XTR300: XTR300

Hi Junzhe,

Would you be able to provide more information? I have a lot of questions like:

• What is the output signal you are seeing?
• What does the input signal look in relationship to the output signal?
• Can you share the schematic? 
• Where are you planning on adding a buffer? 

Best Regards,

Robert Clifton 

Hi Robert,

This is my schematic, I use current output mode.

And this is my input signal parameter, I want the output generate a square wave  ±10mA at the output.

When I run the transient simulation, the output is not a perfect square wave.

I try to delete C2 which use to protect the device, and run transient again, it looks like this.

It seems better but still not perfect. Is this due to the RC circuit I'm using to simulate the load? How can I make the output better under a 100us pulse width? 

I would like to add a buffer between the output and the load, but the XTR300 output is current mode, so I don't think this is feasible.

Best,

Junzhe

Hi Junzhe,

Thank you so much for sending me the simulation. I'm going to be honest, when I ran the simulations, I didn't get quite the same results as you did but I did see that the output was less of a square wave when I had it configured with a 10nF capacitor rather than a 100nF. 

Here's the results with the 100nF capacitor. 

XTR300 Square Wave.TSC

Best Regards,

Robert Clifton

Hi Robert,

Thanks a lot for your reply and simulation file. I ran your simulation and applied your settings to mine. The results of the two simulations are very close. And the better effect is achieved when the capacitor in the RC load is selected as 100nF rather than 10nF.

But in your simulation, resistor C2 is deleted. In the XTR300 data sheet, C2 and R2 are used to protect the device. Will deleting C2 to achieve a better output result make the chip easy to damage? In addition, will adding a diode for protection affect the output effect?

Best,

Junzhe

Also, why does the waveform get better when the capacitance value of the RC load is increased?

Hi Junzhe,

Junzhe Zhang said:

But in your simulation, resistor C2 is deleted. In the XTR300 data sheet, C2 and R2 are used to protect the device. Will deleting C2 to achieve a better output result make the chip easy to damage? 

If you are referring to C2 on your schematic, I technically do have C2 included in mine as C2 and C1 are two capacitors in parallel. Two capacitors in parallel is the equivalent of a single capacitor with both capacitances added together. As far as if it would make it easier to damage, the capacitor shouldn't more or less damage to the output of the device. 

Junzhe Zhang said:

In addition, will adding a diode for protection affect the output effect?

The diode will have some parasitic properties that the XTR300 will see. We actually have diodes protecting the output on the XTR300EVM User's Guide (Rev. A) that you can use as a reference. 

Junzhe Zhang said:

Also, why does the waveform get better when the capacitance value of the RC load is increased?

It is strange. It could be a result of the limitations of the simulation model or an actual characteristic of the device. I'm not entirely sure to be honest. It would be best to verify this test on an actual device. 

Best Regards,

Robert Clifton 

Hi Robert,

Thank you for your reply. For question 1, In my design, I use R2 and C2 as protection, C1 and R24 are used to simulate a real load (including resistance and capacitance.) In your design, you use R10 as protection  which is clear, but for C1, you treat it as protection together with R10, or you treat it as the capacitance part of the load? In your reply, I understand that you regard C1 as protection together with R10?

For question 3, how do I do this test in practice? An oscilloscope can only display voltage waveforms, and it is difficult for a multimeter to obtain good results with such a short pulse width.

Best,

Junzhe

Hi Junzhe,

Junzhe Zhang said:

In your design, you use R10 as protection  which is clear, but for C1, you treat it as protection together with R10, or you treat it as the capacitance part of the load? In your reply, I understand that you regard C1 as protection together with R10?

I'm treating C1 in my circuit as just the protection scheme. I didn't include a capacitance as part of the load. Sorry for the confusion there. 

Junzhe Zhang said:

For question 3, how do I do this test in practice? An oscilloscope can only display voltage waveforms, and it is difficult for a multimeter to obtain good results with such a short pulse width.

Ah yes. That can be quite tricky to verify. There does exist dedicated current probes for many different oscilloscope models. I'm not sure if you have access to them.

Alternatively, you can look at the voltage output and see if it comes close to what you would expect from the simulations. 

When using a 100nF capacitor the output in voltage looks closer to triangle waves:

When using a 10nF capacitor, the output in voltage looks closer to the original square wave. This makes sense as the filter isn't as aggressive as it is with the 100nF. 

Best Regards, 

Robert Clifton 

Hi Robert,

Thank you for your explanation on question 1. So you assume the load is purely resistive. But my understanding is that the actual stimulus object is rarely purely resistive. So if C1 and R10 are regarded as loads, can this better simulate the actual working effect of XTR300?

But does this mean that the original capacitor used for protection must be cancelled? Because if two capacitors are connected in parallel, the waveform effect will become bad. This corresponds to my first result.

Therefore, in my subsequent design, I only retained the 15ohm protection resistor, and the capacitor and another resistor in the circuit together formed an RC load. The result is fine. I don't know if my understanding is correct,  could you please  check with this?

And thanks for your simulation result for question 3. I will try to test in actual.

Best,

Junzhe

Hi Junzhe,

Junzhe Zhang said:

So you assume the load is purely resistive. But my understanding is that the actual stimulus object is rarely purely resistive. So if C1 and R10 are regarded as loads, can this better simulate the actual working effect of XTR300?

You know you have a fair point there. Even if the load is purely resistive, it's usually at the end of the long cables that have capacitance. That's something that my simulation circuit didn't have.

Junzhe Zhang said:

But does this mean that the original capacitor used for protection must be cancelled? Because if two capacitors are connected in parallel, the waveform effect will become bad. This corresponds to my first result.

I disagree with this conclusion. I say keep the C1, and whatever the expected parasitic capacitance that is expected from the cables, represented as C2 as shown below: 

I don't see the same level of distortion as you are seeing from your tests. 

Best Regards,

Robert Clifton 

Hi Robert,

Thank you for your reply. I built the circuit same as yours, but I put 2 current probes. One after DRV (ILoad), another between C1 and C2 (AM1). 

After I run the simulation, the green one is from ILoad which is almost same as your result. But the red waveform from AM1 takes on the shape of a triangle wave. I am confused about this. Does this mean that at high frequencies, the original output current of the XTR300 becomes different after passing through R2 and C1?

Best,

Junzhe

Hi Junzhe,

Junzhe Zhang said:

Does this mean that at high frequencies, the original output current of the XTR300 becomes different after passing through R2 and C1?

Yes, that is correct.  In your circuit R2 and C1 act as a low pass filter that will distort the high frequencies. I chose C1 at 100nF as a starting point, but you might find that you will need to adjust it to better work in your system. Then of course prove it in the real circuit. 

Best Regards,

Robert Clifton 

Hi Robert,

Thank you for your reply. But the value for C1 is 100 nf, R2 is 15ohm. The cut off frequency for this low pass filter should be about 106KHz. The current signal output is only 5KHz. I don’t think it will be like a triangle wave.

Best,

Junzhe

Hi Junzhe,

No need to speculate what is happening here. You can test it in the simulation by doing an AC analysis. Ctr+Alt+A is the shortcut to start the simulation. 

I went ahead and ran a simulation 

It looks like that the cutoff frequency is occurring sooner than what a simple RC lowpass circuit would do. This tells me that the XTR300's internal output impedance is also playing a factor in the total resistance of the R value. You will also see that I marked the 5kHz value showing it's well past the cutoff frequency for the filter response. You will need to lower the capacitance on the output to keep getting a square wave on the output. I recommend tinkering with the simulation to determine what will be required. 

Best Regards,

Robert Clifton 

Hi Robert,

Thank you for your reply. Does that mean that if I want to use XTR300 to provide accurate current output at very short pulse width, both the RC combination value for protection and the RC value of the load need to be carefully considered? Because even if the output current of the DRV pin is good, the current received by the load will be different from the current output by DRV due to the influence of the subsequent circuit?

However, since I want to use the XTR's current output for bioelectric stimulation, I can't predict the exact RC value for an unknown load object, and I may only know the approximate range. Will this cause the current received by the stimulated object to be inconsistent with what I want to achieve? Does this also mean that the XTR300 chip is not suitable for bioelectric stimulation applications?

Best,

Junzhe