Pulse-width modulation (PWM) is widely used in modern electronics to control the power
delivered to electrical loads. Typical applications include LED brightness control and motor speed
control.
In PWM, the voltage applied to a load is rapidly switched on and off.
When PWM is used to control an LED, a sufficiently high switching frequency prevents the human eye from perceiving the individual switching events. Instead, only the average brightness is perceived.
The brightness of the LED is determined by the duty cycle of the rectangular signal, which is defined as
the ratio of the on-time to the signal period.
A constant supply voltage is available for the experiment. This raises the following questions:
To answer these questions, two operational-amplifier circuits are investigated:
the Schmitt trigger and the integrator. First, both circuits are analyzed separately.
Afterwards, they are combined to form an oscillator that generates the required PWM signal.
Finally, the circuit is modified to enable brightness control of the LED.
The operation of an OPV in the linear operating range can be enforced by means of circuitry by feeding back the output signal, i.e., returning it to the inverting input (- input). In the circuit shown, the negative feedback is provided by a capacitor.
Analysis of the circuit:
Negative feedback
$\Rightarrow u_\mathrm{d} = 0 \Rightarrow i_R = \frac{u_\mathrm{e}}{R}$
$i_R=i_C$ (because OPV input current $i_\mathrm{n} = 0$)
$u_\mathrm{a}=-u_C=-\frac{1}{C}\int i_\mathrm{C}\,dt=-\frac{1}{RC}\int u_\mathrm{e}\,dt$
The integrated input voltage appears at the output. The product of resistance and capacitance has the character of a time constant:
$T_\mathrm{i}=RC$
The figure shows the output voltage of an integrator with a square wave voltage at the input. The output voltage at the start $u_\mathrm{a}(t=0)$ depends on the charge state of the capacitor when switched on.
To analyze the behavior of the integrator, the following circuit is used:
Supply voltages (from power supply unit):
$UCC=+3~V, UEE~=-3~V$
Values of the components used:
$R1.3=10~kΩ, C1=10~nF$
The operation of an OPV in the saturation range can be enforced in the circuit by feeding
back the output signal, i.e., returning it to the non-inverting input (+ input).
In the circuit shown, the feedback is provided by resistor $R_\mathrm{2}$.
The output voltage $u_\mathrm{a}$ can only take two values, either $\mathrm{+}U_\mathrm{sat}$
or $\mathrm{-}U_\mathrm{sat}$. It switches between these two values when no voltage drops
at the differential input of the OPV, ($u_\mathrm{d}=0$).
Then $i_\mathrm{R2}=\frac{u_\mathrm{a}}{R_\mathrm{2}}$ and
$u_\mathrm{e}=-R_\mathrm{1} \cdot i_\mathrm{R1}$.
Since the OPV input current is $i_\mathrm{p}=0$, the currents through the
resistors must be equal, and the following relationship applies for the switching points:
$u_\mathrm{e}=-\frac{R_\mathrm{1}}{R_\mathrm{2}}\cdot u_\mathrm{a}=\mp\frac{R_\mathrm{1}}{R_\mathrm{2}}\cdot U_\mathrm{sat}$
The image shows the input/output characteristic curve of the circuit known as a
“non-inverting Schmitt trigger.” The reaction to any input signal is shown on the right.
The circuit only switches to $\mathrm{+}U_\mathrm{sat}$, when the positive switching
threshold at the input is reached. It only switches to $\mathrm{-}U_\mathrm{sat}$, when the
negative switching threshold at the input is reached.
To analyze the behavior of the schmitt trigger, the following circuit is used:
Supply voltages (from power supply unit):
UCC = + 3V, UEE = - 3V
Values of the components used:
R2 = 20kΩ, R3 = 27kΩ
Ue = 3 V (amplitude), f = 1 kHz
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
The circuits previously analyzed individually are now connected to form a complete system.
The integrator and the Schmitt trigger together form an oscillator. The output signal
of the Schmitt trigger is fed back to the input of the integrator. Therefore, the output
signal simultaneously acts as the input signal of the overall system.
Due to this feedback, the circuit generates a periodic signal without requiring an external
input signal, apart from the supply voltages of the operational amplifiers.
The Schmitt trigger generates a rectangular signal that is integrated into a triangular signal
until one of the switching thresholds is reached. At this point, the output state changes
and the process repeats continuously, producing a stable oscillation.
When the circuit is first powered on, the oscillator starts due to small disturbances such
as noise, offset voltages of the operational amplifiers, or slight asymmetries in the circuit.
These small deviations move the system away from the unstable equilibrium point and initiate
the oscillation.
To analyze the behavior of the oscillator (triangle-rectangle generator), the following circuit is used:
Supply voltages (from power supply unit):
UCC = + 3V, UEE = - 3V
Values of the components used:
R1 = 200 kΩ, R1.3 = 10 kΩ, R2 = 20 kΩ, R3 = 27 kΩ, C1 = 10 nF
C1 = 10 nF, f = ƒmin
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 10 nF, f = ƒmax
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 1 nF, f = ƒmin
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 1 nF, f = ƒmax
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
3. Explain how this circuit works in a few sentences.
4. Why is it useful to use R1 as a potentiometer to vary the frequency rather than R2 or R3?
After combining the Schmitt trigger and the integrator, the circuit generates a periodic signal with a fixed duty cycle. For many PWM applications, however, it is necessary to adjust the duty cycle in order to control the average power delivered to the load.
In the case of an LED, changing the duty cycle directly affects the perceived brightness. Therefore, the oscillator circuit is modified so that the duty cycle can be varied to control the brightness of the LED.
To analyze how to adjust the duty cycle of the PWM-signal, the following circuit is used:
C1 = 10 nF, minimum duty cycle
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 10 nF, maximum duty cycle
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 10 nF, middle position
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 1 nF, minimum duty cycle
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 1 nF, maximum duty cycle
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
C1 = 1 nF, middle position
Channel 1: $\frac {Volt}{Div}=$
Channel 2: $\frac {Volt}{Div}=$
Time basis: $\frac {T}{Div}=$
In the final part of the experiment, LEDs are connected to the circuit and their brightness is controlled using the generated PWM signal. By adjusting the duty cycle, the average power delivered to the LEDs changes, resulting in a corresponding change in brightness.
To analyze how to adjust the brightness of LED-lamps, using a PWM-signal, the following circuit is used:
For this experiment, you should be able to apply and explain the following concepts: