Hi Kieran,
What power mode are you operating in? The LMH6550 model should work for high power mode. We do not currently have an available model or equivalent for medium and low power. Best regards, Sam Burnett
Part Number: LMP8350 Other Parts Discussed in Thread: LMH6550 Could you share with us about spice model for LMP8350? According to the following thread, it seems to be able to use the model of LMH6550 for LMP8350 when it is high power mode. https://e2e.ti…
Hi Ajay
This type of ADC generally has better performance and noise immunity if driven with a differential signal.
Something like the LMH6550 would probably be OK for conversion of SE to differential for that frequency of signal.
Best regards,
Jim B
Other Parts Discussed in Thread: ADC12040 , LMH6550 Please let me know about ADC12040's datasheet P18.
Customer planning,
・ADC12040 input is "Vin+/Vin-:2.5V"
・LMH6550 output is "Vin+/Vin-:2.0V"
Is it no ploblem?
Or, Are input and…
Hello, Loren
Thank you for the reply. I have transformer with 1:1 turn ratio. And I see in LMH6550 datasheet that all measurements were done at RL=500 Ohm (diff load resistance). Will LMH6550 behave properly at RL=20 Ohm? How about linearity? Voltage…
Other Parts Discussed in Thread: LMH6554 , LMH5401 , LMH6550 Hello,
I'm using the LMH6554 in SE to Differential configuration, DC coupled connectivity.
Similar to the scheme at Figure 27 in the datasheet:
The problem I encounter is high Group Delay…
Hello Todd,
The cable vendor should be able to give you the impedance of the cable. For example CAT5 cable is 100 Ohms of impedance.
If you use termination then the cable capacitance is not going to impact the amplifier, because a cable that is terminated…
Hi Kawai-san
I have not had success finding the allowed load current for these pins.
I do know light loading is acceptable, given what is shown in Figure 41 of the ADC10040 product datasheet. Since the Vcm pin of the LMH6550 can be connected to the Vcom…
One additional note...
It looks like similar to the LMH6550 datasheet where the equation for Rin (pg 16) can be simplified to Rin = 2 * Rg * (1+Av) / (2+Av) when Rm << Rg, such as the 0 ohms you suggested. In that case, for an Av of 2, Rin would be 1k…