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I've been staring at the datasheet for NXP's TDA1308 headphone amplifier IC for a while, and I'm just feeling confused.

Their recommended application circuit (Figure 4 in the datasheet) looks like this:

TDA1308 Figure 4

Notice that Vdd, pin 8, isn't connected to anything labelled as a positive power supply. Instead, it's shown connected through what I believe to be a voltage divider, to ground. In fact, I can't see any positive supply anywhere in the figure.

So, the chip looks like it's not being powered, which is very confusing.

Might this be a schematic error? Perhaps the line that has the ⌜ corner in the top left of the figure should continue to a positive supply?

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It makes sense that pin 8 is connected to +Vdd. The pin is labeled with that in the datasheet and notice the 100uF (in parallel with 100nF for high frequencies) capacitor to decouple power supply from pin 8 to ground.

R1/R2 form a voltage divider that biases both amplifiers at half the supply voltage so you can suffice with a single power supply (and in- and output capacitors in the signal way).

It seems that the author of the diagram just forgot to include the power supply itself.

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  • \$\begingroup\$ Thanks. So, where in the schematic do you think the missing connection ought to be? Am I correct in expecting it somewhere in the top left corner, so that one side of R1 is connected to +Vdd? \$\endgroup\$ – unwind May 25 '13 at 17:03
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    \$\begingroup\$ It's a schematic diagram, not a physical layout---wherever you connect the power supply, as long as you connect it to the same wire, implies the same circuit. \$\endgroup\$ – The Photon May 25 '13 at 17:04
  • \$\begingroup\$ Physically, the best place to connect a power input is as close as possible to pin 8 of the IC. The resistor divider needs very little current so doesn't need a specially low-impedance connection. \$\endgroup\$ – The Photon May 25 '13 at 17:05
  • \$\begingroup\$ @ThePhoton Okay, thanks. I still find it weird that it's not included in the schematic (other schematics in the same datasheet show a connection). \$\endgroup\$ – unwind May 25 '13 at 17:07
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    \$\begingroup\$ Yes, it's a mistake by the person who drew the schematic. \$\endgroup\$ – The Photon May 25 '13 at 17:08
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I wouldn't say that it was a necessarily a mistake that Vdd was not drawn. There is a possibility it was a mistake that this sub-circuit was published as it is. Large schematics often have implicit (and global) power pins and sometimes don't even show the pins at all. So they may have drawn it wrong or they may have simply snipped it out of a larger schematic (with implicit rail connections) and forgot to add the rail connection back in.

Sometimes you'll get a whole page of decoupling caps which are meant to be sprinkled throughout the design (which I hate) but is common practise. As one example of schematic standards that are weird.

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