18 hours ago •
I saw this article on Instructables website:
I converted the circuit in the link above to audio amplifier that can be made from general purpose transistors (not needing VHF transistors that might cost more money) that can be used for audio applications. You can make audio amplifier with power amplifiers and operational amplifiers. However, most of those IC (integrated circuits) need voltages of at least 4.5 V. This is why I was interested in the circuit shown in the link above.
Step 1: Design the Circuit
The design calculations are already explained in the link and I do not want to copy those calculations.
My circuit is useful for microphone and Line In amplification, will cost less money and will last longer because I increased the resistor values to reduce the power dissipation.
Step 2: Simulations
I used the old PSpice software to simulate the circuit.
You can see that my circuit has about 10 times higher gain because I increased the values of Rc resistors by a factor of 10.
I am thinking of making this circuit in a few weeks when I find the time and money for electronic parts.
2 days ago •
This article describes a simple transistor Darlington pair voltage follower circuit. A voltage follower can be made with just one transistor. However, the circuit presented in this article has a higher input resistance.
There are standard Darlington pair component transistors, two transistors (Darlington pair) in one component. A good example is the TIP122 power transistor.
I made this circuit by cutting a lid of a plastic container:
You can see my circuit working in this YouTube video:
The following video shows transistor working as a touch switch:
Designing the Circuit
I drawn the circuit via online https://easyeda.com software.
The input value should not be above:
VinMax = Vs + Vbe
If Vs = 3 V (shown in the circuit diagram) then:
VinMax = 3 V + 0. 7 V = 3.7 V
We can calculate minimum Rd value by deciding on the maximum LED current:
Rd = (Ve - Vled) / IledMax = (Vi - Vbe1 - Vbe2 - Vled) / IledMax
= (12 V - 0. 7 V - 0.7 V - 2 V) / 10 mA
= 8.6 V / 10 mA
= 860 ohms
We can easily calculate the circuit input resistance:
Ri = (Re || (Rd + Rled) || (Ro + Rl))*(Beta + 1)*(Beta + 1)
Where: Beta is the transistor current gain.
Lets assume the Q2 Ve voltage is 12 V.
Rled = Vled / Iled = Vled / ((Ve - Vled) / Rd) = 2 V / ((12 V - 2 V) / 1000 ohms) = 2 V / 10 mA = 200 ohms
Ri = (100 kohms || (1 kohm + 200 ohm) || (1 kohm + 10 kohm))*(Beta + 1)*(Beta + 1)
= (1 / (1 / 100 kohms + 1 / (1 kohm + 200 ohm) + 1 / (1 kohm + 10 kohm)))*(20 + 1)*(20 + 1)
= (1 / (1 / 100,000 + 1 / 1,200 + 1 / 11,000))*21*21
= 472,040.220564 ohms
= 472.04 kohms
Now we calculate the power dissipation to see why this circuit is not perfect:
Maximum power dissipation occurs at half supply voltage. The collector current is approximately equal to the emitter current. If supply voltage is 12 V, the output is shorted, the LED is taking 10 mA (assuming that we reduce the Rd value to just 400 ohms) then the emitter and collector current will equal to approximately:
Ie = Ve / Ro + Iled + Ve / Re = 6 V / 1000 ohms + 10 mA + 6 V / 100,000 ohms = 16.06 mA
Q2 Transistor Power Dissipation = Ie * Vce = 16.06 mA * 6 V = 0.09636 W = 96.36 mW
Q1 Transistor Power Dissipation = Ie * (Vce - Vbe) = 16.06 mA / (Beta + 1) * (6 V - 0.7 V) =
= 16.06 mA / 21 * 5.3 V = 0.00405323809 W = 4.05323809 mW
For some general purpose transistors the maximum power rating is 100 mW and this is with the heat sink. Thus 96.36 mW power dissipation is just under the limit. However, if you reduce the supply voltage from 12 V to just 6 V, the power dissipation will be half that amount. Another option is to use a power transistor or put additional transistors in a parallel.
2 days ago •
This article explains how you can build a small signal motor driver. A signal is entering the input of the circuit and is amplified with three transistors to drive a low current motor.
This circuit can be connected to an AC amplifier. The AC amplifier can be connected to a sensor that could be a:
- infrared sensor,
Thus this circuit can cause the motor to move due to:
- ultra-sonic waves,
- infrared remote control transmission,
- radio waves.
I thought of the circuit fin this article after looking at the following Instructable:
It is important to mention that the circuit can be implemented with just one MOSFET transistor. However, MOSFETs are still rare and cost more money and you might need to look for a MOSFET in your workshop inventory. Also, the MOSFET transistor would not allow you to control the current magnitude that controls the motor speed. In the circuit shown in this website, the greater is the magnitude of input signal the faster the motor will move. However, you can still control the speed of the motor with MOSFETs and Pulse Width Modulation (PWM) inputs: https://diyelectronics.webs.com/circuits.htm
Step 1: Design the Circuit
Calculate the maximum output current at worse case scenario. At worse case scenario transistor current gain, Beta = 20.
Ib1 = (Vin - Vbe3) / Rb1 = (3 V - 0.7 V) / 4.7 kohm = 2.3 V / 4700 ohms = 489.3617 uA
Ic1 = Ib1 * BetaMin = 489.3617 uA * 20 = 9.78723404 mA
Ib2 approximately equals to Ic1 because very little current is flowing through Rc1 resistor.:
Ic2 = Ic1 * BetaMin = 9.78723404 mA * 20 = 0.19574468085 A
Ic3 = Ic2 * BetaMin = 0.19574468085 A * 20 = 3.91489361702 A
Rc3 is 1 ohm and the transistor does not need to saturate. The maximum current is actually not 3.91489361702 A, but Vs / Rc3 = 3V / 1 ohm = 3 A < 3.91489361702 A (occurs when the motor is a short circuit if its movement is mechanically impeded or stopped), because the transistor saturates. However, you can achieve 3.91489361702 A output current if you raise the supply voltage to 4 V.
Now we can calculate the other resistors.
Rb2 = (Vin - Vbe2) / Ib2Max = (3 V - 0.7 V) / 9.78723404 mA
= 2.3 V / 9.78723404 mA = 235 ohms < (Rb2 || Rc1) = (270 ohms || 4700 ohms)
Rb3 = (Vin - Vbe3) / Ib3Max = (3 V - 0.7 V) / 0.19574468085 A
= 2.3 V / 0.19574468085 A = 11.75 ohms approximately 10 ohms
Rc1 and Ri resistors ensure that the circuit and the motor is OFF when the circuit is disconnected or when no input signal is applied.
Step 2: Simulations
I drawn the circuit in PSpice software to reduce circuit drawing time:
You can see that the transistor still saturates. The minimum input signal is 1 V.