DcDc buck regulator current

[Rajinder1268]

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?

 

[KlausST]

Hi,

Oms law:
You tell us your load resistor is 5 Ohms
You say the voltage is 3.3V
--> then there must be 660mA. Must!

There are so much wrong informations in this thread, thus do yourself a favour and post a photo of your circuit including voltmeter and current meter...

Klaus

 

[d123]

Hi,

Maybe your question has already been answered. An additional path to explore is power source output impedance vs dcdc input impedance, and considering frequency may affect it. Rule-of-thumb is at least 1:10. There's also a well-documented case where one of output or input impedance becomes a negative impedance. Problem may just be principal power source can't provide 12V x 600 (+ circuit operating current) mA in Watts but can muster the 5V (+ circuit operating current) x 600 mA.

 

[Rajinder1268]

Hi,

Maybe your question has already been answered. An additional path to explore is power source output impedance vs dcdc input impedance, and considering frequency may affect it. Rule-of-thumb is at least 1:10. There's also a well-documented case where one of output or input impedance becomes a negative impedance. Problem may just be principal power source can't provide 12V x 600 (+ circuit operating current) mA in Watts but can muster the 5V (+ circuit operating current) x 600 mA.

Hi the DcDc is rated at 1A, we are looking to use it at around 0.5A to 0.6A max. So there is plenty of headroom. So I can't understand why it can't deliver 0.5A with a 12V input?

 

[KlausST]

Hi,

I'm really annoyed now, because you do nothing to make any progress. No effort from your side.
No drawing, no sketch, no photo. No clear and consistent informations.
Just the same questions without caring about our posts.

Helping can only work when you contribute.

I'm out.

Klaus

 

[Rajinder1268]

Hi,

I'm really annoyed now, because you do nothing to make any progress. No effort from your side.
No drawing, no sketch, no photo. No clear and consistent informations.
Just the same questions without caring about our posts.

Helping can only work when you contribute.

I'm out.

Klaus

Hi
Here is a block diagram. So 12V from car battery going to Buck DCDC converter LM53600. 5V from DCDC output feeds into LDO which generates 3V3 MCP1826. This 3V3 powers the Micro, Bluetooth and WiFi .
To simulate current of 0.5A, I put a 5R load across the 3V3 output as the WiFi Bluetooth prototype is still being built. The WiFi Bluetooth can peak to around 0.5A.
The 12V from the car battery will not drop to say 5V. It will remain around 12V, the spec is that the circuitry should operate with a car battery voltage of 9V minimum.
So my question is why I can't draw 0.5A or even 0.66A with a 12V input, as I see only 280mA being drawn at the 3V3 output, when it should be 0.66A. is there something wrong with the circuit?

 

[KlausST]

Hi,

eventually some information one can rely on.
But still far from complete.

Now please take a photo of the circuit that exactly shows the 280mA problem. From LM53600 input to load resistor including amperemeter.

Also use the exact configuration and do three voltage measurements at:
* 12V node
* 5V node
* 3.3V node

Klaus

 

[betwixt]

Thank you, it makes more sense now. Originally you were saying the current was from the 12V, in other words the whole thing wasn't drawing the current you expected, now you are saying the current in the load is too low.

Again, back to basics, if you connect a 5R load across 3V3 the current WILL be (I=V/R) 3.3/5 = 0.66A and there is nothing in physics that can change that. So if your current is lower it can only mean the resistance is higher than 5 Ohms or the voltage is lower than 3.3V. This is why Klaus is asking you to take actual voltage measurements. If any part of the circuit is 'overloaded' it will tend to make the voltages lower and as a consequence the current will be lower too.

Take the voltage measurements with the 5R load connected!

Brian.

 

[Rajinder1268]

Hi,

eventually some information one can rely on.
But still far from complete.

Now please take a photo of the circuit that exactly shows the 280mA problem. From LM53600 input to load resistor including amperemeter.

Also use the exact configuration and do three voltage measurements at:
* 12V node
* 5V node
* 3.3V node

Klaus

The 12V is from a PSU and measures exactly 12V current limit of PSU set to 1A. 5V output is measuring at 5.001V. The 3V3 LDO is measuring 3.285V under load.

 

[KlausST]

Hi,

I really don´t understand why it´s so hard to shoot a photo.
Isn´t it your aim to find a solution?

The 3V3 LDO is measuring 3.285V under load.

There is no way that there are 3.285V and just 280mA on a 5 Ohms load. Ohm´s law will apply.
--> either you hide some information or you do something wrong.

Klaus

 

[betwixt]

Agreeing with Klaus, the voltages are all within acceptable limits but if you put 3.285V across a 5 Ohm resistor there is no way anything other than 657mA can flow through it. Either your 5 Ohm resistor is really 11.73 Ohms or you are measuring wrongly.

Brian.

 

[Rajinder1268]

Agreeing with Klaus, the voltages are all within acceptable limits but if you put 3.285V across a 5 Ohm resistor there is no way anything other than 657mA can flow through it. Either your 5 Ohm resistor is really 11.73 Ohms or you are measuring wrongly.

Brian.

Hi
I will get a picture out asap. Thanks for your help so far. This is what I can't understand it should be around 660mA but is 280mA.

 

[Rajinder1268]

Hi all,
Sorry for the late reply. Been off from work. I re-measured the current on the load 5R, by using a metre in series and the current was at 620mA. So everything is fine.