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Originally Posted by Herandu
Assume a signal input square wave with dV/dt of 250V/ms and 2V (i.e. CD output level) peak amplitude. The input waveform will reach 2V/250V/ms or
8ns, while the output will have changed (8×10-3) only 56mV. The
differential input signal is then (VIN-VO) RI/RI+RF or approximately
1V. The diode limiter will definitely be active and output distortion will
occur; therefore, VIN has to be <0.6V at all times.
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If you have a slew rate of 0.25V/us (or 250V/ms), it will take 8
us for it to change 2V. And where does the 56 mV come from?
Let's take another example. Assume that we have a delta sigma DAC capable of 192 kHz output (like the AK4396). The maximum voltage the DAC can swing is 5V, and the minimum time it can do this is 1/192000, or about 5us. This gives a maximum slew rate of 1V/us. In reality it will be perhaps 10% of this
at most, but this is the theoretically highest slewrate the DAC can achieve. Now let's look at the NE5534, it has a slew rate of 13V/us.
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Originally Posted by Herandu
If as you claim that under normal working conditions the differential input level is always 0V, then the DAC is not giving out any sound.
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Do you know how an opamp works? We are working with audio frequencies of up to 20 kHz. Compare this to the 10+ MHz bandwidth of an opamp. For all intents and purposes, the differential input voltage (that is, the difference between the inverting and non-inverting input) will be 0V.