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This circuit is meant to produce logic level outputs to HIT1 and TRIG1 when an input in detected at X1.

I need to explain the operation of the circuit, calculate the unknown resistor values, explain their roles and determine if there are some components missing, however I'm struggling. Any help/hints would be appreciated in understanding how it works.

I know that OA1 is the Op-amp and CP1/2 are comparators, where the output of the comparator goes into a digital scoring meter, why are two outputs required however?

Just any help understanding would be appreciated, thanks in advance.

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We are not here to do your homework for you. However, we can help with specific questions.

What have you done to try to undestand the circuit yourself? What part specifically are you stuck on? It would help to have a clear spec for what the circuit is supposed to do. This includes the voltage expected to be produced by X1 in response to the stimulus you are trying to detect and some idea of the frequency. Without that it's impossible to say what some of the values are supposed to be.

Do you undesstand the purpose of each of the components? For example, what does the first amp do and how do R1 and R2 effect that process? What is the purpose of D2, C1, and R3 put together? Why are there two comparators and what is the chain of resistors R4, R5, and R6 there for?

Added:

It looks like you have now given at least some thought to the circuit on your own, so I think it's OK to go into more detail about how it works.

Look at just OA1, R1, and R2 for now. You should be able to recognize those form a classic non-inverting amplifier. R1 and R2 form a voltage divider that attenuates the output signal to make the feedback signal. The gain of this stage will be the attenuation ratio. For example, if R1 is 20 kΩ and R2 10 kΩ, then the divider will attenuate by 3 (have a gain of 1/3), which means this opamp stage will have a gain of 3.

X1 is a piezo sensor, which puts out a voltage when subjected to strain. It is unclear what the designer intended for D1. It will clamp the voltage from X1 to one silicon junction drop below ground. That is a good thing to do since piezo transducers can produce quite high voltages when subjected to high strain, like when being hit with a hard object or dropped onto the floor. This effect is exploited in some barbecue ignitors to create a spark. While normal operational voltages may be quite small (sound waves in air aren't going to strain the piezo crystal much), the voltages from unintended shocks can easily damage OA1. I have no idea why the designer thought of this for negative voltages but not for positive. As is too often the case, just because someone draws up a circuit and posts it on the web doesn't make it a good design. I would put a second diode just like D1 in parallel with it but with reverse polarity. That will keep the piezo voltage to within one junction drop either side of ground, which should not damage the amp even when it is not powered. Normal piezo voltages from sounds waves will be much less than that, so this shouldn't interfere with normal operation.

So in normal operation the output of OA1 is the piezo signal after some voltage gain and buffered to a low impedance. Now consider D2, C1, and R3. Note that the output of that section only drives two comparator inputs, which we can consider inifinite impedance for our purpose at this point. Think of just D2 and C1 first. When the opamp output goes high, it will charge up C1 thru D2. When the opamp goes low, D2 will be reverse biased and whatever voltage is on C1 will remain. Basically this is a maximum value circuit. It will capture the peak voltage (minus the D2 forward junction drop) and hold it on C1. A high enough captured peak level will ultimately trigger the outputs. However, you don't want a single event to trigger the outputs forever. That's where R3 comes in. It causes the voltage on C1 to exponentially decay back to 0 over time. When C1 is in Farads and R3 in Ohms, then C1*R3 will be the time constant in seconds. So the negative input to the two comparators is the recent peak input level, with how recent adjusted by the C1*R3 time constant. Such a circuit is called a detector because of its early use in AM radios to detect the carrier amplitude. It captures (or "detects") the low frequency amplitude of a high frequency AC signal.

Now look at the resistor chain R4, R5, and R6. This is just a two-tap voltage divider. The top voltage and the resistances are given, so you should be able to figure out the two tap voltages. These become the threshold voltages at which the two comparators will trigger. When the detected piezo signal amplitude exceeds the low threshold, HIT1 will go low. When it exceeds the higher threshold, TRIG1 will go low. The two outputs do the same thing except that HIT1 is more sensitive than TRIG1. It takes a louder signal or stronger whack to assert TRIG1 than to assert HIT1. In this case, we're assuming negative logic for these, so "assert" means to go low in this case.

We don't know how these signals are used, but consider what you get if they drive the inputs of a XOR gate. The XOR output will be true only when the detected voltage is between the two thresholds. Such a circuit is called a window comparator because it only asserts when the input signal is within a certain range (within a voltage "window").

The above analysis is assuming that the opamp and the two comparators are properly powered. Unfortunately this is not shown, so we can't tell. The TL074 needs a few Volts headroom on each end, so will need a negative supply since it is expected to operate on signals around 0V. The TL07x also need some minimum supply voltage. I don't remember off the top of my head how much, but I'm pretty sure just 5V isn't enough. Running them from +9V and -5V is probably sufficient, although they can go rather higher too.

As always, you need to check the datasheets for the parts you are using to see what special requirements or restrictions they have and to make sure they are run within specs.

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  • \$\begingroup\$ The circuit is supposed to detect impact on the sensor from a ball, the voltage produced from the sensor is "at least 1V" at Va. The rising edge of the TRIG output is required 1.5s after the ball hits the sensor. As far as I can tell the HIT1 comparator is used nowhere in the spec, so I'm unsure why it's there. I understand that R1 and R2 act as a potential divider, whereby they provide negative feedback back into the Op-amp, I assume to reduce distortion and unwanted signals, so that only 1 signal is sent whenever a ball hits the sensor. \$\endgroup\$ – Chloe Jan 14 '12 at 16:56
  • \$\begingroup\$ R3 and C1 I think are to provide a delay between the signal reaching the comparator from the Op-amp, it is meant to take 1.5s to reach TRIG1 so can the values of them be anything where 1.5s = R*C? The diode D2 acts as a valve, preventing negative signals from going through the signals. I don't understand what the chain of resistors R4, R5, R6 are for however. Sorry for the vagueness of the question before, I just wanted some kind of confirmation that I was going along the right lines, without actually saying what I thought the circuit was doing. \$\endgroup\$ – Chloe Jan 14 '12 at 16:56
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    \$\begingroup\$ @Chloe: Unfortunately no what? I was suggesting another diode as a voltage clamp. A zener would clamp too but at a higher voltage. That may be fine when the opamp is on, but keeping the input to one silicon junction drop from ground is a good idea when the opamp is not powered. D2, C1, and R3 form a detector. There is no delay in reacting to a signal. It will decay back down after the signal according to its time constant. \$\endgroup\$ – Olin Lathrop Jan 14 '12 at 23:01
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    \$\begingroup\$ FWIW I'm learning a lot from this question. I really enjoyed trying to figure out the answers to Olin's questions before they appeared in these comments. \$\endgroup\$ – drxzcl Jan 15 '12 at 0:13
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    \$\begingroup\$ @abdullah: You could put a resistor divider on the piezo, and then clamp with the two diodes. \$\endgroup\$ – Olin Lathrop Jan 15 '12 at 14:47

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