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CD74HC08-EP: CPD spec per gate is Higher than other similar TI parts.

Michael Pollock
Prodigy 10 points
Part Number: CD74HC08-EP

Hello TI Forum,

We are looking for alternative parts for our products, one of the parts we are considering full part number is CD74HC08QM96EP, but the data sheet is CD74HC08-EP.

For the test conditions this part shows the CPD to be 37 pf per gate and quiescent supply current at 40 micro amps, and propagation delay at 135 ns.

For another part SN74HC08DT, the CPD shows to be 20 pf per gate and quiescent supply current at 20 micro amps, and propagation delay at 125 ns.

Our circuit is just logic like turning on an enable line, so not a high frequency circuit, seems like this part would dissipate more heat than the other parts.  We are not sure 

this would be a good replacement, and not sure why this almost same part has higher values than other parts.  Any advice would be helpful,

Thanks,

Mike Pollock.

  • 0
    TI__Guru* 96862 points

    Hi Mike,

    Michael Pollock said:

    For the test conditions this part shows the CPD to be 37 pf per gate and quiescent supply current at 40 micro amps, and propagation delay at 135 ns.

    For another part SN74HC08DT, the CPD shows to be 20 pf per gate and quiescent supply current at 20 micro amps, and propagation delay at 125 ns.

    Michael Pollock said:
    seems like this part would dissipate more heat than the other parts.

    If your circuit is low frequency, the Cpd will have no practical impact on heat operation. Any CMOS logic gate will draw very little power. Here's the equation for transient power consumption (from [FAQ] How do I Calculate Power Consumption or Current Consumption for my CMOS Logic Device? ):

    P_T = C_pd * V_cc^2 * f_I * N_SW

    If you have 8 inputs, all operating at less than 1 Hz (typical for enable signals), 5-V supply, your power calculation would be:

    P_T = 37E-12 * 5^2 * 1 * 8 = 7.4 nW

    Even with an awful R_thetaJA = 400 C/W, that would still be less than 0.000003 C increase.

    The constant current draw of 40uA will add 200uW (P = I * V = 40E-6 * 5), which results in less than 0.08 C increase with the same assumed R_thetaJA. You should note that 40uA is the maximum (ie worst case) rating of the device, not the power that would typically be used.

    -

    Why are you using the -EP product instead of a standard commercial device? Are you operating at extreme temperatures?