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LM5104: Uneven propagation delay

Part Number: LM5104

I'm doing my initial testing with the LM5104, and observing some odd asymmetry in the propagation delay during IN rising versus IN falling.

My configuration in this test:

  • Vdd = Vhb = 12V
  • 10ohm Gate resistors for HO and LO
  • RT = 10k for a nominal delay time of 90ns (58ns min, 130ns max according to datasheet table 6.5)
  • Cb = 470nF
  • External load FETs: IRFR1205PBF
  • IN: 100kHz, 0 to 3.3V, variable duty cycle
  • My output half bridge supply voltage is 20V
  • Very light load attached to my output, ~1kohm.

I can see that the specified propagation delays are typically as follows below:


  • IN rising to LO falling: 25ns typical
  • IN falling to HO falling: 25ns typical
  • Trt should be ~90ns typical

Here is my capture:

CH1 (Blue) = IN
CH3 (Red) = HO
CH4 (Pink) = LO

So on the capture we see IN rise, and ~50ns later we see LO fall.
About 100ns after IN rises, we see HO go high, which seems rather fast considering Trt is supposed to be 90ns.
As for the ripple at the top of the HO trace I believe this may be in imperfect probing issue. My half bridge supply voltage is 20V, which is why we see HO pump up to about (20+12)=32V.

Then, on IN falling, we see HO go low about 50ns later, which is in line with the specs.
We however see LO rising roughly 350-400ns later!

Can someone help me to understand the significant mismatch in propagation delay of IN rising to HO rising (~100ns) versus IN falling to LO rising (~350-400ns)?

Is it possible the external load is too light to adequately discharge the HS pin and that is causing some issues?

  • Update - it does appear that my external load was too light to adequately discharge the HS pin.

    I've changed the external load to ~10ohm and now observe that there is still a mismatched propagation delay, but it's now closer to 200ns as opposed to 400ns:

    CH1 (Blue) = IN
    CH3 (Red) = HO
    CH4 (Pink) = LO

  • Hi Jim,

    My first recommendation would be to connect all probes to the same signal in order to ensure that the probes are reading with the same timing.

    If this shows probe consistency, we can move to further investigation of this behavior.


    Daniel W

  • Thanks for looking at this Daniel - I've confirmed that my other probes show the same signal and timing on the LO signal.

  • I poked around my circuit a bit more and removed a bypass I had on some power inductors in the load - my captures have cleaned up significantly, but still show the curious mismatch in propagation delay between IN and LO:

    CH1 = IN, CH2 = HS, CH3 = HO, CH4 = LO

    On IN rising, the delay to LO falling is ~50ns. The delay of IN rising to HO rising is ~100ns.

    On IN falling, the delay to LO rising is ~250ns. The delay of IN falling to HO falling is ~100ns.

  • Hi Jim,

    I would recommend a couple things in order to get accurate timing measurements, HO should be measured with respect to HS. Additionally, the timing specs in the datasheet are measured with no load on HO and LO, which will increase the delay. With these 2 considerations in mind, it appears that the device is functioning properly. The following estimations are made from 50% fall to 50% rise as is done in the timing screenshot you posted from the datasheet.

    IN rising to LO falling appears <50ns which is within expectations.

    LO falling to HO rising appears to be ~90ns which is the expected dead time.

    IN falling to HO falling appears to be ~50ns which follows expectations.

    HO falling to IN rising appears to be ~150ns which follows expectations

    If you have any further concerns let me know.


    Daniel W

  • Hi Daniel - thanks for looking so closely at these captures.

    I think I was confused about measuring HO with respect to HS, as this is not how the datasheet describes it in 6.6 "IN falling to HO falling":

    My main question is on your final point:

    "HO falling to IN rising appears to be ~150ns which follows expectations"
    Do you mean HO falling to LO rising?

    To show it more clearly, here is my confusion in the datasheet section 8.2.3 Figure 16:

    According to this diagram,

    1. LO falling to HO rising is tp+Trt = 50ns + 90ns = 140ns
    2. HO falling to LO rising is tp+Trt = 50ns + 90ns = 140ns
    The datasheet is suggesting that these two transition delays should be equal.

    Below, I've taken a fresh capture with no load connected to align with the no load condition of the datasheet. I understand that loaded will be different. I tried to zoom in to 100ns/div to get a better picture of the whole story:

    CH1 = IN, CH2 = HS, CH3 = HO, CH4 = LO

    Above we see LO falling to HO rising = ~50ns, HO falling to LO rising - it really depends on where you measure from HO falling.
    Since my bridge supply voltage is 20V, the high side Vgs will be ~0V at the time it crosses 20V, which would mean HO falling to LO rising would be over 100ns.

    Is the datasheet time diagram above incorrect in making it look like these 2 delays are the same? In my capture we observe 50ns versus over 100ns.

    Thanks for helping me to understand this situation!

  • Hi Jim,

    In for the HO-HS measurement, the measurement is made with HS grounded, so the most accurate way to replicate this in the system would be to measure HO wrt HS.

    Additionally, the no load condition is no FETs on LO and HO, not in reference to the HS load. 

    Also, the timing listed in the datasheet for the dead time delay has a max of 130ns for the 10k DT resistor.

    However, this is all with HS grounded and no load on LO or HO, so some more delay is expected with HS charging and discharging and loads on HO and LO


    Daniel W

  • Thanks Daniel, this absolutely makes sense now that no load in fact means no external devices at all.

    The asymmetry still does not make sense however.

    Does it make sense that Trt based on my static 10k DT resistor appears different on the IN rising side versus the IN falling side?

    1. LO falling to HO rising: ~50ns
    2. HO falling to LO rising: >100ns

  • Hi Jim,

    This can be explained by the fact that HS is not grounded, and so the HS charge affects the dead time. In order to see a more accurate dead time, I would recommend taking a scope shot of HO wrt HS, however I would still expect it to be uneven.


    Daniel W