Hi everybody
I have a DMM it's VC9805. For measuring the output current of this Buck module, I used it this way:

Then I connected the red test lead to 20A and set it to DC mode(it is in DC mode itself) and connected the test leads to the output of the module. and when I turned the supply on, BOOOOM!

Burnt out.
Why? (I think there is a big load in DMM to draw current)

I want to measure the current on my board. I want to know Is my MCU(STM32F103) drawing current or not(want to know it's functioning/working or not). but I'm worry that maybe it would burnt out my debugger(it's J-Link) or my MCU. Should I be worry?

 

Can you draw a diagram of exactly how you connected the meter?

You know the meter should be connected in series with the load if you want to measure current?

 

I just connected the red test lead to +output and the black to the -output.
Is it incorrect?

 

I just connected the red test lead to +output and the black to the -output.
Is it incorrect?

Nope.

 

Nope.

?

 

Hello,

On what range setting was the DMM?
What is damaged?
The powersupply or the DMM?

Bertus

 

Nope.

But probably it is!
if it's not, then why it's burnt out?

 

Hello,

On what range setting was the DMM?
What is damaged?
The powersupply or the DMM?

Bertus

1- 20A
2- LM2596
3- LM2596 on the Buck module

 

But probably it is!
if it's not, then why it's burnt out?

This is not how you measure current; you shorted-circuited the output of the buck converter.

The meter should be in series with the load.

 

Hello,

In the link is given that the Buck module can deliver 3 Amps.
The 20 A connection will act as a short.
The buck converter is likely NOT short circuit protected.

Bertus

 

Hi everybody
I have a DMM it's VC9805. For measuring the output current of this Buck module, I used it this way:

Then I connected the red test lead to 20A and set it to DC mode(it is in DC mode itself) and connected the test leads to the output of the module. and when I turned the supply on, BOOOOM!

Burnt out.
Why? (I think there is a big load in DMM to draw current)

I want to measure the current on my board. I want to know Is my MCU(STM32F103) drawing current or not(want to know it's functioning/working or not). but I'm worry that maybe it would burnt out my debugger(it's J-Link) or my MCU. Should I be worry?

First you need to check for short circuit condition.
This is accomplished by putting a light bulb in series with your circuit.
The light bulb should be the same wattage you expect from your circuit.
If light is about half brightness, then you do not have a dead short.
If light is on full brightness, then you have a dead short.
If light is off, then you have a open circuit.

Best Regards

 

sounds like the DVM was connected directly across the supply in which case it is a short circuit. If the supply blew up then it did not have a short circuit protection capability.

 

I just wanted to know can it give me 3A really.
This is the schematic of my DMM:

How could I measure the max current at the output of my Buck module?

 

Hello,

Did you read the specs of the buck converter in the link you provided?

• Input: DC 3V to 40V (input voltage must be higher than the output voltage to 1.5v above can not boost)
• Output: DC 1.5V to 35V voltage continuously adjustable, high-efficiency maximum output current of 3A.
• Features: All SANYO solid capacitors, the 36u thickening circuit boards, high-Q inductance with output value of high-power LED indicator
• Dimensions: 45 (L) * 20 (W) * 14 (H) mm (with potentiometer)

It states a maximum output current of 3 A.

Bertus

 

You need a circuit like this:



The resistor sets the current and must have a value of

R = (Expected Voltage) / (Expected Current)

And this resistor must have a power rating of (Expected Voltage) x (Expected Current) x 5

 

But probably it is! if it's not, then why it's burnt out?

By design, a voltmeter has a very high input impedance or resistance. It tries to look like an infinitely large resistor that drawn no current.

However, a current meter is the opposite. It tries to look like zero ohms. If you are measuring the output current of a power supply that has built-in current limiting, the supply will protect itself and the meter will show the short-circuit current. But most buck regulators do not have this kind of protection built in.

When the manufacturer says the regulator is good for 3 A, he means that you can draw more than 3 A if you want, but it might damage the regulator. That is what happened in your case. If you want to confirm that the device can supply 3 A, you need to connect a load resistor that is calculated (with Ohm's Law) to draw 3 A at whatever the output voltage is, and put your meter in series with it.

ak

 

If you fried your DMM then R20 (10mΩ) and pcb track around it have probably been destroyed .

 

The safest. Use a clamp on ammeter add-on to your meter.
Max.

 

Thanks guys

Did you read the specs of the buck converter in the link you provided?
It states a maximum output current of 3 A.

Yeah, sure! but I would see it actually.

 

First you need to check for short circuit condition.
This is accomplished by putting a light bulb in series with your circuit.
The light bulb should be the same wattage you expect from your circuit.
If light is about half brightness, then you do not have a dead short.
If light is on full brightness, then you have a dead short.
If light is off, then you have a open circuit.

Best Regards

I would use a fuse for the maximum amperage you would expect in your circuit.

Best Regards Again