DAC3484 Power Consumption

Hello,

I have a question regarding power supply of DAC3484.

DAC3484 need voltage in a range from 3.3V to 1.2V. On my board I have 5V voltage supply, which I will convert into 3.3V and 1.2V. 3.3V is an analog power supply. There are three different 1.2V pin types, which should be separated from each other - CLKVDD, DACVDD, DIGVDD. That means I need to take different regulators for each of them.

I want to produce a Signal of 10MHZ, Will it influence the quality of my signal if I take 5V convert it with the LDO Regulator to 3.3V, which I will use as analog supply and then take the same 3.3V and convert them with 3 different regulators (parallel) to 1.2V?

So the chain would look like this: 5V 3.3V ANALOG SUPPLY
1.2V CLKVDD
1.2V DACVDD
1.2V DIGVDD


Or should I convert the 5V to 3.3 (or 3.4V dependent on Dropout Voltage of LDO) and then use 4 separate regulators? (So 5 in total)
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also about pin current consumption:

I have rather simple question for all with experience of reading datasheets:
In the datasheet of DAC3484, page 12, under power consumption we can find current consumption values of different pins. My question:
Are all currents - IAVDD , IPLLAVDD, IOVDD, need 123 mA (Typ) or each of them? (123mA * 3 = 369 mA? )
And all currents of one type need the value which is written in the datasheet? For example MODE1: ALL IDACVDD pins need 35 mA, not that each needs 35 mA.
Do I understand it right?

best regards

Den

For the power supply design you might refer to the evaluation board schematic. It e.g. uses a 1.2V switcher for 1.sV digital and a common 1.2V LDO for clock and DAC VDD in the default configuration.

There's a very clear current consumption specification for each supply node in the datasheet, in case of doubt check with the total power that's also specified.

FvM,
thank you for reply,

I don't want to use switching regulators, they produce too much noise. And I have the schematics. What I want to do, again, is to drop the 5V voltage to 3.3V and use it for analog supply AND use those 3.3V for 1.2V supply. So my question,if it will be counted as isolated Analog vs Digital power supply? Since Dig.Power Supply generates the unwonted noise and it goes after analog supply anyway, so I though it would be ok.

an one more question. Can I chain LDOs lwith different current value? like drop 5V to 3.3V with LDO which can deliver 200mA and then drop it again to 1.2V with LDO which can deliver 1A? Will it deliver 1A?

best regards

den

Not using switching regulators is O.K. but will cause respectively higher power dissipation. I think, it should be sufficient to have one regulator for each voltage level and decouple the individual supply nodes by ferrite beads, as suggested by the evaluation board schematic.

You can chain LDOs, but why? It won't change the total power dissipation if all regulators are of linear type. If the higher voltage level can source 200 mA, then the sum of all supply currents drawn from derived voltage nodes must not exceed 200 mA.

hmm the second question is stupid, forget it. (about 1A after 200mA)

- - - Updated - - -

FvM,
thanks again.

If I calculate all Pds i get Pd(analog sup) = 229.5mW , Pd(clkvdd) = 361 mW, Pd (dacvdd) = 190mW - they are all ok, nothing to worry about. This chip would be good solution TPS79912-Q1. BUT, then comes Pd(digvdd) and this jerk dissipates = 2470 mW. even if I divide it by two LDOs = 1235mW is a lot. I really don't want to use switchers. Is there a way to solve it?

Also about those ferrite beads, i've never encountered them before, since I'm new to PCB design. How do they work? In schematics there are two types of them the one has Resistance and other Capacitance. for example in DAC3484 EVM Schematics, page 7, FB34 has 1kOhm and FB30 has 27uF. What is the difference?
And one more, if i connect 1kOhm on the LDOs output the whole voltage will fall across this ferrite bead or does it work somehow different? How should I read its parameters? for example this one - http://www.mouser.com/ProductDetail/Murata-Electronics/BLM41PG102SN1L/?qs=%252bi%252b2JQhhbn3KYzONFOUA/g== ( 81-BLM41PG102SN1L)

Danke in advance =)

Den

I don't see a perfect solution for the Vdd supply. You can either use a switcher or accept the power dissipation of a linear regulator.

Ferrite beads are lossy inductors, the specified resistance value, e.g. "1k" for the type used on the TI evaluation board, is an impedance at 100 MHz. The 3 terminal parts with "µF" specification are low-pass filters. Also not a bad idea, but less commonly used than regular ferrite beads.

Ferrite beads have a low DC resistance, for supply lines with higher currents, it's essential to use "power" ferrite beads with a respective current rating, e.g. 0.5 .. 2A. Otherwise too high voltage drop and inductor saturation must be expected. You can review the evaluation board BOM for the exact ferrite bead type used therin.