I have a circuit with logic gates that implements a function in order to get 3 logic control signals. This logic works with 12V or 0V for high and low states. Logic input pins will be connected to other board pins through a connector. This board connector pins can be at GND or high impedance (no connected) states/levels. Last stage needs more current that logic outputs can offer, so driver will be used.

In order to "translate" high impedance state to high logic state ('1') I have connected a pull up resistor to 12V. So I can get 12V when input board pin is at high impedance. In case that input board pin is at GND level, GND also is at logic gate input pin, so this will be a '0' logic state.

I have done some protoboard tests doing it at two stages. At first, I connected the input board pin with pull resistor to the logic gate input pin and this pull up configuration is working correcly.

At second stage I need to bring the same input board signal pin directly to an input driver pin but also keeping this pin connected to the input logic gate pin. At the picture an only one connector pin is drawn, it's called D.

enter image description here

This stage is not working well because I'm not able to keep 12V at logic input gate (VIN1). I may have the explanation for this, due the driver is built with two transistors. They both are off with 0V input, but they are ON with bigger voltage, as ie: 12V. So the net is carrying current throught resistors.

I use restistors for current limitation. But I have noticed that removing driver resistor, the input gate voltage is reduced from 12V to less than 4V. So, logic is not working well. But if I put a 30K resistor voltage is only reduced until 10V. As R = 40K is a value that offers the input current that my application needs to keep 70mA at driver output, I keep this as my solution.

I need the pull up configuration for logic gate input, but the same signal must arrive to the driver input. I need to keep 12V, there (logic gate input pin). Now I have get 10.5V using R = 30K.

My question is how would you solve the situation I have? Is there a way for improving this solution?

Notes: I'm using ULN2803A device for driver and cd4072B device for NOR.

  • 2
    \$\begingroup\$ It might be better to draw a schematic with the tool, the block diagram is unclear: 1) The logic gate and drive do not have part numbers. 2) What is the switch? \$\endgroup\$
    – Voltage Spike
    Commented Sep 26, 2019 at 15:15
  • \$\begingroup\$ I put the part numbers at the end of the post. Switch is on the other board. But related behavior is based on my test sceneario and there is no switch connected. It is only a not connected wire. Some times I put wire to gnd and the rest of times it is not connected to anything. \$\endgroup\$
    – Suvi_Eu
    Commented Sep 26, 2019 at 15:27
  • 1
    \$\begingroup\$ You can just reduce the value of the pull-up to a lower value such as 1k. \$\endgroup\$ Commented Sep 26, 2019 at 15:32
  • \$\begingroup\$ I have thought before about using two free inverters that I will have on the board for doing this, but I have no experience about what was better solution. Do you think I could use the inverters for buffering this signal to the driver? I' m using a 6 inverters device but I'm not using the whole number of available inverters. \$\endgroup\$
    – Suvi_Eu
    Commented Sep 26, 2019 at 15:39
  • \$\begingroup\$ @KevinWhite this R finally works, I have just the voltage that I expected! but I first used 4.5K because of input current must be less than 10mA and 12V/4500 = 2.6mA at first stage no driver connected. \$\endgroup\$
    – Suvi_Eu
    Commented Sep 26, 2019 at 15:58

1 Answer 1


Look in your ULN2803A datasheet. You'll see that it contains a 2K7 base resistor into the Darlington transistor pair and parallel pull-down resistors to disable the transistors when the input is floating.

So an input voltage above something like 1.4 V (two base diode drops in series) is being loaded by the 2K7 resistor in series with two forward-biased diodes in series - 1.4 V-ish. Your 4K5 pull-up resistor provides the top half of a potential divider with the load 2K7 and diodes at the bottom, leading to your 4 V found.

Try using a driver with a high input impedance instead of the input-current-thirsty ULN2803A. Look for driver ICs, there are plenty far more modern than this part. Or you may be able to use an op-amp connected as a voltage comparator that transitions at 6 V. It's what suits your load current for it.


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