Unit Buffer for circuit isolation

[Ian_Ren]

I am trying to acquire a signal from a VR speed sensor which is a part of an automotive engine sensor system. To minimize the effect of attaching the data acquisition system to the engine electrical circuit, I use unit buffers (voltage followers).

This works great until the input voltage (i.e., signal from the VR sensor) exceeds the power supply voltage to the op-amp, then there is clearly an impact on the engine electrical circuit and the engine shuts down.

Is this behavior expected from an unit buffer? I use LM348 op-amp.

A VR speed sensor can output very high voltage. My initial thought was that the output voltage from the unit buffer would be trancated to the supply voltage and anything outside the range of the power supply would be lost.  I did not expect the input side to be affected too.

Can someone give a tutorial on this and tell me what I can do to overcome this issue.

Thanks.

Ian     

[johnsold]

Ian,

I can probably help. It would help me to help you if you would post the specifications of the speed sensor and the schematic of the unit buffer you are using.  That way I do not need to make a bunch of guesses about what might be happening.

A general comment: When the input voltage on a standard op amp exceeds it power supply voltage, the input protection circuit begins conducting. The currents can be much higher than the input current when the amplifier is operating in its normal range (a few mA rather than pA or nA).

Designing a circuit to accept a signal with a range much larger than the power supply and maintaining a very high input impedance can be non-trivial.

Lynn

[Ian_Ren]

Lynn

Thank you so much for your response. I think your comment "A general comment: When the input voltage on a standard op amp exceeds it power supply voltage, the input protection circuit begins conducting. The currents can be much higher than the input current when the amplifier is operating in its normal range (a few mA rather than pA or nA)." is right on the problem I am experiencing.

I cannot get into too much details of the project. But basically the speed sensor is a 2-wire variable reluctance (VR) sensor pointing to a steel rotating gear. When a gear tooth passes the sensor, it creates a pulse similar to a sine wave. The magnitude of the signal increases with gear rotating speed. In this case, at maximum speed, the peak voltage from the sensor is around +/-35V.

The two wires of the sensor are connected to engine ECU via engine electrical harness. I am "Teeing" a channel of my data acquisition system (USB 6009) to these two wires. Since the signal range is more than the range of the AI channel, I have to reduce it. I plan to do it by using a voltage divider, but putting in a voltage divider directly lowers the impedance of the whole electrical system, and connecting and disconnecting the DAQ creates a change in impedance, leading to some undesirable responses from engine ECU, including shutting down the engine.

What I want to do is to isolate my DAQ system so that whatever happened on my DAQ side would not affect the ECU side to impact the performance of the engine.

I am using unit buffers created from LM348 op-amp powered by 4x9V batteries, so the unit buffer has a +/- 18V power supply. Because the signal input range can be as high as +/-35V, I have a problem. I can live with the truncated +/- 18V signal as long as it does not lower the overall circuit impedance and impact the performance of the engine.

Is there a different method or op-amp you can suggest? The solution must be very "cost effective" (the reason of using 9V batttery and USB 6009) and I need the raw analog signal (so a chip converting the pulse train to a TTL signal does not work for me).

Your help is much appreciated.

Thanks.

Ian

[johnsold]

Ian,

The LM348 is not the best choice for this application because of the relatively large input bias currents. Something like the TL074 has similar gain and bandwidth characteristics and three orders of magnitude smaller input bias current.

A circuit like this should work for you. It has an input impedance greater than 1 megohm, which should be high enough to avoid loading the sensor or affecting the ECU. The peak voltage at the input of the op amp is about 3.2 V when the input is at 35 V. The protection diodes (D1 and D2) will not conduct until the input voltage reaches approximately 200 V. With a 35 V input the output is ~10.2 V.

If you use the LM348 in this circuit, the worst case input bias current of 200nA results in about 20 mV at the op amp input and about 65 mV at the output. With the TL074 the offset due to the bias current will be smaller than the input offset voltage of the device.

You could also go to higher input impedance with the TL074, although you would need to check the effects on bandwidth due to input and stray capacitance.

Lynn

[Ian_Ren]

Lynn

Thank you so much for your help. I will give it a try. Two more questions.

1) Do you think there will be noise issue when you put a R1=1 MW impedance there? 

2) Does it make any difference if I flip connection to V2 and V3 (i.e., R3 to V3, and the joint of R4 and R5 to V2)? I think it does not matter since V2=V3, but I would like you to confirm it.

Thanks.

Ian

[johnsold]

Ian,

1) The thermal noise in a 1 MΩ resistor and a 1 kHz bandwidth is about 4 uV. So that is not an issue.

2) The circuit I posted is wrong. The inverting and non-inverting inputs should be interchanged as you suggest.

Lynn

[Ian_Ren]

Lynn

While I am waiting for my order of TL074 to build a circuit that you suggested, I also tried one more thing simply by putting a 500kΩ resistor between the VR sensor and the unit buffer. Everything seems to be working. I do not have problem in shutting down engine any more. When the voltage from the VR sensor exceeds the power supply of the unit buffer, the output is truncated, but that's fine with me.

Is there any reason I should NOT use this simpler configuration? BTW, I do prefer keeping the gain to 1:1 if possble because the DAQ system already has a built-in 11:1 voltage reduction (voltage divider) which I cannot change.

Thanks.

Ian

[johnsold]

Ian,

The biggest disadvantage of the series resistor approach is that you are running current through the input protection circuitry in the op amp on every cycle of the speed sensor signal when the magnitude exceeds the power supply. With the large value resistor, you limit that current to what is probably a safe value, but the protection circuits are not designed to be used continually.

I have seen op amps have subtle damage when the inputs are overloaded. Although probably not an issue here, the input offset voltage, bias currents, and noise all increase, sometimes by large amounts.

That is why I included explicit clamp diodes (D1 and D2) and a second series resistor (R3) in the circuit I posted.

If you make the op amp gain match the voltage divider attenuation, the overall circuit has unity gain.  If you want to clip the output above ~10 V, it is better to use a circuit which reduces the gain when the output exceeds a certain value.  This prevents the op amp from going into saturation. Some amplifiers are very slow to come out of saturation, and some can even reverse the polarity of the output when the inputs are out of the specified common mode input range.

Lynn

[Ian_Ren]

Lynn

The circuit should give me a 1:1 gain if I use R2=200kW and R4=50kW, but will there be an issue if the input voltage is more than the power supply of +/-18V? 

I can live with a truncated signal, but not anything that can damage the circuit. 

Thanks.

Ian

[Ian_Ren]

I think it should be fine with protection from D1, D2 and R3, am I right?