The voltage divider formula is as follows: Vout = Vsource * (R2/R1+R2). And if you do the calculations you may see that if R2 closer or smaller than R1, your Vout will be a fraction of Vsource. Take a look at the Simulation in I setup in LTspice.
As you can see, of R2=R1=1K, you get 2.5V at the input of the amplifier stage. You basically waste energy for no reason.
In this case, you get a voltage very close to the Source voltage which is visible in the schematic. You get 4.95V at the input of the amplifier stage.
Therefore, one must pay attention to the input and output impedances of the circuits. If the signal source has a low output impedance, the amplifier should have a high input impedance.
]]>Let’s consider a case where you set unrealistic expectations (these are: assuming there won’t be any difficulty, it will take only a couple days, you will get good very quickly, there won’t be any pain or disappointment), unaware of the unconscious flaws in your thinking that I explained earlier. You set yourself a goal, you want to learn how to draw and you want to be really good at it. (You just watched Bob Ross got inspired by him, and you are fully motivated at least for now). Now you believe you will be able to draw like Bob Ross in a couple of days, or you believe that it won’t require much effort (This is literally what I have thought in the past). After trying it for the first time, you face the harsh reality that you can’t even draw a proper straight line yet alone a painting. You realize that it will take significantly more time and practice than you originally expected. Noticing how far reality falls short of your expectations, your motivation is destroyed and you quit in the first couple days.
Learning anything meaningful takes far more time than you expect. And not only is it a long journey, it also involves pain, disappointment and a lack of gratification for a prolonged period. But if you begin with these things in mind, if you accept that it will not be easy and your progress will be very slow, you will eventually get there.
]]>We use decoupling capacitors to provide current almost intaneously to the part in the microcontroller where it is required. Current must be delivered in nanoseconds, but this cant be done by the power supply, because the wiring that connects power supply to the power terminal of the microcontroller has some inductance, and that resists to the sudden changes of current. This is where capacitors come into play, we connect it as close as possible to the power terminal of a microcontroller, and the current is delivered almost instantenously.
Lets look at two examples, one where we dont have a decoupling capacitor, and one we have a decoupling capacitor.
We connect power supply to our microcontroller, represented by a current source that requires 100mA when its turned on. The power line connecting the supply to the circuit has some inductance at nH level. Even if it seems small, it effects the circuits function significantly. See the schematic and the simulation results. The voltage at the pin of the current source oscillates a lot before it can stabilize at 2V, and it takes microseconds to stabilize, which is huge for a 100 Mhz microcontroller. This is caused by the inductance of the wiring, it resits sudden current changes. The power source wont be able to supply enough current in time, therefore microcontroller will malfunction.
Only difference now is we connect a capacitor between the power supply and the ground, differences are huge. When a sudden current is needed, capacitor can supply this current almost instanteosuly for a short amount of time until the power supply can start delivering current at required speed. See the differences in the simulation results, oscillations at the microcontroller terminal is almost non-existent.
In the inital state, the BJT is turned on. The gate of the high side MOSFET is at 0V. Bootstrap capacitor charges to VCC through the diode. The load is connected to ground, so the source of the MOSFET is at ground potential.
When the BJT is turned off, capacitor discharges to the gate through the 1k resistor and turns the MOSFET on and we get 12V at the source terminal. Since the capacitor and the source terminal are connected, we get a total of 24V at the gate terminal which turns the MOSFET fully on.
]]>I chose the GPIO pin PA6 and set it to TIM_CH1, meaning Timer 3 and channel 1.
After activating the timer channel you set the pin to, configure the timer source to “Internal Clock” (Unless you want to use an external clock.) then set the Channel1 to “PWM Generation CH1”
Use the following formula to get your desired PWM frequency :
My clock source is running at 72MHz, I set the values so I get 80 percent and 50 percent duty cycles. These are the results on my oscilloscope:
How to solve this? Of course with a bandpass filter!
I designed the circuit using inductors and capacitors, created the schematic and also desinged the PCB as below:
Even though soldering these incredibly small smd parts were too difficult, I have managed to solder them properly.
The pcb looks like this now with sma connectors on both ends:
The S parameter shows that the filter is working as intended, having the lowest S11 point at around 400-450 Hz which is exactly what I needed.
To make sure, I connected a 1.3GHz signal and than a 433MHz signal to it to see if it is working properly, and saw that I now have a very low amplitude coming from the 1.3GHz signal which was exactly what I have wanted:
