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I'm trying to understand how BTL driver on TI's DRV8662 Piezo Haptic Driver (datasheet) works. What I am confused about is how a 200Vpp output signal can be generated from a driver that only has rails at GND and 100V. If I understand that datasheet correctly the driver looks something like this:

enter image description here

The output according to the datasheet is this:

enter image description here

What I am confused about is that the differential outputs have a common mode voltage of 50V (as shown in figure 8). So how is the voltage difference in between Out+ and Out- signals leads to a 200Vpp swing (as in figure 10) when figure 8 clearly shows that the maximum difference between these two voltages is 100V? How can the output swing to -100V if no negative rail is ever supplied to the part?

I tried reading these documents to clear up my issue but it didn't help.

BTL Speakerdrive Reference Note (link)

Audio Design Note (link)

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If you connect the +100V rail to the "positive" plate of the piezo, you get +100V over it. If you connect the +100V rail to the "negative" plate, you get -100V. So the achievable voltage swing is 200V p-p.

Think of the differential output as two half-bridge outputs driven independently. If you drive one output to a fixed voltage of +50V and the other one to the sine wave having 100V p-p, you will get sine wave having 100V p-p at the piezo. If you drive both outputs to the sine wave with DC offset of 50V, swing of 100V p-p and opposite phases, you double the voltage produced - that is 200V p-p.

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  • \$\begingroup\$ Ok I understand how you can get a 200Vp-p but you can never drive 200V across the Piezo because you are applying 100V to one end or the other. So the only advantage of this setup to a single ended driver is to avoid common mode noise? \$\endgroup\$ – EasyOhm Mar 31 '14 at 2:34
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    \$\begingroup\$ None of the diagrams you supplied suggests the chip ever drives 200V across the piezo. A sine wave is symmetric, so 200V p-p means 100V amplitude. This setup allows you to drive the piezo to an amplitude equal to the supply voltage, while a single-ended setup would allow at most half-supply-voltage amplitude. \$\endgroup\$ – motoprogger Mar 31 '14 at 2:36
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As opposed to single-ended signals that are ground-referenced, differential signals are self-referenced. This means that a differential signal made up of two single-ended lines that swing between 0 and +V can actually measure up to 2*V difference.

If we look at Figure 8 in your question, when the positive line is at 100V and the negative line is at 0, this is a difference of 100V (100V - 0V). When the positive line is at 0V and the negative line is at 100V, this is a difference of -100V (0V - 100V). This results in a total peak-to-peak voltage difference of 200V (100V - (-100V)).

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