LMX2594: CPOUT (PIN 12) & VREGIN(PIN 10)

THANKs for the reply Dean, what is the I/O standard for the input & outputs. The block diagram shows the outputs to be terminate to VCC through a resistance

For the SPI, we might have some resistors, but this is just in case you driving it with more than 3.3V, so these resistors are not necessary.

As for the RFoutA and RFoutB, yes, you need a pull-up component. This sets the impedance, so we use 50 ohm. You can also use an inductor for more power, but then you might want to follow it with a pad for better match.

thanks for the reply Dean, i can wait till the we get the current nos.

Siddharth,

I think that what may be propmting the question about the currents through individual pins is series resistance we show in the schematic. However, these resistors are all zero. We had resistors here to allow the possibility of tinkering with them to try to optimize spurs, but they did not have much of an impact. So as they are zero ohms, the drop accross them will be zero.

It would be nice when we get the time/resources to do a full analysis at every pin, but that might be a while. For now, I can take a ballpark estimate -- not fully verified or an actual measurement).

Expect the majority of the current to go through VccBUF, VccPLL, and VccVCO. The output buffer is on the order of 95 mA, the PLL and VCO in the 50-100 mA range (ballpark estimate). VccDIG and VccMASH I would expect to be low (<10 mA, but not sure), and VccCP is the charge pump supply, so something on the order of the charge pump current.

Regards,
Dean

Siddharth,

The outputs are open collector and are pulled to Vcc.  The output voltage varies with Vcc, OUTA_PWR setting, and frequency and is expressed in dBm for some of the typical performance charts.  At higher frequencies, this resembles a sine wave, but closer to a square wave at lower frequencies.

There is no Vcc Pin.  There is VccDIG, VccMASH, and VccCP.  I would expect VccCP and VccDIG to have higher current.

As for the series resistors, note that there are not any in the typical application schematic on section 8.2. It recommends shorting the supplies together and this is why we don't have individual current numbers per pins.  On our EVM instructions, figure 12, we do show series resistors, but they are all 0 ohm.  I thought this was what was prompting your questions.

Regards,

Dean

Siddharth,

Neither. Let's approach this another way:

For a rough estimate, forget about the pin names for a second and think of it by block. TICSPro is useful for analyzing this.
Also, realize that current varies on many factors not included, but this is a ballpark idea.

Core Current (PLL Locked, both outputs off, output divider Disabled):
Icc: 248 mA
Supply Pins:
VccVCO, VCCVCO2, VccMASH, VccCP, VccDIG

For VccVCO and VccVCO2, these are shorted intenrally. I though the VCO took about 80 mA
This leaves about 168 mA, which seems a lit high for the PLL (I was expecting 80), but the PLL is supplied by VccMASH, VccCP and VccDIG. VccMASH is on the order of 5-10 mA if in fractional mode, VccCP is on the order of 10 mA. I don't think

Output Current (one output ,OUTx_PWR=31):
Icc: 96 mA
Supply Pin: VccBUF, and Pull-up Component

Output Divider Current:
39 mA


Although you would probably like a list of pins and currents for each, we don't have this and these pins should be shorted together anyways. That being said, this should be sufficient information to figure out what LDO to use for this device. In general, one typically powers the whole device off of one LDO. In some cases, the pull-ups can be on another supply.

Regards,
Dean

Pins 22,23, 18, and 19 are open collector outputs, which means that the output is switching between Vcc and ground. However, depending on the OUTx_PWR setting, it coudl be a lot less and also, it probably doesn't switch all the way to ground, but closer to 1V. In any case, these outputs must be AC coupled.

There is no special supply sequencing required for the supplies.

Regards,
Dean

Siddharth,

No, this seems off.

Datasheet says one output power up is about 340 mA total.
Of this current, about 95 of this is the output buffer, leaving 255 mA.
Of this current, about 80 of this is the VCO leaving, 175 mA
So that leaves 175 mA for the Input path (OSCin) of the PLL, PLL charge pump, PLL Mash Engine, and pull-up resistors. There might also be some muxes somewhere burning some of this current.

If you download our TICSPro software, you can configure it and it will estimate the current based on the configuration.

Regards,
Dean

One output powered up means that one of the following is true:
OUTA_PD=0 and OUTB_PD=1

or

OUTA_PD=1 and OUTB_PD=0

Regards,
Dean

TICSPro software is at www.ti.com/.../TICSPRO-SW

You can reduce current consumption a little, but not anywhere down to 104 mA. If you turn out an output, that saves you 95 mA, but device is useless with both outputs powered off.
You could certainly go 100 mA more if you turn on another output.

REgard,s
Dean

Siddharth,

Yes.  This is correct.

Regards,

Dean

thanks Dean. I was trying to understand your previous reply on current draw. So at no load the current draw will be 248ma right? is this no. certain or is there any doubt in this as well?

Also it appears from your reply that its not certain how the no load 248ma is appearing? because the constituent supplies VCC_MASH, VCC_DIG don't consume more than 10ma as per you. Am i correct?

Siddharth,

"no load"  is to vague.  I mean OUTA_PD = OUTB_PD = 1.   The 248 mA is fairly predictable, but it does vary based on setup conditions charge pump, phase detector).  Remember in this state, the PLL and VCO are still both powered up and running.  A lot of the current will be going through VccVCO/VccVCO2, then next could be VccCP.  The PLL current can be divided amoung several pins.

What is prompting all these questions about current per pin?   The supplies are shorted on our EVM.

Regards,

Dean