watt load.

Once, I parallel it with another same resistor, and i get my Pout/Pin, which is efficiency for 1.0watt load and so on.

Am I right?

SgtWookie

Joined Jul 17, 2007
22,221
Erm, basically, I am not very understand with the 0.5watt resistor and 0.25 watt, with my own understand, I assume it to be a resistor with consume 0.5watt when voltage is applied.
No. A resistors' power rating in Watts means that the resistor is rated for up to a specified power rating; if you exceed that rating the resistor will overheat.

A general "rule of thumb" is to use a resistor with twice the power rating requirement. If you wanted 10mA current across 47 volts, then R=E/I, so R=47v/0.01A = 4,700 Ohms or 4.7k Ohms or 4K7 Ohms.

The power rating would then be P=EI, or Watts = Voltage * Current, or 47*.01 = 0.47 Watts. In this case, doubling the result gives you 0.94 Watts; the closest standard value of wattage is 1 Watt.

However, I found it is strange when :

there is a output which is 50V, and it drives a 0.5watt resistor (100K ohms). According to Ohm's law, it the current should be 50V/100k = 500uA. And P = IV, thus P = 0.025watt. How do I explain this phenomenon?
The resistor is merely RATED for up to 1/2 Watt; it does not dissipate that much power unless sufficient voltage is applied across the resistor.

For the method you mentioned above, i explain what I've understood here, and please kindly correct me if i am wrong.

For output voltage 47 and connected to 47K,0.5watt. Then i measure the current, and get Pout. After this, I calculate Pin/Pout. Then, the result of the efficiency is for 0.5watt load.
No, I said 4.7k Ohms - but if you want to start with 47k Ohms, that is OK. Adjust the output for 47V across 47k Ohms. You don't need to measure the current through the resistor, as you already know that 47v/47k = 0.001A, or 1mA, and the power dissipated in the resistor will be 47mW.
Measure the input current and voltage to get the input power.

Then go to 4.7k, 1 Watt resistor. With 47V out, there will be 10mA current through the resistor and 470mW power dissipation in the resistor.
Measure the input current and voltage to get the input power.

Then place two 4.7k 1W resistors in parallel on the output. You know by Ohm's Law that with 47v out you'll get 20mA current through the pair, and 940mW power dissipation.
Then measure the input current and voltage to get the input power.

• xw0927

xw0927

Joined Dec 19, 2010
114
No. A resistors' power rating in Watts means that the resistor is rated for up to a specified power rating; if you exceed that rating the resistor will overheat.

A general "rule of thumb" is to use a resistor with twice the power rating requirement. If you wanted 10mA current across 47 volts, then R=E/I, so R=47v/0.01A = 4,700 Ohms or 4.7k Ohms or 4K7 Ohms.

The power rating would then be P=EI, or Watts = Voltage * Current, or 47*.01 = 0.47 Watts. In this case, doubling the result gives you 0.94 Watts; the closest standard value of wattage is 1 Watt.

The resistor is merely RATED for up to 1/2 Watt; it does not dissipate that much power unless sufficient voltage is applied across the resistor.

No, I said 4.7k Ohms - but if you want to start with 47k Ohms, that is OK. Adjust the output for 47V across 47k Ohms. You don't need to measure the current through the resistor, as you already know that 47v/47k = 0.001A, or 1mA, and the power dissipated in the resistor will be 47mW.
Measure the input current and voltage to get the input power.

Then go to 4.7k, 1 Watt resistor. With 47V out, there will be 10mA current through the resistor and 470mW power dissipation in the resistor.
Measure the input current and voltage to get the input power.

Then place two 4.7k 1W resistors in parallel on the output. You know by Ohm's Law that with 47v out you'll get 20mA current through the pair, and 940mW power dissipation.
Then measure the input current and voltage to get the input power.
Thank you for your very kind explain, you teach me a lot which are not being taught in the class. I do understand how the things work right now.

Thank you very much again.

p/s: I wanted to write 4.7k Ohms, it was a typo error xw0927

Joined Dec 19, 2010
114
I've tested the circuit with the method you taught me...
As I have only 10k resistor,
but the result is kinda of weird:

when only 1 resistor(10k)
my efficiency is around 55%;

when there are 2 resistors (10k)
my efficiency is around 60%;

when there are 3 resistors (10k)
my efficiency is around 67%...

I thought it should be constant??? Why would it be like this?

SgtWookie

Joined Jul 17, 2007
22,221
At light loading, less power is delivered to the load. There is "overhead" for operating the 555 timer (charging/discharging the timing cap) and charging/discharging the MOSFET gate, amongst other losses (like in the timing resistors).

At heavier loading, more power is delivered to the load, so the losses in the 555 circuit become less significant.

At some point, the efficiency will start to go down again and/or the boost converter won't be able to maintain the set output voltage.

• xw0927

xw0927

Joined Dec 19, 2010
114
At light loading, less power is delivered to the load. There is "overhead" for operating the 555 timer (charging/discharging the timing cap) and charging/discharging the MOSFET gate, amongst other losses (like in the timing resistors).

At heavier loading, more power is delivered to the load, so the losses in the 555 circuit become less significant.

At some point, the efficiency will start to go down again and/or the boost converter won't be able to maintain the set output voltage.
Oh I see. The cct is running with constant output if connected to fixed resistor.

AT 2W, its efficiency goes up to 75%. After that, I try to increase resistor again, but HISS sound is coming, so I assume it is FULL load at 2W. Am I correct? Thanks a lot very much

SgtWookie

Joined Jul 17, 2007
22,221
Hiss sound? Is the inductor or MOSFET warming up?

At 2W for input power, that's an average of 167mA current in the inductor - but probably more like 666mA peak with a 25% duty cycle.

Perhaps the base frequency of the boost regulator is too low for the size (uH) of the inductor.
The easiest way to increase the base frequency is to reduce the size of C1.
You might try 2.2nF (0.0022uF)

Last edited:
• xw0927

xw0927

Joined Dec 19, 2010
114
Hiss sound? Is the inductor or MOSFET warming up?

At 2W for input power, that's an average of 167mA current in the inductor - but probably more like 666mA peak with a 25% duty cycle.

Perhaps the base frequency of the boost regulator is too low for the size (uH) of the inductor.
The easiest way to increase the base frequency is to reduce the size of C1.
You might try 2.2nF (0.0022uF)
If no mistake, the inductor is heating, does it mean, it reaches it full-load condition?

SgtWookie

Joined Jul 17, 2007
22,221
If no mistake, the inductor is heating, does it mean, it reaches it full-load condition?
If the inductor is getting hot, then it won't last very long. You'd need to reduce the load until it cooled back down.

• xw0927

xw0927

Joined Dec 19, 2010
114
there are 2 sets of experiments I have done.

1st: I used 4.3k resistor at 46V.
Then,when 5resistors in parallel, a significant hiss sound produce.
when only 4 resistors in parallel,its output current is abt 42.8mA. Output WATT = 2W. Efficiency = 75%

2nd: I used 10k resistor at 45.
When 7 resistors at parallel, my efficiency is 81%.
when 8 resistors at parallel, my efficiency is also 81%.(I have eight 10k resistors)
but at the time, the current of output is only 36mA.

Could I say that 42mA is my FULL LOAD CURRENT?

SgtWookie

Joined Jul 17, 2007
22,221
You need to figure it out by yourself.

• xw0927