[SOLVED] A pull-up resistor increases the low side of a signal, how can I counteract it?
[SOLVED] A pull-up resistor increases the low side of a signal, how can I counteract it?
I have an ebike computer that has two inputs that accept square wave signals. The range is 0-5V. One of the inputs has no pull-up/down resistors on it. When I hook up a device producing signal, the
Solution
Using an open drain buffer like the 74LVC1G07 did the trick. Note that this part only works for inputs with a pull-up resistor. There are other parts from the same lineup that can work with any input.
Problem
I have an ebike computer that has two inputs that accept square wave signals. The range is 0-5V. One of the inputs has no pull-up/down resistors on it. When I hook up a device producing signal, the signal is detected correctly. Both highs and lows are detected correctly. The other input has a built-in pull-up resistor. When I hook the same signal to it, only the high side is detected. Measuring the voltage of the low side, it's a bit higher than when hooked to the other input and I'm guessing it isn't crossing the low threshold the computer expects in order to register the low.
Am I right in thinking that the pull-up resistor is increasing the lows?
Is it possible to counteract the internal pull-up resistor by hooking a pull-down resistor to the input?
I'm a software guy with decent soldering skills and minimal knowledge in electronics so please answer like I'm not the brightest electronic bulb on the tree.
Thank you in advance!
Cross-posting my question from StackExchange.
Problem you are having could be
- output impedance of the signal generator you use to generate the square wave. When you set it to low, output impedance of the signal generator builds a voltage divider together with the internal pull up, and the device ends up sensing a higher voltage than "low". This is something you can see if you have an oscilloscope, try to hook up a probe to the input and ensure whether you get what you set in the signal generator.
- some signal sources have no or limited capacity to drain current. And when you set it to low, this is exactly what it us supposed to do, drain current from the internal pullup.
As you mentioned in another comment, solution is a simple buffer. This could be an opamp, but even a simple nmos transistor should suffice (open drain as you said). But you need yo be careful with current ratings of the transistor, which you can easily calculate by dividing 5V by the pull up resistor. Send a message if you need help.
Before these, I suggest you yo use an oscilloscope or multimeter to measure the voltage when it is supposed to be low, and see that in fact the problem is that voltage at the input doesn't go "low".
Thanks for the explanation!
I measured the signal generator alone, not connected to the input. It goes to 4.5V high and 0.001V low. Then I connected it to the input and measured at the input. I got 4.5V high and 1.1V low.
Also I ordered the 74LVC1G07 buffer along with a breakout board which should allow me to hook it up inline and test. I'll report back when I do.
Do you have a good low impedance ground connection between the devices? This sounds like there is something else going on.
Looks like it. I measured the cable and I got .1-.2 Ohms on every pin, ground included. I replaced both the computer and the sensor device in question with new units and the behavior is the same.
I got an interesting solution proposed in StackExchange - to use a buffer instead of messing with the pull-up. I ordered some parts to try it out.
Do you have any idea what "open drain" means in the context of buffers?
Open drain is the same as open collector. They both require an external pull-up resistor to function. There are a couple of uses like level shifting the output down (connect the pull-up R to the lower voltage) and connecting multiple outputs together with no extra circuits as the outputs can't fight. Look up the I2C bus as that uses open drain to work for more info.
Potentially silly question, but many function generators have a serial output impedance (resistance) of 50ohms which will make a potential divider with any input pull-up. A buffer would eliminate this problem as your controller would only see the really low output impedance of the buffer.