Measure ferrite bead

[knicklicht]

Hello,
I have a couple of SMD ferrite beads of unknown impedance. However, googling only lead me to solutions involving expencive equipment the cheapest being a NanoVNA as impedance is measured at 100MHz. I would like to know if there is a cheap circuit I can build to determine the values? I found someone who briefly mentioned a voltage divider and an oscillator but did not go into details. Can someone help me?

 

[dick_freebird]

You could use a signal generator and two 'scope probes,
and a load resistor (whose frequency response you know).
Dial the frequency and record datapoints for input from
sig gen and output amplitude into the load.

I don't imagine you need a whole lot of points, and the
post-analysis seems like Excel.

 

[BradtheRad]

Starting with a ferrite bead I made a 'joule thief' following instructions at:

bigclive.com/joule.htm

You turn the bead into a transformer by looping two thin wires through it. The circuit oscillates at a frequency determined by the ferrite's characteristics. (50 kHz is typical).

 

[knicklicht]

You could use a signal generator and two 'scope probes,
and a load resistor (whose frequency response you know).
Dial the frequency and record datapoints for input from
sig gen and output amplitude into the load.

I don't imagine you need a whole lot of points, and the
post-analysis seems like Excel.

Could you go into more detail please? I am very new to analog circuts. Is the type of wave important? Could I use a 100MHz square wave? How would the probes be connected? From the measured voltages how do I get impedance?

 

[KlausST]

Hi,

Square wave is a mixture of sinewaves of different frequencies. Thus you can't get a result for a dedicated frequency.
From the measurement view you can't compare a 1MHz sinewave with a 1MHz square wave.

Impedance:
Just use Ohm's law: impedance = V / I.
Impedance is for AC. V and I should be RMS values.

How to connect?
Any circuit you can think of. You just need to be able to measure:
* the voltage across the DUT (device under test)
* the current through the DUT

It's the same for any (linear) device, like resistor, capacitor, inductor or any combination of them.

Example circuits:
Do an internet search, or read
* some datasheets for test conditions or test circuits...
* or according application notes.

All information is available for free. If I had to design it I'd do internet search, too.

Klaus

 

[betwixt]

For simplicity and to overcome wave shape problems, maybe a Wheatstone bridge might be an alternative solution, at least if lower frequencies can be used.

Brian

 

[c_mitra]

Could you go into more detail please? I am very new to analog circuts. Is the type of wave important? Could I use a 100MHz square wave? How would the probes be connected? From the measured voltages how do I get impedance?

You may need a sine wave signal generator where you can select the frequency. You may measure the impedance at 10kHz, 20kHz, 50kHz and 100kHz.

You will also need another good resistor, says 1k or 10k (depending on the output voltage of the signal generator.

I guess you have the SMD beads (they have no holes) and look like resistors.

If you have the beads with holes, they there will be some small difference.

you put the known resistor in series with ferrite bead. then apply the connection from the signal generator. note down the frequency.

Measure the voltage across the known resistor and the ferrite bead using a multimeter. You can also use a scope (if you have one).

Note down the two voltages and the frequency.

Calculate the current from the resistor value and the measured voltage.

Calculate the impedance from the measured current and the measured voltage.

Do that for the 4-5 diff freq. plot the result in a graph.

Decide the usefulness of the ferrite bead for your application.

 

[KlausST]

Hi

a Wheatstone bridge might be an alternative

Does it work properly with phase shift in bead and square wave excitation?

Klaus

 

[betwixt]

I would guess it probably would. The bridge (assuming the other three arms are resistive) would still give a null out when R=Z. It may not be a perfect null but pretty close. The purely resistive element of the inductor can be measured at DC with a DVM. I confess the last time I tried this was about 30 years before SMD parts hit the market!

Brian.

 

[FvM]

In a first order estimation, all SMD ferrite beads on the market are behaving as inductors up to at least 10 MHz. If you eleminate the DC series resistance in measurement respectively measure only the inductive part of the Ls + Rs series circuit, you can measure it at any frequency up to 10 MHz, depending on the frequency and inductance range of your LCR instrument or measurement circuit. Unfortunately, the inductive impedanz at 10 MHz doesn't tell much about the impedance at 100 MHz, it's rarely factor 10 higher, more typical 3 to 5.

This means, if you want to sort the ferrites according to its specified 100 MHz impedance, you need to measure at 100 MHz. If you want additionally to distinguish the ferrite type, you need measurements at multiple frequencies, in case of doubt with a VNA.

 

[KlausST]

Hi

In a first order estimation, all SMD ferrite beads on the market are behaving as inductors up to at least 10 MHz.

I thought these beads use "lossy core material", so at the given frequencies the energy is disspated as heat.
If so, it´s not that inductive with respect to phase shift.
But I´m not sure.

Klaus

 

[FvM]

Look at a typical ferrite bead complex impedance curve

 

[KlausST]

Hi

I see that Z rises with frequency. This is what we know from inductance.
But here the R rises, while with inductors X rises.
Energy becomes dissipated as heat, thus one does not get high Q resonance. I expect about no ringing.

Klaus

 

[c_mitra]

I thought these beads use "lossy core material", so at the given frequencies the energy is disspated as heat.
If so, it´s not that inductive with respect to phase shift.

You are right.

If you see the graph in post #12, impedance is dominated by X upto 40MHz. At 100MHz, R is twice as X. At 1000MHz, only skin conduction is important. At <10MHz, R is rather small and it is almost pure reactance

 

[FvM]

At 1000MHz, only skin conduction is important

The curves don't tell you if R is caused by skin effect or ferrite loss factor, generally you have a combination of both effects. Notice that above 500 MHz X is negative, in other words parallel capacitance becomes dominant.

Below shown is an empirical equivalent circuit for Murata BLM03AG601SN1, fitted to the published S parameters. The fact that losses are modelled by frequency independent parallel resistors doesn't mean that there's no skin effect, it's an equivalent circuit, not a physical model. The three curves shows a comparison of Z, R and X, measured S-parameter (blue) versus model (magenta).

--- Updated Nov 7, 2021 ---

Here the measured complex impedance values in a single diagram

 

[c_mitra]

Thanks for the explanation! I was confused initially because Z and R are decreasing at high frequency.

The model is also interesting: two parallel LCR resonant tanks in series. The LC values are 0.945 and 0.216 (10^-18)

The shunt resistors are nearly equal: 548 vs 515. The series resistor 1.5 can be safely ignored.

That is why the R is closely following Z and even though we have two LCR in series, we have only one peak in Z. X shows two peaks as expected.

Are these beads recommended for use below 10MHz? At 100MHz 50% signal will be gone apparently.

 

[FvM]

The model is also interesting: two parallel LCR resonant tanks in series.

As said, its an equivalent circuit fitted to the impedance curve. Modelling empirical impedance curves by ladder circuits is common approach, used e.g. for batteries or lossy capacitors. Physically, there are no separate LC tanks. It's just so that ladder two elements give a good approximation in this case, in other cases, more may be neccessary.

Are these beads recommended for use below 10MHz?

A characteristic parameter for ferrite beads is the crossing frequency of R and |X|. It can be below 10 MHz for some SMD ferrites, different from 40 to 60 MHz observed with the previously discussed types. This one gives better damping of resonances at lower frequencies.

 

[D.A.(Tony)Stewart]

If You can measure ferrite resistance like 50k with a multimeter and then they're VHF beads, otherwise is more like insulators then they are HF beads. The difference is the Zn Mg content