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RE: AD1955 and transformer I/V stage

Posted by Dave Davenport (M) on June 10, 2007 at 12:50:59

In Reply to: RE: AD1955 and transformer I/V stage posted by SunRa on June 10, 2007 at 05:04:47:

Hi Florin,

I'll try to keep this simple to give you the concept. I will use examples and first-order approximations so don't hold the exact numbers, just the concepts. I’ll use the PCM1794 numbers because I am more familiar with them. It is not clear from the AD1955 specifications if their currents are specified the same way – apples to apples comparison.

The difference between the maximum current and minimum current is 7.8mA, which means that a maximum sine wave will have a 7.8mA peak-to-peak signal. We will keep signals at peak-to-peak in this discussion.

Let’s say for this example that you choose to have a 4V p-p maximum output signal and the gain of your amplifier is 4. Therefore you need a 1V p-p input signal.

The maximum 7.8mA signal is transferred through the 1:10 LL9206 and becomes 0.78mA at the output. Since you need 1V, Ohms law says 1/0.00078 = 1282Ohms for the I/V resistor. This 1282Ohms is transferred back across the transformer as the square of the turns (100) and becomes 12.82Ohms. This does not take into account the resistances of the windings.

At this point I don’t worry about the added resistance of the transformer because there are other variables that may come into play to make the 4 times gain of the amplifier an approximation. Now build the circuit using a 1K or a 1.2K or a 1.5K resistor for the I/V resistor. Put a test CD on and measure the output. You probably will have something between 3V and 5V output. The ratio of the measured output to the desired output is the same as the ratio of the I/V resistor that you used to the value of your final I/V resistor.

So, there are three factors: the impedance that the DAC sees, the gain in the system, and the output level. You get to choose two of the three and the third will be determined by the other two.

Dave

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