Equivalent to LM358 but very high impedance output needed - Page 1
Forum section: electronics / projects

author: rbm
author_rank: Regular Contributor
timestamp: June 22, 2014, 12:06:41 pm
content: I have a motorcycle project that I'm having difficulty to get working properly. I've replaced the speedometer on my motorcycle with an aftermarket unit. This new device reads road speed by counting pulses as the wheel rotates. It expects a magnetic reed switch input, which gets grounded each time the magnet passes (creating a negative going pulse).The motorcycle has a builtin speed sensor but the output from the sensor is wrong. I built a wave shaping circuit using a LM358 op-amp. Very simple; it is a comparator that outputs a negative going 5V pulse for each time the input crosses a threshold set by the voltage divider on the non-inverting input. The schematic is below.The circuit works properly as I have bench tested it and verified it operation on the motorcycle using a scope. However, when I attach the LM358 output to the speedometer, the needle of the speedo does not respond properly. It will register equivalent to 80 KPH while the motorcycle is on the centerstand and in neutral.The only reason for this behaviour that I can surmise is that the output impedance of the LM358 is too low to simulate the open circuit impedance presented to the speedometer if I was using the reed switch.Is there an equivalent op-amp to the LM358 that offering extremely high output impedance? The device has to accept a wide power supply range between 10-20V . Thanks for pointers or ideas.

author: dannyf
author_rank: Super Contributor
timestamp: June 22, 2014, 12:13:05 pm
content: Increase output impedance is simple, just connect a serial resistor to it.your problem sounds somewhere else, however.

author: onlooker
author_rank: Frequent Contributor
timestamp: June 22, 2014, 12:27:02 pm
content: Serial a diode on the output to have a open circuit during no pusle. You will still have the needed low impedance when output is low (during pulse). A pull up resistor post the diode can adj the open circuit impedance if needed. Though the low state will not be 0V,but 0.3 to 0.6 V. You may also try to use 3 state buffer / line driver for it.

author: Paul Price
author_rank: Super Contributor
timestamp: June 22, 2014, 12:59:23 pm
content: The problem is the circuit, not the LM358.Your circuit c probably oscillating.First try inserting a series 476 ohm resistor in series with the TACH input from the LM358 output.You probably need a 1-k resistor paralleled with a small ceramic cap(.01) to ground at the input from the sensor.Also: the positive feedback resistor creates 0-rpm offset and the output of a LM358 is always not quite zero when it should be when powered with a single supply, but this is not likely the problem here.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: June 22, 2014, 01:20:34 pm
content: Where does it say the speedo expects a 0 - 5 volt square wave? Or is it rectangular.

author: wraper
author_rank: Unknown
timestamp: June 22, 2014, 01:35:31 pm
content: Use LM393 comparator IC with open collector output. How high is open circuit voltage on open circuit reed input? If it is much higher or lower than 5 volts that is your problem. Open collector output will remain just open or closed as reed switch and won't try to pull output to 5V voltage.

author: Fank1
author_rank: Regular Contributor
timestamp: June 22, 2014, 03:02:50 pm
content: Use a 10k Resistor and a 10n Cap across the sensor.Use the other half of the 358 to make a 10x amplifier.Use it's output to drive the comparator after changing the reference to 1 volt.

author: Marco
author_rank: Super Contributor
timestamp: June 22, 2014, 03:30:25 pm
content: You're using the wrong terminology, you don't want a high impedance output ... you want an open collector or open drain output. The reed relay pulls down, the opamp output is push pull. The speedo won't necessarily see 5 volt as high.

author: alanb
author_rank: Frequent Contributor
timestamp: June 22, 2014, 07:20:46 pm
content: The circuit shows a coil as the input device bit the text talks about a reed switch. A reed switch would suffer from contact bounce which would lead to incorrect readings.Sent from my iPhone using Tapatalk

author: rbm
author_rank: Regular Contributor
timestamp: June 22, 2014, 08:38:00 pm
content: Quote from: alanb on June 22, 2014, 07:20:46 pmThe OEM sensor is a variable reluctance transducer which outputs a rough sinusoidal wave with a frequency of six cycles per wheel rotation. I am not positive the makeup of the VR sensor but it is probably a coil of sorts; it definitely is passive. The Aftermarket speedometer wants a square wave input between 5-12 volts. It triggers on a negative going edge, I believe. I can't say for certain because the manufacturer does not specify exactly how the unit works. This circuit I built is attempting to change the output of the OEM sensor into an input compatible with the speedometer.Quote from: alanb on June 22, 2014, 07:20:46 pmYou may be quite right. However, programmable parameters in the speedometer firmware impart inertial to the servo motor that drives the speedometer needle. As one changes this parameter, the needle becomes less responsive to small changes caused by bounce.

author: rbm
author_rank: Regular Contributor
timestamp: June 22, 2014, 08:39:41 pm
content: Thank you everyone for adding to this discussion. I'm going to try out the suggestions that have been given to solve this problem. There are many that I didn't know of or think of.

author: rbm
author_rank: Regular Contributor
timestamp: June 22, 2014, 08:47:20 pm
content: Quote from: HackedFridgeMagnet on June 22, 2014, 01:20:34 pmThe circuit is being interfaced with a Motogadgets Classic. The input is specified by the manufacturer as being a signal that gets grounded "X" times per wheel revolution. They have designed the speedometer to accept a passive reed switch input or an active Hall sensor input. I am making some assumptions about the required input based on the typical output of a three wire Hall sensor.

author: kc9qvl
author_rank: Regular Contributor
timestamp: June 23, 2014, 03:17:30 am
content: With 5 v supply That opamp does what 1volt to 4 volt swing. And the tach wants an open then grounded pulse to count revs. Maybe add a Schmidt trigger or logic level fet to switch the tach signal.

author: eetech00
author_rank: Regular Contributor
timestamp: June 23, 2014, 04:49:30 am
content: Quote from: rbm on June 22, 2014, 08:47:20 pmHi1. It sounds like the sensor is switching between ground and either open or some voltage. Is it a passive device or is it active? DC pulse output or AC output?2. The circuit you've shown cannot generate a negative going output signal to the tach without a negative supply voltage. Your circuit would need dual supply op amps and dual supplies. The tach probably reads a negative "edge triggered" signal as opposed to "negative going"signal.3. What type of signal is the tachometer designed to accept? What voltage level?Whatever device you decide to use (seems like Hall effect would be easiest) please post a part numberso it can be looked up..I think what is needed is a circuit to condition the signal so it can be read by the Tach. The signal chain probably should be something like this:Hall effect sensor -> Op amp buffer -> comparator -> Tach.But at this point I'm just guessing with out more info.eT