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# DcDc buck regulator current

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#### Rajinder1268

##### Full Member level 3
Hi all,
I am using a LM53600 DCDC buck regulator. The output is 5V fed into a LDO to give 3V3.
The input to the DCDC is 12V. I have a 5R load across the 3V3 LDO. This equates to 600mA.
When I operate at 12V input to the DCDC converter I get 140mA. When I lower the voltage to the DCDC to say 5V I get 600mA.
Why is this the case?

Power in = Power out, plus the extra to overcome losses.
5 * 0.6 = 3W output power
12 * 0.14 = 1.68W input power
5 * 0.6 = 3W. input power

What is happening is the internal 'switch on' period is longer as the input is lowered so the average current increases. At 12V I would have expected the input current to be nearer 250mA.

Brian.

Power in = Power out, plus the extra to overcome losses.
5 * 0.6 = 3W output power
12 * 0.14 = 1.68W input power
5 * 0.6 = 3W. input power

What is happening is the internal 'switch on' period is longer as the input is lowered so the average current increases. At 12V I would have expected the input current to be nearer 250mA.

Brian.
I have checked it and it is 280mA at 12V. I need it to be 600mA on 12V and at 5V. Is there anything I can do?

Hi,

first I recommend to put all in the correct order. From left to right means input to output.
hopefully I did understand correctly:

1)
12V --> [LM53600 DCDC buck] --> 5V --> [LDO 3.3V] --> 3.3V --> [5 Ohms] -- GND

so now from right to left:
* 3.3 Ohms at 5 Ohms is 660mA, not 600mA
* the LDO just drops the voltage, but the current does not change (indeed it adds a bit for the contol loop)
* so let´s calculate with 10mA added, so the input current is 670mA at 5V making a power of 3.35W
* a buck converter keeps the power constant (indeed it adds a bit for the control loop). Let´s say it adds 0.5W, so the input power now is 3.85W. 3.85W on 12V is 320 mA

Your 140mA can´t be correct at all.

Klaus

Why do you want the 12V current to be 600mA? It makes no sense at all.

Power in = Power out, plus the extra to overcome losses.
5 * 0.6 = 3W output power
12 * 0.14 = 1.68W input power
5 * 0.6 = 3W. input power

What is happening is the internal 'switch on' period is longer as the input is lowered so the average current increases. At 12V I would have expected the input current to be nearer 250mA.

Brian.
Hi
I have checked again at 12V the current output is 280mA. Is there anyway I can get this to 600mA at 12V?
Hi,

first I recommend to put all in the correct order. From left to right means input to output.
hopefully I did understand correctly:

1)
12V --> [LM53600 DCDC buck] --> 5V --> [LDO 3.3V] --> 3.3V --> [5 Ohms] -- GND

so now from right to left:
* 3.3 Ohms at 5 Ohms is 660mA, not 600mA
* the LDO just drops the voltage, but the current does not change (indeed it adds a bit for the contol loop)
* so let´s calculate with 10mA added, so the input current is 670mA at 5V making a power of 3.35W
* a buck converter keeps the power constant (indeed it adds a bit for the control loop). Let´s say it adds 0.5W, so the input power now is 3.85W. 3.85W on 12V is 320 mA

Your 140mA can´t be correct at all.

Klaus
Sorry I remeasured it and it is 280mA. So how can I get this circuit to deliver 660mA at 12V and 5V input?
--- Updated ---

Hi,

first I recommend to put all in the correct order. From left to right means input to output.
hopefully I did understand correctly:

1)
12V --> [LM53600 DCDC buck] --> 5V --> [LDO 3.3V] --> 3.3V --> [5 Ohms] -- GND

so now from right to left:
* 3.3 Ohms at 5 Ohms is 660mA, not 600mA
* the LDO just drops the voltage, but the current does not change (indeed it adds a bit for the contol loop)
* so let´s calculate with 10mA added, so the input current is 670mA at 5V making a power of 3.35W
* a buck converter keeps the power constant (indeed it adds a bit for the control loop). Let´s say it adds 0.5W, so the input power now is 3.85W. 3.85W on 12V is 320 mA

Your 140mA can´t be correct at all.

Klaus

Why do you want the 12V current to be 600mA? It makes no sense at all.
We need to be able to run from 12V, the devices are capable of drawing up to 0.5A i.e. Bluetooth and WiFi devices

Last edited:

Hi,

again: Why?

I can only see a couple of reasons why to draw 600mA from 12V:
* you want to heat your room
* you want to increase your electricity bill
* you want to shorten batter lifetime

non of these I want to do. So why do you want this?

Klaus

Hi,

again: Why?

I can only see a couple of reasons why to draw 600mA from 12V:
* you want to heat your room
* you want to increase your electricity bill
* you want to shorten batter lifetime

non of these I want to do. So why do you want this?

Klaus
We are running off a 12V car battery and the electronics that is connected can run up to 0.5A, Bluetooth and WiFi devices. What would be the solution? Run at a lower input voltage? Or use a voltage regulator? Can you assist with any other solutions?

Hi,

I still don´t get it.
What would be the solution?

But usually neither bluetooth nor WiFi is supplied by 12V. So why are you talking about the 12V current.

I don´t want to bore you with the same questions. Thus I come back after you provided a complete schematic/sketch of how you want to connect your deivces.

Klaus

Hi,

I still don´t get it.

But usually neither bluetooth nor WiFi is supplied by 12V. So why are you talking about the 12V current.

I don´t want to bore you with the same questions. Thus I come back after you provided a complete schematic/sketch of how you want to connect your deivces.

Klaus
Yes I will show how I want to connect the devices. Basically the 12V is stepped down to 5V via the buck. This 5v is fed into a 3V3 LDO, this supplies the 3V3 for the micro, Bluetooth and WiFi devices ( which according to the datasheet can consume upto 0.5A). The 12V will always be there from the car battery, hence I need 12V at 600mA.

Everything makes sense up to "hence I need 12V at 600mA".
Where do you get the figure 600mA from?
Surely you want the current to be a small as possible for whatever load you place on the output, and it will be different as the load changes.

Brian.

hence I need 12V at 600mA.
Hence you don't.
The device takes the current it needs to generate the desired power out.
It makes absolutely no sense to try to increase that.

5V @ 0.5A is 2.5W.
If you have a 80% efficient buck converter, the input will be 2.5W / 12V / 0.8 = 260mA.
Due to the efficiency of the converter the current in is less than the current out.

You need to learn about power in and power out, which you seem to have no understanding of.

Hence you don't.
The device takes the current it needs to generate the desired power out.
It makes absolutely no sense to try to increase that.

5V @ 0.5A is 2.5W.
If you have a 80% efficient buck converter, the input will be 2.5W / 12V / 0.8 = 260mA.
Due to the efficiency of the converter the current in is less than the current out.

You need to learn about power in and power out, which you seem to have no understanding of.
That's true I don't really fully understand the principles. So what happens when my circuit requires more current from the 3V3 line? Do I need to choose a better efficient DcDc,?

You start any design by finding out what you want to achieve. The second stage is investigating different solutions and the final stage is building it. Design work is 90% inspiration and 10% perspiration!

When you say "what happens when my circuit requires more current" you should already have decided what the maximum would be and designed to be able to provide it. If you want to travel at 100KPH you don't design a bicycle then work out how to make it go that fast. You start with a specification then plan how to meet it or exceed it.

Brian.

You start any design by finding out what you want to achieve. The second stage is investigating different solutions and the final stage is building it. Design work is 90% inspiration and 10% perspiration!

When you say "what happens when my circuit requires more current" you should already have decided what the maximum would be and designed to be able to provide it. If you want to travel at 100KPH you don't design a bicycle then work out how to make it go that fast. You start with a specification then plan how to meet it or exceed it.

Brian.
Hi really thought that it was designed for that purpose i.e. to be able to provide 0.5A with an input of 12V. Seems like this is not the case

Hi

how do you come to "Seems like this is not the case...".?

when it clearly already supplies 600mA (or 660mA)

0.5A is 500mA which is smaller than 600mA.

I don´t see the problem.

Hi

how do you come to "Seems like this is not the case...".?

when it clearly already supplies 600mA (or 660mA)

0.5A is 500mA which is smaller than 600mA.

I don´t see the problem.
I am really sorry, perhaps it's my understanding of some principles. When I have 12V input fed to the buck 5V out, then this 5V to power the 3V3 I get 280mA with 5R load.

When I reduce the 12V input to say 5V, this fed into the buck 5V, then the buck 5V fed into the 3V3 LDO again with a 5R load I get 0.66A.

My question is why I do not see 0.66A at 12V input? I know you mentioned power conversion but for most of the time the electronics will be powered by a 12V input, which gives 280mA, not 0.66A.

I think I do not understand this clearly. Can anyone advise please

Hi,

current... at which point?

If you talk about your load, which is connected to 3.3V then focus on the current at these 3.3V.
If you are interested in the 5V supply then focus on 5V current
If you are interested on 12V then focus on the 12V current.

But you can´t talk about a 3.3V supplied device that draws 500mA (from these 3.3V) ... but then continue to talk about the 12V current.

Why are you concerened about the 12V current when the devices are supplied with 3.3V?

Klaus

Hi,

current... at which point?

If you talk about your load, which is connected to 3.3V then focus on the current at these 3.3V.
If you are interested in the 5V supply then focus on 5V current
If you are interested on 12V then focus on the 12V current.

But you can´t talk about a 3.3V supplied device that draws 500mA (from these 3.3V) ... but then continue to talk about the 12V current.

Why are you concerened about the 12V current when the devices are supplied with 3.3V?

Klaus
I apologise. At 3V3 output the current is 280mA and 660mA when powered from 12V (which it will be all the time) or 660mA when the 12V is reduced to 5V, this second case is highly unlikely.

Perhaps I am not good at explaining. But I want there to be 3V3 @660mA when the input is 12V.

Back to basics:
Voltage is what you want to appear at the output, 5V, 3V3 or whatever...
Current is how much you ARE drawing from it.

So the design has to provide the voltage you want while the current can be anything from zero to a maximum you specify. Except in special circuits called constant current sources, your load decides the current and all you have to do is make sure the source is capable of providing it. You don't design the load current, you just have to be prepared to provide it.

I'm guessing your question is more like "why does the current drawn from the battery increase if the battery voltage is lower" which would appear to be contrary to Ohms Law. If I'm right, the answer is the buck regulator stores the power in an inductor so if you put less voltage in, you have to turn the current source on for longer to store the same amount of power. Nearly all 'switch mode' supplies draw current in short bursts and the length of the burst is adjusted to regulate the output, longer bursts mean the average current is seen to be higher.

Brian.

Back to basics:
Voltage is what you want to appear at the output, 5V, 3V3 or whatever...
Current is how much you ARE drawing from it.

So the design has to provide the voltage you want while the current can be anything from zero to a maximum you specify. Except in special circuits called constant current sources, your load decides the current and all you have to do is make sure the source is capable of providing it. You don't design the load current, you just have to be prepared to provide it.

I'm guessing your question is more like "why does the current drawn from the battery increase if the battery voltage is lower" which would appear to be contrary to Ohms Law. If I'm right, the answer is the buck regulator stores the power in an inductor so if you put less voltage in, you have to turn the current source on for longer to store the same amount of power. Nearly all 'switch mode' supplies draw current in short bursts and the length of the burst is adjusted to regulate the output, longer bursts mean the average current is seen to be higher.

Brian.
Yes that is what I could not understand. It is contrary to my understanding of ohm's law.

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