Can I test the pulse directly with a light bulb!?

It is very unintuitive and unscientific to see its frequency by connecting the light bulb.

Clause. 1. Whether the power generated by the pulse signal generator can bring up the power consumed by a light bulb;

Second, the frequency below 80Hz can be roughly estimated by the naked eye, but with the increase of frequency, it is difficult for us to perceive the specific frequency with the naked eye;

Clause. Third, the analysis from the pulse source is not scientific, only from the signal generator to produce the type of pulse, divided into: sine wave, moment tooth wave, triangle wave, square wave, and the first several kinds of compound, can we use the bulb to estimate?

If you are referring to the up-and-down pulses of the CPU and PCI bus of the computer motherboard or between the memory and the CPU, it will make people laugh when measured with a light bulb, and you must borrow a high-frequency scanner. Ha ha.

Your question is not detailed enough, and I don't make it clear which part to ask, but in any case, it is still recommended that you string a small capacitor of the PF level in the signal to load the signal attenuation and load it on the scanner for observation, which is very intuitive.

It is dangerous to connect the light bulb directly:

If you receive a frequency above the ultra-high frequency, close to or reach the microwave level of the frequency, we learn in DC and 50Hz AC The characteristics of resistance and capacitance everything changes, only the distributed capacitance and induced current can break down the bulb instantly, and if you don't get it right, it will affect your body, high-frequency jet is very dangerous, it's best not to play;

In short, don't do things that you are not sure about!

Unless it is a very low frequency, it is not visible to the naked eye.

It is to use the high and low level of the pulse to control the light on and off! The high pulse light is off, and the low pulse light is on! I don't know if I understand it correctly!

1. In Figure B, a refill is connected in series, and Figure C is connected with two refills. Therefore, the length of the conductor has changed, and the conductor temperature and cross-cutting area have not changed. It shows that the longer the conductor length, the greater the resistance, under the condition that the conductor temperature and cross-cutting area remain unchanged.

2. In Figure B, a refill is connected in series, and Figure C is connected with two refills in parallel. Therefore, the cross-section of the conductor has changed, and the length and temperature of the conductor have not changed. It shows that the larger the conductor cross-cutting area, the smaller the resistance, when the conductor temperature and length remain unchanged.

According to your question A, use 2 batteries and a small light bulb in series with wires, assuming that the brightness is a

Figure B: On the basis of Figure A, a section of pencil lead is connected in series, and the brightness is dimmed at this time, assuming that the brightness is B at this time.

Figure C connects two pencil leads in series on the basis of Figure A, and the brightness of Figure B is darkened compared with Figure B, assuming that the brightness is C. at this time

Figure D: If the second pencil lead is not connected in series, but is connected in parallel with the first pencil lead, you will see that the light is brighter than when only one pencil lead is connected. Suppose the brightness is d at this point. (Contrast of their brightness: a d b c).

In the comparative experiment of Figure A and Figure C, the same experimental condition is the voltage, but the experimental condition is different in that the circuit of Figure A is connected in series without resistance. A resistor is connected in series in the C circuit. This comparative experiment illustrates an increase in loop resistance, a dimming of the bulb, and a decrease in current.

In the comparative experiment of Figure B and Figure D, the experimental conditions are the same, and the voltage and resistance are both connected in series. The experimental conditions are different in that there is one resistor in Figure B, and two resistors are connected in parallel in Figure D. One resistor lamp is darker and the current is smaller, and the two resistors are brighter in parallel, and the current is larger.

This comparative experiment illustrates that the resistance in parallel becomes smaller.

The effect is not noticeable because the pencil lead resistance is very small.

It is possible to average multiple measurements, but you have to pay attention to one fact:

If you are asking for a cold state resistor, the above method is fine;

If you are looking for the resistance of the bulb in the working state (hot state), it cannot be measured with a meter or a multimeter! In this case, the only way to calculate this is to measure the current and voltage and calculate it according to Ohm's law.

OK. However, the results are the same for multiple measurements. It doesn't make much sense.

Because the lamp resistance changes with temperature, it makes no sense to average it.

The demand is not clear, the specific width of the pulse width, the order of magnitude is clear. It is critical to the implementation.

Here's how:

1. If the single-chip microcomputer has the capture function, the timer time is recorded after capturing on the edge, and then the time is calculated, and the output signal can be compared.

2. If not, access the interrupt pin of the single-chip microcomputer (the upper and lower edges are interrupted, if not, immediately change the next interrupt trigger mode after entering, and switch the rising and falling edge interrupts), and the timer time is recorded in the external interrupt service of the single-chip microcomputer, and then the time is calculated after the next interruption comes, and the output signal can be compared.

3. If there is none, the function is single, the pulse width is relatively wide, the requirements are not high, and the hardware is not on the interrupt pin, you can consider the main program to query and record the time.

5ms, very wide, if the function is single, the timer can be dispensed with, directly use the single-chip microcomputer command cycle to calculate the time, define a long integer variable, constantly monitor the pins, add one after monitoring the variable, and then compare.

However, I hope to use the method 2 I mentioned, which is more versatile. Think about the specific implementation yourself, and use it carefully by another friend, there are many problems, and I won't explain them one by one.

If the crystal oscillator frequency is 12MHz, the pulse width, and the measurement is 65535us, it is still very easy.

However, the question requires that the minimum value of the pulse width be measured!

You know, the pulse to be measured can't be too small.

If you say: less than or equal to, then output a signal, such as the light on, greater than, the light off.

It doesn't work out. Hehe.

Observe the rated voltage of the small bulb, and then connect the circuit, the general voltmeter range is 3V, the ammeter is on the line, the switch should be disconnected when the circuit is connected, and the sliding rheostat should be adjusted to the position of the maximum resistance.

There are three main steps to do the test, 1. Adjust the voltage at both ends of the bulb to the rated voltage, record the current and observe the brightness of the bulb. 2. Adjust the voltage to about times the rated voltage, and record the current and the brightness of the bulb. 3. Adjust the voltage to be smaller than the rated voltage, and record the current and the brightness of the bulb.

Finally, the power of the bulb was calculated for each of the three experiments and the changes in brightness were compared.

The plate of the small bulb!

In the side process! Look at the voltmeter, measure the small bulb and the ammeter.

p=ui

Nameplate of the bulb: Rated voltage.

Junior Physics Experiment Measure the electrical power of a small light bulb.