Another.....FOR BEGINNERS ONLY, tutorial.
This is another circuit design tutorial
For beginners, to learn how to implement different transistor types into one ciruit.
And come up with a working circuit, that can be used with other circuits to build syustems.
This excersise will be explained to great detail as possible, so anyone starting out with a basic knowledge of OHMS law, can grasp a hold of these concepts.
The design is for a Voltage Peek detector, that has the output coupled to a LED, for a display.
This same circuit was breadboarded and tested for clarity, so I know it works properly.
DESIGN:
Refer to the schematics:
Below.
First I chose VCC to be 12v.
Then I chose RCQ1 to be 470 ohms.
Then I chose a standing (idle) output voltage for Q1 of 8v. (VCQ1)
I chose a base voltage of 5v. for Q1. (VBQ1)
Assuming a Vbe of 0.7v. VEQ1 will be 4.3v. (for calculation purposes only)
Now solving for the current through RCQ1 ~= 8.5mA. = ICQ2
And taking that current divided into VEQ1 ~= 505 ohms.
That establishes the front end of Q1, but before I calculate the base resistors of Q1 I need to consider the bias of Q2 stage,
Now lets move on to Q2,. for a moment.
Q2 will have its collector direct coupke to the base of Q1, so VBQ1 = VCQ2, and VCQ2 needs to be greater than the base voltage of Q2 (VBQ2,) but VBQ2 will be the voltage input to be detected. So looking at the schematic, the base of Q2 is not biased but rather just an input terminal, to the detector circuit.
Now the resistance for the emitter of Q1 is a split resistance.
The emitter of Q2 couples to the intersection of this network. This is to cause a feedback that will make Q1 turn off almost like a switch, rather than a linear response to the signal coming from Q2's collector.
So now since the VCQ2 = VBQ1 = 5v. then in order for Q2 to operate as a transistor, the base VBQ2 has to be a lower voltage than the collector VCQ2, so I'll choose VBQ2 to be 3.5v.
This 3.5v. is actually the peek voltage signal to be detected.
Now that the base input signal voltage is established than VEQ2 is 0.7v. less or around 2.8v.
So now it's time to solve for REQ2, by taking (VEQ2 / ICQ2) to give me around 330 ohms for REQ2. Now the original 505 ohms total emitter resistance earlier calculated can subtract this 330 ohms from it to give around 160 ohms for REQ1.
Now the emitter resistors are solved for Q1 so now it's time to return back to the base resistors of Q1. I will choose to make RAQ1 to be around 2 times the total emitter resistance of Q1 which was 505 ohms. So RAQ1 can be 1K ohms.
Now dividing this resistor into the VBQ1 = 5mA.
Solving for the RBQ1 gives a resistor of 1.5K ohms.
So Now the Volt. Pk. Detector, is finished at this point, and looking at the schematic you will see that a base input voltage that is around 3.2v. does NOT trigger the circuit. Vout is standing 8v.
But when the input volt. rises to around 3.5v. the circuit switches into conduction with full 12v. output.
Now with a standing (idle) voltage output of 8v. does not do any good for driving a LED load. (LED would be lit with around 1.8v. across it.
So now it's time to design a output stage that will be used as a comparator, so when the output voltage is idle, the comparator is off, but when the detector is triggered the comparator compares this new voltage against a reference voltage and sends an output signal to a LED display.
Alright the output standing volt. = 8v. So the best way to handle this is with a voltage device MOSFET, rather than a current device BJT.
Feed the output to the gate of the mosfet. .And Bias the source term. to equal the gate voltage under idle conditions. So source is biased at 8v. (this is the voltage divider network consisting of RS1Q3 and RS2Q3)
Now lets move on to a LED driver stage, since the output of the mosfet will be a negative going signal, than a PNP would be a good choice for this application..
I'll choose a 470 ohm resistor for the emitter of the PNP transistor. This is used as a current limiting resistor for the LED.
I'll make the base resistor 10 times this emitter resistor, at 4.7K. (RAQ4)
How it all works:
Now looking at the schematic you'll see that when there is a standing voltage of 8 volts at the collector of Q1, than the gate of Q3 is at 8v. and the source is biased at 8v. (reference voltage) So the mosfet is considered off. The base of the PNP is sitting at 12v. as well as the emitter, at 12v.
So this PNP is off and the LED is off.
When a high enough signal comes into the base of Q2 to cause it to conduct it begins to steal current away from the emitter of Q1, as well as dropping the base voltage of Q1 simultaneously, causing Q1 to go into a rapid change over into a switching off mode (acts like a switch rather than an amplifier) when Q1 switches off the collector voltage rises to 12V. through RCQ1 which places a 12v. signal voltage at the gate of Q3, and because the source is held at 8v. the mosfet essentially compares the gate voltage 12v. to it's reference voltage of 8v. and conducts heavily the resistance across the mosfet drops considerably so that the reference voltage at the source term. of the mosfet is now felt at the base of the PNP which lowers this base voltage below the 12v. felt at it's emitter thereby causing the PNP to conduct heavily sending current through the LED, to light it up.
When the input goes below a certain threshold voltage to the point where the Q2 base becomes much lower than its emitter than the circuit switches back off as the Q1 steals current away from the emitter of Q2, causing a change back in the state of the entire circuit, making the LED shut off until another input signal of proper poaqrity and voltage amplitude is applied again.
Study it very carefully and get a good undrstanding how these components are implemented into this circuit. And there function, and why and how the values were chosen, for the resistors.
This takes you away from designing a transistor as a pure amplifier, and seeing how to use it in a switching mode. You can see how the biasing is used in a whole new way rather than linear amplification, but instead the biasing was used for setting reference voltages, throughout.
As always have fun with it...
This is another circuit design tutorial
For beginners, to learn how to implement different transistor types into one ciruit.
And come up with a working circuit, that can be used with other circuits to build syustems.
This excersise will be explained to great detail as possible, so anyone starting out with a basic knowledge of OHMS law, can grasp a hold of these concepts.
The design is for a Voltage Peek detector, that has the output coupled to a LED, for a display.
This same circuit was breadboarded and tested for clarity, so I know it works properly.
DESIGN:
Refer to the schematics:
Below.
First I chose VCC to be 12v.
Then I chose RCQ1 to be 470 ohms.
Then I chose a standing (idle) output voltage for Q1 of 8v. (VCQ1)
I chose a base voltage of 5v. for Q1. (VBQ1)
Assuming a Vbe of 0.7v. VEQ1 will be 4.3v. (for calculation purposes only)
Now solving for the current through RCQ1 ~= 8.5mA. = ICQ2
And taking that current divided into VEQ1 ~= 505 ohms.
That establishes the front end of Q1, but before I calculate the base resistors of Q1 I need to consider the bias of Q2 stage,
Now lets move on to Q2,. for a moment.
Q2 will have its collector direct coupke to the base of Q1, so VBQ1 = VCQ2, and VCQ2 needs to be greater than the base voltage of Q2 (VBQ2,) but VBQ2 will be the voltage input to be detected. So looking at the schematic, the base of Q2 is not biased but rather just an input terminal, to the detector circuit.
Now the resistance for the emitter of Q1 is a split resistance.
The emitter of Q2 couples to the intersection of this network. This is to cause a feedback that will make Q1 turn off almost like a switch, rather than a linear response to the signal coming from Q2's collector.
So now since the VCQ2 = VBQ1 = 5v. then in order for Q2 to operate as a transistor, the base VBQ2 has to be a lower voltage than the collector VCQ2, so I'll choose VBQ2 to be 3.5v.
This 3.5v. is actually the peek voltage signal to be detected.
Now that the base input signal voltage is established than VEQ2 is 0.7v. less or around 2.8v.
So now it's time to solve for REQ2, by taking (VEQ2 / ICQ2) to give me around 330 ohms for REQ2. Now the original 505 ohms total emitter resistance earlier calculated can subtract this 330 ohms from it to give around 160 ohms for REQ1.
Now the emitter resistors are solved for Q1 so now it's time to return back to the base resistors of Q1. I will choose to make RAQ1 to be around 2 times the total emitter resistance of Q1 which was 505 ohms. So RAQ1 can be 1K ohms.
Now dividing this resistor into the VBQ1 = 5mA.
Solving for the RBQ1 gives a resistor of 1.5K ohms.
So Now the Volt. Pk. Detector, is finished at this point, and looking at the schematic you will see that a base input voltage that is around 3.2v. does NOT trigger the circuit. Vout is standing 8v.
But when the input volt. rises to around 3.5v. the circuit switches into conduction with full 12v. output.
Now with a standing (idle) voltage output of 8v. does not do any good for driving a LED load. (LED would be lit with around 1.8v. across it.
So now it's time to design a output stage that will be used as a comparator, so when the output voltage is idle, the comparator is off, but when the detector is triggered the comparator compares this new voltage against a reference voltage and sends an output signal to a LED display.
Alright the output standing volt. = 8v. So the best way to handle this is with a voltage device MOSFET, rather than a current device BJT.
Feed the output to the gate of the mosfet. .And Bias the source term. to equal the gate voltage under idle conditions. So source is biased at 8v. (this is the voltage divider network consisting of RS1Q3 and RS2Q3)
Now lets move on to a LED driver stage, since the output of the mosfet will be a negative going signal, than a PNP would be a good choice for this application..
I'll choose a 470 ohm resistor for the emitter of the PNP transistor. This is used as a current limiting resistor for the LED.
I'll make the base resistor 10 times this emitter resistor, at 4.7K. (RAQ4)
How it all works:
Now looking at the schematic you'll see that when there is a standing voltage of 8 volts at the collector of Q1, than the gate of Q3 is at 8v. and the source is biased at 8v. (reference voltage) So the mosfet is considered off. The base of the PNP is sitting at 12v. as well as the emitter, at 12v.
So this PNP is off and the LED is off.
When a high enough signal comes into the base of Q2 to cause it to conduct it begins to steal current away from the emitter of Q1, as well as dropping the base voltage of Q1 simultaneously, causing Q1 to go into a rapid change over into a switching off mode (acts like a switch rather than an amplifier) when Q1 switches off the collector voltage rises to 12V. through RCQ1 which places a 12v. signal voltage at the gate of Q3, and because the source is held at 8v. the mosfet essentially compares the gate voltage 12v. to it's reference voltage of 8v. and conducts heavily the resistance across the mosfet drops considerably so that the reference voltage at the source term. of the mosfet is now felt at the base of the PNP which lowers this base voltage below the 12v. felt at it's emitter thereby causing the PNP to conduct heavily sending current through the LED, to light it up.
When the input goes below a certain threshold voltage to the point where the Q2 base becomes much lower than its emitter than the circuit switches back off as the Q1 steals current away from the emitter of Q2, causing a change back in the state of the entire circuit, making the LED shut off until another input signal of proper poaqrity and voltage amplitude is applied again.
Study it very carefully and get a good undrstanding how these components are implemented into this circuit. And there function, and why and how the values were chosen, for the resistors.
This takes you away from designing a transistor as a pure amplifier, and seeing how to use it in a switching mode. You can see how the biasing is used in a whole new way rather than linear amplification, but instead the biasing was used for setting reference voltages, throughout.
As always have fun with it...
