Parallel RC to ground

What is the purpose of a parallel RC from output of LM35 to ground. The LM35 is feeding 6 comparators.
R=56k
C=1uF

Is this some kind of filter? If yes, how can I calculate its 3dB?

The 56k is probably to provide a DC path to ground to bias the comparator inputs, and the capacitor is to filter noise. The LM35 may not be stable with a 1uF cap connected directly from its output to ground. With questions like this, it's best to provide a schematic.

Pjdd said:

The 56k is probably to provide a DC path to ground to bias the comparator inputs, and the capacitor is to filter noise. The LM35 may not be stable with a 1uF cap connected directly from its output to ground. With questions like this, it's best to provide a schematic.

Click to expand...

Attached is the schematic if you would like to have a look. Why is it important to bias the comparator input with a resistor to ground? Does the capacitor value depend on the frequency of an close EMI's? What is the purpose of the 1N4148 diode?

Of course, it is necessary to provide a bias current to the LM339 input transistors (as indicated by Pjdd already).
The capacitor in parallel has the task to ground all sudden and unwanted input peaks that shall not activate the comparators.

The circuit is fed with a varying voltage, possibly an audio signal?

C1 charges to the peak of the instantaneous voltage. R6 is to discharge C1 slowly. Each l.e.d. will light up when the voltage across C1 reaches or exceeds the voltage at its driver's non-inverting input.

It's a peak indicator as may be used on audio equipment.

Hi,

D5 acts as a simple rectifier.

Without C1, the voltage on R6 will follow the positive part of the input signal (minus the forward drop of the diode, about 0.5V here). In this case and even if the signal peak (supposed being sinusoidal) at pin 6 is higher than of pin 7, the LED D1 will light close to the signal peak only. LED 1 will light almost half the time (during the positive half cycle of the input signal). This may be a desirable effect

Adding C1, the voltage on R6 will be smoothed and will not drop to zero volt during the negative cycle. But if C1 is made very large, the capacitor voltage will not drop fast enough (by discharging via R6) to follow the variations of the input signal. Anyway, when adding C1, the LEDs will be brighter and if the ripple on C1 is relatively small, each LED will be fully on or off (at each instant, unlike the case when C1 is removed for which the maximum duty cycle is close to 50%).

Since R5 << R6 and with C1, the diode acts more as a positive peak detector.

Kerim

^ The OP said that the input is fed by an LM35 which is a Centigrade temperature sensor.

If the purpose of the circuit is to have a visual indication of temperature, the presence of D5 will degrade accuracy. This is because because the forward drop
1) has tolerance, that is, it varies between individual samples
2) is not constant even for the same sample. It's a function of the diode V-I curve
3) varies with temperature

If the diode is included with the intention of raising the minimum scale, that could easily be done by using apprpriate values for the voltage divider chain R8-R12.

Pjdd said:

^ The OP said that the input is fed by an LM35 which is a Centigrade temperature sensor.

Click to expand...

Silly of me to have overlooked that. :roll:

Now there are two

Yes, unfortunately the whole circuit was not shown. Again an example for an uncomplete and misleading question.