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How to measure high speed output voltage of LDO

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Hi,

I want to measure the output voltage of LDO voltage regulator circuit.
However, the output ripple is very sharp, so I think it has high frequency components.
My oscilloscope only supports up to 200MHz bandwidth, so I am worried about distortion from measurement.
I can borrow high bandwidth oscilloscope, but the problem is that it comes with only 50-ohm termination.
If I connect my output to 50-ohm termination, then output would be ground.
I don't know how to measure high speed voltage signal, which has relatively higher impedance than 50 ohm.

I thought of using decoupling cap but I also need to measure DC signal in order to get load regulation.
Is there any general way of measuring this type of signal?

Thank you in advance.
 

Read this article on scope probing.

You should be using a active probe with the high speed scope, you should not be using a passive probe on that scope.
 

Hi,

In very most cases an LDO has a fast capacitor at it's input and one at it's output.

At least 100nF...now at a frequency of 100MHz this gives 16mOhms of impedance.
If there is no output capacitor, then add one.
If there already is an output capacitor, then I doubt your measurement.

Klaus
 

Its rather surprising that such fast pulses can pass through any linear regulator.

Its much more likely that there is a grounding problem, and these pulses are originating from somewhere other than the linear voltage regulator.
 

Thank you for your help.
Actually, I have budget issue for buying active probes, so that I asked about bypassing this problem.
I will consider this.
 

Hi Klaus,

Actually I have an output capacitor, however, which size is sub-nF.
This is my target, so I can't put additional capacitor on output.

Thank you.
 

Hi Tony,

fast pulses are not passing through this LDO.
During regulation, the ripple is generated on output voltage, so I don't think it is from noise on ground.

Thank you.

- - - Updated - - -

I actually thought of introducing high bandwidth unity gain buffer for 50ohm termination by using commercial OP AMP IC (LTC6252).
The specification says it has relatively high input impedance (several kilo-ohm for differential and several mega-ohm for common mode signal) and low output impedance.
However, the output impedance varies with frequency, ranging from 0.02 to 10ohm, which would generate distortion. (I'm not sure this output impedance is measured in open loop or closed loop).

Do you guys think this is not a bad idea?
 

It think you're going off on a tangent with this. The problem is not the edge rates of the ripple, it's that you have ripple in the first place. Maybe you've got stability problems or the LDO doesn't have enough load when measuring the output or something else.

I'm not in any way an expert or even a power supply engineer, but I would think that posting your regulator design schematic and the parts used would help someone debug this. As it is you are now chasing a different problem, trying to measure the edge rate of some ripple. I'm not entirely sure what is meant by "However, the output ripple is very sharp, so I think it has high frequency components." as this gives no indication of what you consider as sharp and why you are convinced this is the root of the problem.
 

Hi ads,

Let me clear this.

I did simulation with my LDO design, and I found by dft of transient response that my output voltage has high frequency component. That is why I want to use high bandwidth oscilloscope.
The ripple happens only at the beginning of the transient response, so I don't think I have stability problem.

Actually, this problem can be interpreted as "how to measure the node voltage which has fast (high bandwidth) transient response with oscilloscope".

Thank you for your help.
 

Hi,

Again...
The input as well as the output of an LDO should be near DC. There should be capacitors blocking AC.
The higher the frequency, the lower the capacitance impedance, the better the attenuation.
Sharp edges need/generate high currents at a capacitor. Please do some calculations.

"Sharp" signal waveforms tell me there is something really wrong...either your LDO circuit or your measurement.

Klaus
 

Hi,

Thanks for your reply.

The higher the frequency, the lower the capacitance impedance, the better the attenuation.
Click to expand...
I don't get this part.

Let's just say that I want to measure some node which have high frequency and relatively higher impedance than 50ohm. How do I measure the node voltage with 50ohm oscilloscope.

and FYI, my measurement procedure is that when load current changes a lot and fast, then measure the voltage droop.
Sharp signal waveform does not necessarily mean something wrong. I intentionally put a small output capacitor as an experiment.
 

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