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I have TD62783 powered from 12V rail, inputs are driven from 5V rail by HC540. Each output has bunch of leds on the other side along with constant current sink (multiplexed 7 segment display).

Schematic 1 Schematic 2 Current sinks

IC18/BCR420 is not installed. Q6,Q5 = AO3400, R33 = 330Ohm

When HC540 provides hi level signal, TD62783 output turn on as it should. But if then signal goes low, output does not turn off as quickly as it should. It shuts down very, very slowly, so in effect i have large "crosstalk" between LED displays. Oscilloscope shows this:

enter image description here

Yellow signal is TD62783 input signal, blue is output signal.

Could someone tell me what may be causing this?

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  • \$\begingroup\$ Please include the relevant portions of the schematic. How are the outputs connected? \$\endgroup\$ – Edgar Brown Feb 23 at 10:53
  • \$\begingroup\$ Your question says HC450. The schematic says 74HC540. Add links to the datasheets. Make it easy for your readers. \$\endgroup\$ – Transistor Feb 23 at 10:58
  • \$\begingroup\$ @EdgarBrown: I've updated post with more complete schematic. \$\endgroup\$ – Kaworu Feb 23 at 11:05
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There are quite a few issues with your circuit, but the main one for me is how can it be working in any way at all.

  • The TD62783 is a poor choice for a LED driver, as it is a Darlington emitter-follower configuration. In circuits where I have seen this done, the large voltage drop and output impedance leads to low and variable intensity displays.

You, however, are using a 12V supply, so this drop should not be too much of an issue, as it is in the 5V circuits I have seen.

  • You are linearly driving individual LEDs off of 12V. That’s a lot of wasted power.

LEDs forward voltage is less than 3V, every single volt above that is just wasted power.

  • You are using 12V to drive individual LEDs without any current limiting. This would bring out the magic smoke in any LED and those drivers.

The TD62783 can push more than 500mA, at 12V this would be 6W between a LED and the driver, and that is for just one channel. Popcorn comes to mind, however:

  • The TD62783 is an open-emitter driver, it does not actively drive a zero.

This is why nothing has blown up yet. You are trying to pull currents out of the LED cathodes with a device that cannot pull current at all.

So, the real question in my mind is what can possibly be pulling that node down at all, how can these be going down so fast, and how are those LEDs lighting up?

The only possibility I see is leakage currents, enough leakage to pull 10V down to zero in 1ms. Assuming 1000pF in the LED nodes, that would be ~10uA. Which would not be unreasonable for the back-polarized protection diodes of the TD62783 (and less than the 100uA maximum specified in the datasheet). But this should be nearly unnoticeable in an LED.

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  • \$\begingroup\$ It is working, because You've missed few details. TD62783 is connected to display common anodes. On the segment/cathode side, there is 8 constant current sinks created by IC19, Q6, R33. Maximum current is limited by OpAmp input voltage divided by R33, 3.3V/330Ohm = 10mA. Actual current is set by DAC's voltage output. This pretty much works and I have brightness control this way. I was going to use TBD62783 (DMOS version of above) but I couldn't find in in any reasonable shops (farnell/tme.eu). So... You are saying, using push-pull config for cathodes/segments should solve problem? \$\endgroup\$ – Kaworu Feb 23 at 17:19
  • \$\begingroup\$ @Kaworu according to your diagram. MD_G is connected to a segment’s anode while MD_CA4 to the segment’s common cathode. So it’s not me who missed anything. \$\endgroup\$ – Edgar Brown Feb 23 at 17:24
  • \$\begingroup\$ Not quite. MD_G is connected to pin "G", and "MD_CAx" is connected to "COMmon electrode" it dosen't say it's display anode or cathode. \$\endgroup\$ – Kaworu Feb 23 at 17:40
  • \$\begingroup\$ @Kaworu I was just giving you the benefit of the doubt. It’s a seven-segment display with segments called A through G (which you connected to a high-side driver) and one pin called COM. So, there are only two choices, EITHER it’s a common cathode display OR you have no idea of what you are doing. Which one is the correct choice? \$\endgroup\$ – Edgar Brown Feb 23 at 17:46
  • \$\begingroup\$ I may have no idea what I'm doing (I'm not very good in analog electronics), but nevertheless it's working as designed. Except for that ghosting thing. So... pull segments/cathodes to 12V when they are disabled? \$\endgroup\$ – Kaworu Feb 23 at 18:06

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