Designation of clamper circuit:
Input Square Wave is described as follow:
V max : 20V
V min : -15v
Frequency : 10kHz
Design a clamper circuit to shift its referecence point so that its reference point is -10V.
Specify all component values and part numbers(if any) used in your design. Show all calculation clearly. Justify any assumption made.
That was the question given. I want to know that in negative clamping this wave so that its reference point goes down to 10v, what should be done? like what should increase or what should decrease? additional capacitors? don't have to give me the specific value but please help me by telling me the rough idea of where i should start and what i should be looking for or notice. Please help me with this as i am totally lost. thank you so much!!
I would use a capacitor to clamp the output. The output capacitor will have a 5v charge on it, adding to the wave form. The time constant for discharging should be very long, so as to not affect the shape of the waveform. You can use a zener diode to stabalize the capacitor voltage.
To shift the reference point of the input square wave down to -10V, you can use a negative clamper circuit.
In a negative clamper circuit, a capacitor and a diode are connected in series with the input waveform. The capacitor charges to the peak voltage of the input waveform while the diode blocks the voltage from discharging through the capacitor during the positive half-cycle of the waveform. During the negative half-cycle, the diode conducts and allows the capacitor to discharge, effectively shifting the reference point of the waveform.
To design a negative clamper circuit, you will need to select appropriate components - a capacitor and a diode - based on the given input waveform characteristics and the desired reference point.
Here's a general approach to design the negative clamper circuit:
1. Calculate the time period (T) of the input waveform:
T = 1 / Frequency = 1 / 10kHz = 0.0001 seconds
2. Determine the capacitor value (C):
The capacitor value can be determined based on the following equation:
C ≥ (T / R) * (Vmax - Desired Reference Voltage)
Where R is the resistance connected to the capacitor (ideally a high resistance in series with the capacitor). For a rough estimate, you can assume a large resistor, such as 1 megaohm (1MΩ).
For example, if the desired reference voltage is -10V, the equation becomes:
C ≥ (0.0001 / 1M) * (20 - (-10))
3. Select a capacitor value:
Based on the above calculation, you would need to select a capacitor value that satisfies the requirement. It's important to note that the selected capacitor value should be commercially available. Capacitor values are usually specified in microfarads (uF) or picofarads (pF).
4. Choose a diode:
An ideal diode has negligible voltage drop in the forward direction and infinite resistance in the reverse direction. In practice, you can use a standard silicon diode (such as 1N4148 or 1N4001) which has a small forward voltage drop (around 0.7V).
5. Circuit implementation:
Connect the selected capacitor in series with the diode, and then connect this combination in parallel with the input waveform source. The negative terminal of the capacitor should face towards the input waveform source.
It's important to note that these are rough guidelines to get started with the design. In practice, you might need to fine-tune the capacitor value and consider other factors such as diode characteristics, leakage currents, and circuit impedance.
Remember to double-check your calculations and verify the component values according to your specific requirements before finalizing the design.