So I have an ultrasound system operating in the 1-10MHz range:

Pulser -> Tuning circuit -> Transducer -> Tuning circuit -> Voltage amplifier

The goal is, I'm trying to optimize the tuning circuits to obtain maximum signal at the voltage amplifier, as measured by an oscilloscope. So far, I've tried varying impedance matched and non-impedance matched tuning circuits, and what I've found is that the impedance matched circuits obtain the best results. That is, when Zi=R-jX and Zo=R+jX. The tuning circuits I'm using are basic L-match circuits.

What I'm wondering then is, do my experimental results that impedance matching obtains maximum signal fall in line with theory? I'm aware that impedance matching creates maximum power transfer, but maximum voltage is when Zi=0 and Zo=inf, so I'm confused on why maximum power transfer obtains max voltage in my experiments. I would have thought that the tuning circuit which maximized Zo in relation to Zi would result in maximum voltage. In our system, the traces are short enough that reflections are not an issue.

An example circuit on the receive chain would be like this: Transducer -> Tuning circuit -> Voltage amplifier input impedance Which might look like this


simulate this circuit – Schematic created using CircuitLab

So then Zi is the series combination of the Rsource & 2 caps, and Zo is the parallel combination of L and R2.

  • 1
    \$\begingroup\$ Show the positions of Zo and Zi on your diagram \$\endgroup\$ – Andy aka Apr 26 '17 at 17:10
  • \$\begingroup\$ "...Zo in relation to Zi..." - are Zo and Zi the 'tuning' circuit output and input impedances? What do the 'tuning' circuits consist of? What signal do you want to maximize? \$\endgroup\$ – Bruce Abbott Apr 26 '17 at 17:29
  • \$\begingroup\$ Check the edits. \$\endgroup\$ – WHL Apr 26 '17 at 17:53
  • \$\begingroup\$ Your transducer model is almost certainly wrong, usually they model as a series RLC network (with typically a small C, the motional capacitance) plus a rather large shunt C. \$\endgroup\$ – Dan Mills Apr 27 '17 at 10:10

Well, I don't know if the schematic you uploaded is what you have in your case, or it is only an example. Checking it, it has a mistake, and maybe it is the origin of the issue.

Let me explain it from the begining:

  1. For maximum output power, you need to match the external components of the Source to be its conjugate. That means that the real part seen from the Source must be 10 ohm, and the imaginary part must be the oposite than the produced by the 2nF internal capacitance.

  2. As you have 100ohm Rload, you need to convert it to an equivalent 10ohm. The only way to reduce the equivalent value of a resistor is adding another in parallel. That is why you add an inductor in parallel to R2. The L value is calculated to obtain 10 ohm as real part of this parallel equivalent reactance.

  3. For obtaining a Requivalent of 10ohm, you use a inductor that introduce a complex inductance, so you need to add a capacitor in serie (that works as a negative imaginary value) to remove this complex part.

  4. When you calculate the C needed to remove the imaginary part, you must take into account the existing C on the source.

To solve that, there is a few formulas used, that includes the Q value concept as explained in previous answers.


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The 1,06nF required is not what you need to place outside the Source, this is the equivalent C needed in all the LC network. As you have an internal 2nF capacitance, you need to calculate the outside C to get an equivalent value of 1,06nF.

The equivalent capacitance of 2 capacitors in serie es calculated as:

enter image description here

enter image description here

So what you need is an 2,255 nF capacitor instead of the 1nF that you have placed outside the source.


Consider this LowNoiseAmplifier:


simulate this circuit – Schematic created using CircuitLab

Z(2,000 pF at 10MHz) = -j7.5 ohms

You need to match.

  • \$\begingroup\$ I don't understand how this explanation illustrates why it is necessary to match impedances. \$\endgroup\$ – WHL Apr 26 '17 at 18:36
  • \$\begingroup\$ The Zin is capacitive, and very low. Thus energy is not transferred into the base, to be amplified, but wasted in the Cmiller. \$\endgroup\$ – analogsystemsrf Apr 27 '17 at 3:45

Firstly I think your transducer model is wrong, they usually look like a series RLC plus a shunt C, not just a series RC.

Next, for the receive side you probably actually want to match for minimum noise, not maximum power, these are usually not the same impedance.

An impedance step up L match will produce more voltage at its output then there was at the input, by a factor of its loaded Q and for a resistive load more power in the load means more voltage across the load.

Sure you can design a broadband high Z amp and hang it directly on the transducer (Actually a fairly common approach in wideband systems, but surprisingly tough to get right, and cable between the transducer and the amp can be problematic), but in a narrowband system matching to the amplifier noise impedance can get you significantly lower overall noise.

  • \$\begingroup\$ The impedance was measured from an impedance analyzer, so I believe this simplified model will suffice. Holding the load resistor constant and changing the tuning circuit (always ensuring the imaginary impedances cancel out), I have experimented with instances where Zi=Zo and where Zo>Zi, and the one that gets the most signal is when Zi=Zo. Now if I increase the load resistor and retune, I do get more voltage, so at least that makes intuitive sense, but there are limits to how high the load resistor can be as it is dependent on the voltage amplifier settings. \$\endgroup\$ – WHL Apr 27 '17 at 13:28
  • \$\begingroup\$ Watch what happens to be bandwidth as the Q rises. In the sonar game we often found that adding deliberate resistance to lower the network Q was necessary to get the desired bandwidth. As to the simplified model, maybe, but to get optimum bandwidth and power I usually found a 5th order match was the way to go at least on transmit. The magic usually involved a tuned transformer with the secondary inductance resonant with the fixed capacitance of the transducer and a L network (often with deliberate resistive damping) to match the power amp. \$\endgroup\$ – Dan Mills Apr 27 '17 at 14:39

Since you are transferring power, matched impedances always yield maximum output power. This always applies to linear systems.

Short answer

You have maximum power transfer when the voltage is 50% of no load voltage source. So you are never trying to maximize voltage but achieve 50% of the no load source with a conjugate matched impedance.

Other misc ticky tacky info on MPPT

In the case of nonlinear source such as diodes or current sources, the maximum power transfer occurs when the incremental source impedance matches the incremental load impedance. For non reactive Z, we call this ESR and for nonlinear PV arrays, the MPPT usually at the cube root of 50% Voc or the Voc/Isc load line. but since Isc changes with solarity from 0 to some Imax the incremental source impedance has his effect of changing from the square root of 50%=70% for low sun power to the 4th power root of 50% = 84% for highest sun power. (from my research, sorry no ref's)


However source loss will also equal load loss, so minimizing both where possible improves on the maximum power for any given supply voltage.

You are converting DC power into mechanical piezo crystal motion and there will always be a conversion loss defined by the load crystal ESR (effective series resistance at f). Therefore minimizing cable and switch losses ensures you get the most output.

Piezo ultrasound devices like all crystals and ceramic resonators are defined by at least a R-L-Cs//Cp discrete parameters and not a simple resistor, so conjugate impedance matching is implied, ( which is a bigger topic) but the main concept here is to at least match the resistances. To examine your cct. the filter presents an impedance to the load of RLC of (10R+30nF)//1uH which resonates at 3MHz and |X(f)| for ~19 Ohms at 0 deg at LC resonance. The filter BW & rise time can be derived from Q = |X(f)|/R = 19/10 = 1.9 .. where X(f)= 2pi*3MHz*1uH = 19 ohms . Here a low Q gives wide bandwidth or fast rise time , a desirable characteristic.

We don't know Z(f)(piezo) or maximum power it can handle, so I cant judge what is optimal but sometimes, you don't want maximum power transfer but rather lower source impedance so that load variations do not affect the load voltage.

In this case the source is much lower than the load R you show, so there are other factors besides maximum power transfer such as load regulation just as in AC & DC power supplies where the source is always < 10% Z load and in the case of 1% regulation source Z is ~1% of load and in the case of the AC grid it may be 8~10% of the min. load Z but then network design with tap changers accommodate load fluctuations to give much better load regulation.

The reason matching is important is that power reflects back to the source if the interface impedances are not matched resulting in lower forward power. These are explained by an understanding of scattering parameters or s-parms such as s11 s21 and s22 the input return loss, transmission loss and output reflected return loss.

Having no load certainly gives you maximum output voltage but with no work being done and no power and energy being transferred.

Whenever there are time delays in a channel, reflections from mismatched impedances can cause image ghosting. But then your power efficiency is 50% at best. This how RF systems work in order to prevent echoes or ghosting or ringing in pulsed carriers. Since the load also depends on acoustic coupling of the ultrasonic waves to the target, and the target load impedance, this must also be considered when high resolution image artifacts or bogey targets must be avoided. This is one reason why fish finders cannot work close to the boat hull since that interface causes some strong reflections which must be masked. ( gated off)

  • \$\begingroup\$ If I have made any incorrect assumptions, pls feel free to correct them. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 15 '17 at 6:03

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