Hi everyone. I'm probably overthinking this but I am wanting to implement a pulse excitation of a 400PT160 ultrasonic transducer simply by using a I/O pin on a PIC microelectronic which is configured to output a 12.5us 3V3 pulse. The voltage will be amplifed using an IRLZ44 logic level gate drive MOSFET.

I would connect the I/O to the gate via a 100ohm series resistor (if needed). The source will be connected to ground and the transducer connected between 12V and the drain of the FET. I think this should mean that when the FET is off the voltage drop across the transducer will be 0V as both sides will be 12V. Then when the FET is on the voltage drop across the transducer will be effectively 12V as the Rds(on) is about 0.05ohm and the transducer impedance is 1Kohm.

Will this work or am I missing something? It seems much simpler than other excitation circuits I have seen but I can't see why this wouldn't work.

 

The source will be connected to ground and the transducer connected between 12V and the drain of the FET. I think this should mean that when the FET is off the voltage drop across the transducer will be 0V as both sides will be 12V.

No.

Your transducer is essentially a capacitor; while its impedance may be only 1000Ω at its center frequency, its DC resistance is many, many megohms. The first time your FET turns on it will charge this capacitance (≈ 2400 pF) and the capacitor will stay charged after the pulse has ended. You need to provide a discharge path for the transducer, in the form of either a resistor or an active pull-down.

 

No.

Your transducer is essentially a capacitor; while its impedance may be only 1000Ω at its center frequency, its DC resistance is many, many megohms. The first time your FET turns on it will charge this capacitance (≈ 2400 pF) and the capacitor will stay charged after the pulse has ended. You need to provide a discharge path for the transducer, in the form of either a resistor or an active pull-down.

This makes sense. Would a resistor in parallel across the FET drain and source leads work? and if so what kind resistance values are we talking about? I want to have as big a voltage drop across the transducer as possible, but also not dissipate too much across the resistor

 

This makes sense. Would a resistor in parallel across the FET drain and source leads work?

Nope. That's the opposite of what would work: a resistor across the transducer.

and if so what kind resistance values are we talking about? I want to have as big a voltage drop across the transducer as possible, but also not dissipate too much across the resistor

I imagine somewhere between a few hundred ohms and a thousand ohms might be a place to start. But a much better solution would be to drive the transducer with a hefty gate driver IC such as are used to drive MOSFETs that are required to switch fast. One that works well that I've used is the MCP1407. Much better, and far less wasteful of power, than using a passive pull-up.

 

Nope. That's the opposite of what would work: a resistor across the transducer.


I imagine somewhere between a few hundred ohms and a thousand ohms might be a place to start. But a much better solution would be to drive the transducer with a hefty gate driver IC such as are used to drive MOSFETs that are required to switch fast. One that works well that I've used is the MCP1407. Much better, and far less wasteful of power, than using a passive pull-up.

Thanks. Would I just connect the transducer across the two output pins in place of the MOSFET I was going to use?

 

Would I just connect the transducer across the two output pins in place of the MOSFET I was going to use?

No. You would connect the two output pins together and connect them to one end of the transducer, and connect the other end of the transducer to circuit ground.

 

great, thanks so much!

 

Sorry, I have another question about the driver and the current/power draw of the circuit. With a 12.5us on time and 500us off time duty cycle, when charging and discharging this 2400pF capacitive load, how much current would you expect it to draw and what minimum power supply rating should I use?

The datasheet confused me as a graph showed about 10mA draw but also stated it would need over 2A to charge that kind of load in 20ns

 

With a 12.5us on time and 500us off time duty cycle, when charging and discharging this 2400pF capacitive load, how much current would you expect it to draw...

Charging a 2400 pF capacitance to 12 volts will take Q = C * V (in coulombs, farads and volts) = 2.4E-09F * 12 V = 28.8 nC; if this repeats roughly 2000 times a second, that results in an average current of 2000 * 28.8 nC/s = 58 μA.

So simply supplying the current to operate the transducer will take roughly 60 μA; and if you're using a MCP1406/7 to drive it, add another 100-250 μA. So total current would be well under a milliamp.

...and what minimum power supply rating should I use?

Given the above, I don't think power supply rating is a limiting factor here.

The datasheet confused me as a graph showed about 10mA draw but also stated it would need over 2A to charge that kind of load in 20ns

What datasheet-- for the MCP1406, you mean? I don't see any graph showing that, nor any statement about taking 2A to charge that load. Could you point out where you're looking? Be specific.

In any case, driving even a few thousand pF to 12 volts in 20 ns will take a lot of current; I = C * dV / dT = 2.4E-09 * 12 / 20E-09 = 1.44A. But that current flows ONLY while the capacitance is charging (i.e., 20 ns) and, since there is roughly 500 μs between pulses, the average current will be well under a milliamp as I indicated above.

(If you're still concerned about the peak current, you can always reduce it by putting a resistor in series with the transducer; something in the range of 10Ω - 100Ω would probably suffice.)

 

Thanks for the clarification, I'm learning a lot. It was figure 2-14 and section 4.3 (though it was about 18v and 2500pF now I look again)

 

I'm curious about how you got a current draw of 10 mA from Fig. 2-14:



Looking at the graph, I can see that for a capacitive load of 2400 pF, the supply current would be roughly 10 mA at a pulse frequency of 200 kHz; but the frequency you're working at is only 1/100th of that (i.e., your pulse period is about 500 μs). Since supply current is approximately proportional to frequency (usually the case with CMOS devices), I would infer from the graph that at 2 kHz the supply current would be about 100 μA.

 

Hi, Catriona. Are you familiar with LTSpice? Lots of members in this forum use it to simulate circuits. Maybe this is a good moment for you to start learning it. That way you could play with your circuit and explore all of the "what if" scenarios you might be curious about. And best of all, it's free!

 

Hi, Catriona. Are you familiar with LTSpice? Lots of members in this forum use it to simulate circuits. Maybe this is a good moment for you to start learning it. That way you could play with your circuit and explore all of the "what if" scenarios you might be curious about. And best of all, it's free!

I do have it but I haven't had the time to sit down and learn how to use it yet. I'll maybe look into dedicating some time to that.

 

I do have it but I haven't had the time to sit down and learn how to use it yet. I'll maybe look into dedicating some time to that.

I suggest you try to make a simple first draft of what you have in mind and post it here ... then we can take it from there and help you however we can.

 

Worth to add that most of exciters are slightly resonant devices, not so much sharp as crystall oscillator quartz, but anyway. Simply there are a certain frequency, what must be obeyed. And adjusting it one may get the maximum mechanical power. or apply the 4046 for better automated response.

 

Hello everyone, thanks for your help so far but i'm afraid I have to bug you for more advice/assistance (I think this will end up happening at every stage of the hardware design!).

I completed the transmitter circuit using a FOD3182 MOSFET driver optocoupler (since my PIC dev kit is powered via my laptop USB and the rest of the system will be powered via 12V power supply).
I use a 12.5us pulse from the PIC microcontroller as a single excitation pulse for the transducer.


An issue I am seeing is that I have transient spikes at the rising and falling edge of my pulse when measured at the output of the optocoupler (at this point I haven’t checked if these transients are present when the output disconnected from the transducer). Obviously transients are a bad thing but I haven’t seen anyone else who do similar circuit complain of this happening.

My receiver circuit at the moment is just an INA129P in-amp with a second scope probe connected to the output. The gain is set to 10 at present.


My amplifier is also picking up these transient pulses and feeding them through to the output stage. Again, I am not sure at this point if my receiver picks up these transients when not connected to the circuit or it is picking them up on the breadboard. I have attached a picture of my waveforms where the transducers are 200mm apart. Both transmitter and receiver circuits are on the same breadboard and use the same 12V supply.


Is there any advice on how I can surpress these transients, or prevent them occurring in the first place? Is it my design, or the hardware that is causing the issue? (breadboard/scope probes)?

I apologies for my ignorance!

 

Both transmitter and receiver circuits are on the same breadboard and use the same 12V supply.

Uh-oh...

Is there any advice on how I can surpress these transients, or prevent them occurring in the first place?

Use two breadboards and separate power supplies. I suspect that will make a big difference.

 

Uh-oh...

Use two breadboards and separate power supplies. I suspect that will make a big difference.

I can certainly look to do this for prototype testing but when it comes to making PCB will I not see similar issues? Would using using two separate LDOs regulating from a single supply work OK?

Also, there will still be transient spikes that are higher than the 12V supply, should I not try to do something to suppress these even if it doesn't affect other circuitry?

 

I can certainly look to do this for prototype testing but when it comes to making PCB will I not see similar issues?

You will if you don't take care in laying out the board to avoid both capacitive and inductive cross-soupling between the driver circuitry and the receiver.

Would using using two separate LDOs regulating from a single supply work OK?

I don't know whether doing that will take care of the problem all by itself, but it should help.

Also, there will still be transient spikes that are higher than the 12V supply, should I not try to do something to suppress these even if it doesn't affect other circuitry?

You can probably minimize these by judicious use of decoupling capacitors.

 

Here's some reading on the subject: