This is a tutorial in designing complex transistor amplifiers at the HOBBY LEVEL.
In this turoial you will learn how to bias transistors for switching operations, and amplifying operations.
Then you will learn how to DC couple each stage to the next while maintaining testpoint voltages.
Then finally a AC stage will be designed and capacitively coupled to this DC amp , to produce a small signal detector circuit.
Here is a quick video demonstrating it being used as a detector for cell phone activity, and flourescent light noise detection.
This circuit will pick up very small noise interferences in flourescent lightings, and electronic devices, it could be used as a current sensor as well.
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Ok here are the steps taken to design this kind of a circuit.
The object of this game, is to preserve as much as posssible the original voltages at key points (test points) at the output.
First lets choose a transistor to use at the output. I'm going to use a PNP transistor to drive a LED load.
Now lets decide on a current to flow through this LED branch. I'll choose 10mA.
Alright that's it, lets get to designing.
First look at data sheets and do some preliminay test procedures to see how a component works under ceretain conditions.
Testing an LED I found that around 1.7 to 1.8 volts dropped across it is good to use for a current of 10mA. (Test conditions show that.)
Now a very important step to consider in designing transistor circuits.
If you want to use a transistor as a switching device, aim to make the voltage across the transistor (VCE) to be around 0.3V or less.
If you want to use the transistor as an amplifying device then aim to bias the transistor to have more than 1V across it (VCE)
So now the output stage Q1 will be biased to drive an LED with 1.7V across it @ 10mA.
lets do some designation.
IC1 = current flow through Q1 branch. = 10mA.
RE1 is the emitter resistor of Q1.
RBA1 is the pull up resistor for Q1
RBB1 is the pull down resistor for Q1.
Since Q1 is a PNP transistor it gets it's base current drive through the pull down resistor (RBB1).
Note that RBA1 and RBB1 form a voltage divider for the base voltage to be established for Q1. (very important).
Learn these 2 rules.
The PNP transistor is biased on when the collector is more negative than the base, and the base is more neg. than the emitter.
The NPN transistor is biased on when the collector is more positive than the base and the base is more positive than the emitter.
The base to emitter voltages can be anywhere from 0.6V to 0.8V, typically we use around 0.7V for design calculations.
This is known as Vbe.
ONE VERY IMPORTANT STEP:
After each stage of design, build the circuit stage, and check the bias voltages, adjust resistor values(tweak only) as needed to keep as CLOSE as POSSIBLE, the calculated values. You will very unlikely keep the values constant throughout the whole design, as you are stuck to use values that may not be what you calculated, so by using standard values could and will upset voltages elsewhere, as you continue to add on stages, when this happens, you tweak values as you build and test. NEVER, NEVER allow the transistor, to be the cause of the change in bias, when tweaking.
Example: you calculated a value for the base bias divider, and it sits much higher in voltage than what you designed it for, however you discover that when you hook the transistor base to the divider it brings the base voltage down to where you wanted it to be.
Very BAD design, now the transistor becomes part of the base voltage, and it will not be stable, as the transistor parameters change so will the base voltage change. You always design the bias networks to be INDEPENDANT of the transistor parameters, the bias voltages need to remain as stable as POSSIBLE, with respect to the transistor hooked to it.
So we make the bias divider current at least 10 times greater than expected base current, so the transistor has very little influence on it. I like to make the divider current around 1/10 th of the collector current, it seems to work fine that way, for a lot of my designs.
Lets begin the simplistic math work now. All is needed is a calculator
Now lets solve for the values of all these resistors.
Since the resistors have tolerances you try to use values close to your calculated values.
I'm going to write down the step procedures, I will start out with some detail explanations in the calcualtions, but as the succeding stages are designed I will stop the explanations and just show the math sequence, just follow along, and you will begin to see the pattern of approach to solving for these values.
Here we go.
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StageQ1 design:
VLED= 1.7V
VCEQ1 = 0.3volts. I want it to be used as a switch to drive the LED.
Starting from ground (VLED + VCEQ1) = (1.7v + 0.3v) = 2V. the result of adding the two voltage drops desired across both components in series. Now this is the voltage at the emitter with respect to ground.
Now lets continue this branch and solve for the value of REQ1.
Since we have 2V dropped across both the LED and Q1 together, than subtracting that value from the battery voltage of 8V results in (8V - 2V) = 6V left to be dropped across RE1.
Therefor The current through this branch was determined to be 10mA, so RE1 = (6V / 10mA)= 600 ohms. A choice can be made to find values close to that value, or if you have that value than use it. Different manufactures of resistors and there tolerances have differing values to choose from. MY resistor assortment doesn't carry a 600 ohm resistor, the closest values I can use is 560 or 620 ohm resistor. So I need to make a choice and see how it works. I am choosing to use a 560 ohm resistor. (No special reason)
OK so RE1 is 560 ohms.
Now to solve for RBB1. Since I have 2V at the emitter respect to ground, than subtract 0.7Vbe, makes the voltage at the base of Q1 be (2v - 0.7v)= 1.3V. That is the base voltage (VBQ1) I'm aiming for to ensure I'll have my 10mA of collector current in the load branch.
Remember I said earlier we need test points. that is one of the key test points, throughout the rest of this design.>> (VBQ1)!
Now I normally for an amplifying stage want the divider current to be around 1/10th of the collector current, however because I want this transistor to be biased on as a switch, I want the divider current to equal the collector current.
So the divider current will be 10mA as well. Now we can solve for RBB1.
RBB1 = (VBQ1 / 10mA) = 130 ohms. MY supply carries 130 ohms so I'll use that value in the build.
Now the last resistor is RBA1, this is solved by this equation.
RBA1 = {(VCC - VBQ1) / 10mA}>> {(8V - 1.3V) / 10mA} = 670 ohms. the closest value in my collection is 68o ohms. I'll use 680 ohms. (VCC is the battery voltage)
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Stage Q2 design:
Now for stage Q2 that drives stageQ1.
I want Q2 to be biased as a switch. So VCEQ2 = 0.3V.
I want to use an NPN transistor in a CE mode to drive Q1.
I also want this to be direct coupled to Q1 stage.
Now all the parameters are set for Q2 stage design.
Keeping RBA1 in the circuit and substituting RBB1 with Q2 works out as follows.
I know from last step that the current flowing through RBA1 is 10mA, therefor Q2 will supply that 10mA through its collector branch.
A note to remember, Q2 is acting as a current sink in this branch, technically, however using the term supplying current is feasable as well.
So now how to calculate the value of RE2.
Since we know from the above step that VBQ1 is 1.3V, and we want 0.3V to be VCEQ2, than that leaves 1V across RE2 which is (VRE2).
Now RE2 is solved as such:
RE2 = (VRE2 / 10mA) = (1V / 10mA)= 100 ohms.
Q2 is a NPN transistor so its base voltge Vbe is 0.7v higher than its emitter, so VBQ2 is (VRE2 + Vbe) = (1v + 0.7v) = 1.7V for base voltage of Q2.>> (VBQ2).
Again because Q2 is acting as a switch the divider current is the same as the collector current in this case again 10mA.
So now RBA2 =(VBQ2 / 10mA) = (1.7v / 10mA)= 170 ohms, I'll go higher and use 180 ohms closest standard for my collection.
finally again RBB2 = {(VCC - VBQ2) / 10mA}>> {(8v - 1.7v) / 10mA}= 630 ohms my standard closest would be 620 ohms.
So in order to design this as a direct coupled stage, I substituted the RBB1 resitor with Q2 and its bias network.
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Stage Q3 design:
Now that's for these two stages designed to work close to being as switchs, the rest of the design stages are going to be biased as amplifiers, here is how this next stage is calculated.
Note:
Q3 will be biased into its linear region, so arbitrarily I'm choosing to put around 4V across these transistors. (VCE).
Also the base divider current now is going to be 1/10th of the collector current. In this case 1mA.
Now if I use another NPN transistor to drive Q2, than I'll have to design it as a CC stage, but I want voltage amplification, so I need a CE stage, so another PNP will make that possible., This way the load (Q2) is in the collector lead, and voltage amplification stage is aquired.
So a PNP transistor is chosen for this stage, and it will be used to direct couple and take the place of RBB2 previously used.
So keeping RBA2 in place it now becomes the collector resistor voltage for Q3.
Now lets solve for RE3.
VCEQ3 is chosen to be 4V.
So VEQ3 (voltage at the emitter of Q3 from ground is (VBQ2 + VCEQ3) = (1.7V + 4v) = 5.7V = (VEQ3)
Now to solve for RE3 goes like this, (VCC - VEQ3) / IC3. where IC3 is the base divider current for Q2 previously used. Which was 10mA.
Therefor RE3 = {(8v -5.7v) / 10mA } = 230 ohms, my colsest value 220 ohms.
Now again solving for RBA3 is as follows, the voltage at the emitter of Q3 minus the Vbe (remember with PNP subtract), would be the voltage at the base of Q3 (VBQ3). So (5.7V - 0.7v) = 5V for VBQ3.
Remember the divider current is NOW 1/10th of collector current for LINEAR biasing operation of the next stages.
So RBA3 calcualtes as {(VCC - VBQ3) / 1mA}= (8v - 5v) / 1mA= 3K ohms.
And finally RBB3 is solved as (VBQ3 / 1mA) = (5v / 1mA) = 5K ohms, closest value for my collection is 5.1K ohms.
NOTE: after I built and tested the circuit I found that with 220 ohms in the emitter of Q3 raised the voltage at the base of Q1 from 1.2V to around 1.4v, that increased the voltage across Q1, (which I don't want), so I lowered RE2 to 200 ohms and the VBQ1 went to around 1.17V, so now the stages are working together.
The base voltage of Q1, (VBQ1) needs to be as low as possible, to be used in switching mode.
This demonstrates how you need to design and build and test each stage added, if you can keep the voltages at the testpoint close to the same values through out the entire design then the circuit will be a success, if the voltages change at the test points, than going back and redesigning values for the stage that caused it is necessary, as I did in this case.
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Ok from here I will just show the math as I go to the next 3 stages of design.
When I get to stage Q7, I'll be coupling it by a capacitor, to the previous stages and I'll give my design explanations about that part.
Stage Q4 design:
VCEQ4 = 4V chosen value.
VRE4 = (VBQ3 - VCEQ4) = (5v - 4v) = 1V
RE4 = (VRE4 / 1mA) = 1K ohms.
VBQ4 = (VRE4 + 0.7Vbe) = 1.7V= (VBQ4)
Divider current for Q4 base is 1/10th of collector current (1mA / 10 ) = 100uA.
RBA4 = (VBQ4 / 100uA) = 17K (use 16K standard for me)
RBB4 = {(VCC - VBQ4) / 100uA}= 63K (I'll use 62K standard) ohms.
After testing I found that 62K ohms was too high it changed the testpoint voltages, so I lowered it to 56K ohms, and got the test voltage back to normal, NOTE: (Test voltage is the voltage at the base of Q1)
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Stage Q5 design:
I wanted to use another NPN transistor stage, so that leaves now a CC amp to be designed, this is to get the input imprdance up high, at the final stages.
VCEQ5 = 4V.
ICQ5 = 100uA
divider current for Q5 is now 1/10th of that = 10uA.
(VBQ4 + VbeQ5) = (1.7v + 0.7v) = 2.4V > (VBQ5)
RBA5 = (VBQ5 / 10uA) = (2.4V / 10uA)= 240K ohms.
RBB5 = [(VCC - VBQ5) / 10uA] =[(8v - 2.4v) / 10uA]= 560K ohms.
VCQ5 = voltage at the collector of Q5 with respect to ground (VBQ4 + VCEQ5) = ( 1.7v + 4v) = 5.7V = VCQ5
RE5 = {(VCC - VCQ5) / IC5} = [(8v - 5.7v) / 100uA]= 23K ohms. (24K closest standard)
Again after testing at the testpoint (VBQ1) I found I had to raise the RBA5 value up to 270K ohms to keep the test point value close to original value of 1.2V.
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Q6 stage design.
To increase the input resistance further another CC stage with a NPN transistor is used.
VBQ6 = (VBQ5 + 0.7v) = (2.4v +0.7v) = 3.1V VBQ6.
divider current = (10uA /10) = 1uA.
RBA6 = (VBQ6 / 1uA) = (3.1v / 1uA) = 3.1Meg ohms. standard 3.3Meg can be used.
RBB6 = {(VCC - VBQ6) / 1uA}= [(8v - 3.1v) / 1uA]= 4.9Meg ohms.
NOTE: when you get this small of collector and base divider currents, you are now ready to amplify very miniscule signals, meaning that, resistors in the divider network cannot be calculated anymore, the amplification of noise and such comes into play, its at this stage in the design, where you incorporate trimmer resistors into the bias network, here is where you need to really watch the testpoint value (VBQ1), you put a voltmeter reading at the base of Q1, then begin tweaking the base resistors of Q6, until you attain close to value of your voltage of 1.2V or lower, for Q1 base. Once this is worked out your LED should be ON, this ensures your circuit is working properly. Now to make it have an actual use, you need to turn off the LED so that an input signal can trigger it into conduction. So adjust the trimmer pot on Q6 stage until the LED goes off and remains off, at this point the VBQ1 will be well above 1.2V most likely in cutoff, however because of the amplification that is designed into this it won't take much of a voltage input to trigger it into conduction again.
Once this is attained, then the DC portion of this amp is completed, and it should pick up and amplify the noise coming from around the circuits environment, the LED will light up brighly when you touch any leads on Q6 stage, when it does you know your ready for the input stage Q7.
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Stage Q7 design.
Now that the input impedance to this amp is way up in the megohm range, it is nothing to design a capacitive couple CE amp stage to it, and know that the output of this stage will not be bogged down as it is feeding into this Dc amp we just built.
I arbitrarily chose to use 3.9K ohm resistor for the collector load of Q7.
Now using the equation voltage gain = (RC7 / RE7) I decided to use a stage gain of around 10.
So that would make RE7 10 times less than RC7. or 390 ohms.
Now you want to make the voltage drop across the collector resistor be close to half the supply voltage.
So VRC7 = 4V
now the collector current for Q7 is solved as follows (VRC7 / RC7) = (4v / 3.9K) = 1mA
now volt drop across RE7 is (1mA x 390 ohms) = 399mV. >(VE7)
now VBQ7 is (VE7 + 0.7Vbe) = (0.39v + 0.7v)=~1.1V
Now divider current 1/10th of collector current makes it = 100uA.
and RBA7 = (VBQ7 / 100uA) = 11k ohms.
and RBB7 = [(VCC - VBQ7) / 100uA] = 69K (standard 68K can be used)
You can use any capacitor for (CC7). from 0.1uf to 47uf.
Experiment with different size capacitors for (BP7) to get more sensitivity to the circuit performance.
Build and test this stage, however because it has no influence on the other stages due to capacitor isolation, than the test point for this stage would be the collector volrtage VCQ7 of Q7, tweak the values of RBA7 or RBB7 until the VCQ7 is around 4V. or so.
Now again you may need to trim the Q6 syage to get the LED to turn off and remain off until a valid input signal is applied to Q7 stage.
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To test it hook up a long wire through a capacitor to the base of Q7 and see if it will light up the LED, it may take some tweaking but it should pick up noise when electrical appliances are close to the wire antenna.
I hooked it up to a small inductor and hooked up the other end of the inductor to ground, and used it to detect electrical currents flowing through my cell phone and computer peripherals.
It is up to you to design and build circuits to add on to this to turn on different loads, or different input circuits, ect...
This circuit is one building block, for more involved circuit designs.
Enjoy your sucess in designing a complex small signal amplifier.
In this turoial you will learn how to bias transistors for switching operations, and amplifying operations.
Then you will learn how to DC couple each stage to the next while maintaining testpoint voltages.
Then finally a AC stage will be designed and capacitively coupled to this DC amp , to produce a small signal detector circuit.
Here is a quick video demonstrating it being used as a detector for cell phone activity, and flourescent light noise detection.
This circuit will pick up very small noise interferences in flourescent lightings, and electronic devices, it could be used as a current sensor as well.
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Ok here are the steps taken to design this kind of a circuit.
The object of this game, is to preserve as much as posssible the original voltages at key points (test points) at the output.
First lets choose a transistor to use at the output. I'm going to use a PNP transistor to drive a LED load.
Now lets decide on a current to flow through this LED branch. I'll choose 10mA.
Alright that's it, lets get to designing.
First look at data sheets and do some preliminay test procedures to see how a component works under ceretain conditions.
Testing an LED I found that around 1.7 to 1.8 volts dropped across it is good to use for a current of 10mA. (Test conditions show that.)
Now a very important step to consider in designing transistor circuits.
If you want to use a transistor as a switching device, aim to make the voltage across the transistor (VCE) to be around 0.3V or less.
If you want to use the transistor as an amplifying device then aim to bias the transistor to have more than 1V across it (VCE)
So now the output stage Q1 will be biased to drive an LED with 1.7V across it @ 10mA.
lets do some designation.
IC1 = current flow through Q1 branch. = 10mA.
RE1 is the emitter resistor of Q1.
RBA1 is the pull up resistor for Q1
RBB1 is the pull down resistor for Q1.
Since Q1 is a PNP transistor it gets it's base current drive through the pull down resistor (RBB1).
Note that RBA1 and RBB1 form a voltage divider for the base voltage to be established for Q1. (very important).
Learn these 2 rules.
The PNP transistor is biased on when the collector is more negative than the base, and the base is more neg. than the emitter.
The NPN transistor is biased on when the collector is more positive than the base and the base is more positive than the emitter.
The base to emitter voltages can be anywhere from 0.6V to 0.8V, typically we use around 0.7V for design calculations.
This is known as Vbe.
ONE VERY IMPORTANT STEP:
After each stage of design, build the circuit stage, and check the bias voltages, adjust resistor values(tweak only) as needed to keep as CLOSE as POSSIBLE, the calculated values. You will very unlikely keep the values constant throughout the whole design, as you are stuck to use values that may not be what you calculated, so by using standard values could and will upset voltages elsewhere, as you continue to add on stages, when this happens, you tweak values as you build and test. NEVER, NEVER allow the transistor, to be the cause of the change in bias, when tweaking.
Example: you calculated a value for the base bias divider, and it sits much higher in voltage than what you designed it for, however you discover that when you hook the transistor base to the divider it brings the base voltage down to where you wanted it to be.
Very BAD design, now the transistor becomes part of the base voltage, and it will not be stable, as the transistor parameters change so will the base voltage change. You always design the bias networks to be INDEPENDANT of the transistor parameters, the bias voltages need to remain as stable as POSSIBLE, with respect to the transistor hooked to it.
So we make the bias divider current at least 10 times greater than expected base current, so the transistor has very little influence on it. I like to make the divider current around 1/10 th of the collector current, it seems to work fine that way, for a lot of my designs.
Lets begin the simplistic math work now. All is needed is a calculator
Now lets solve for the values of all these resistors.
Since the resistors have tolerances you try to use values close to your calculated values.
I'm going to write down the step procedures, I will start out with some detail explanations in the calcualtions, but as the succeding stages are designed I will stop the explanations and just show the math sequence, just follow along, and you will begin to see the pattern of approach to solving for these values.
Here we go.
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StageQ1 design:
VLED= 1.7V
VCEQ1 = 0.3volts. I want it to be used as a switch to drive the LED.
Starting from ground (VLED + VCEQ1) = (1.7v + 0.3v) = 2V. the result of adding the two voltage drops desired across both components in series. Now this is the voltage at the emitter with respect to ground.
Now lets continue this branch and solve for the value of REQ1.
Since we have 2V dropped across both the LED and Q1 together, than subtracting that value from the battery voltage of 8V results in (8V - 2V) = 6V left to be dropped across RE1.
Therefor The current through this branch was determined to be 10mA, so RE1 = (6V / 10mA)= 600 ohms. A choice can be made to find values close to that value, or if you have that value than use it. Different manufactures of resistors and there tolerances have differing values to choose from. MY resistor assortment doesn't carry a 600 ohm resistor, the closest values I can use is 560 or 620 ohm resistor. So I need to make a choice and see how it works. I am choosing to use a 560 ohm resistor. (No special reason)
OK so RE1 is 560 ohms.
Now to solve for RBB1. Since I have 2V at the emitter respect to ground, than subtract 0.7Vbe, makes the voltage at the base of Q1 be (2v - 0.7v)= 1.3V. That is the base voltage (VBQ1) I'm aiming for to ensure I'll have my 10mA of collector current in the load branch.
Remember I said earlier we need test points. that is one of the key test points, throughout the rest of this design.>> (VBQ1)!
Now I normally for an amplifying stage want the divider current to be around 1/10th of the collector current, however because I want this transistor to be biased on as a switch, I want the divider current to equal the collector current.
So the divider current will be 10mA as well. Now we can solve for RBB1.
RBB1 = (VBQ1 / 10mA) = 130 ohms. MY supply carries 130 ohms so I'll use that value in the build.
Now the last resistor is RBA1, this is solved by this equation.
RBA1 = {(VCC - VBQ1) / 10mA}>> {(8V - 1.3V) / 10mA} = 670 ohms. the closest value in my collection is 68o ohms. I'll use 680 ohms. (VCC is the battery voltage)
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Stage Q2 design:
Now for stage Q2 that drives stageQ1.
I want Q2 to be biased as a switch. So VCEQ2 = 0.3V.
I want to use an NPN transistor in a CE mode to drive Q1.
I also want this to be direct coupled to Q1 stage.
Now all the parameters are set for Q2 stage design.
Keeping RBA1 in the circuit and substituting RBB1 with Q2 works out as follows.
I know from last step that the current flowing through RBA1 is 10mA, therefor Q2 will supply that 10mA through its collector branch.
A note to remember, Q2 is acting as a current sink in this branch, technically, however using the term supplying current is feasable as well.
So now how to calculate the value of RE2.
Since we know from the above step that VBQ1 is 1.3V, and we want 0.3V to be VCEQ2, than that leaves 1V across RE2 which is (VRE2).
Now RE2 is solved as such:
RE2 = (VRE2 / 10mA) = (1V / 10mA)= 100 ohms.
Q2 is a NPN transistor so its base voltge Vbe is 0.7v higher than its emitter, so VBQ2 is (VRE2 + Vbe) = (1v + 0.7v) = 1.7V for base voltage of Q2.>> (VBQ2).
Again because Q2 is acting as a switch the divider current is the same as the collector current in this case again 10mA.
So now RBA2 =(VBQ2 / 10mA) = (1.7v / 10mA)= 170 ohms, I'll go higher and use 180 ohms closest standard for my collection.
finally again RBB2 = {(VCC - VBQ2) / 10mA}>> {(8v - 1.7v) / 10mA}= 630 ohms my standard closest would be 620 ohms.
So in order to design this as a direct coupled stage, I substituted the RBB1 resitor with Q2 and its bias network.
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Stage Q3 design:
Now that's for these two stages designed to work close to being as switchs, the rest of the design stages are going to be biased as amplifiers, here is how this next stage is calculated.
Note:
Q3 will be biased into its linear region, so arbitrarily I'm choosing to put around 4V across these transistors. (VCE).
Also the base divider current now is going to be 1/10th of the collector current. In this case 1mA.
Now if I use another NPN transistor to drive Q2, than I'll have to design it as a CC stage, but I want voltage amplification, so I need a CE stage, so another PNP will make that possible., This way the load (Q2) is in the collector lead, and voltage amplification stage is aquired.
So a PNP transistor is chosen for this stage, and it will be used to direct couple and take the place of RBB2 previously used.
So keeping RBA2 in place it now becomes the collector resistor voltage for Q3.
Now lets solve for RE3.
VCEQ3 is chosen to be 4V.
So VEQ3 (voltage at the emitter of Q3 from ground is (VBQ2 + VCEQ3) = (1.7V + 4v) = 5.7V = (VEQ3)
Now to solve for RE3 goes like this, (VCC - VEQ3) / IC3. where IC3 is the base divider current for Q2 previously used. Which was 10mA.
Therefor RE3 = {(8v -5.7v) / 10mA } = 230 ohms, my colsest value 220 ohms.
Now again solving for RBA3 is as follows, the voltage at the emitter of Q3 minus the Vbe (remember with PNP subtract), would be the voltage at the base of Q3 (VBQ3). So (5.7V - 0.7v) = 5V for VBQ3.
Remember the divider current is NOW 1/10th of collector current for LINEAR biasing operation of the next stages.
So RBA3 calcualtes as {(VCC - VBQ3) / 1mA}= (8v - 5v) / 1mA= 3K ohms.
And finally RBB3 is solved as (VBQ3 / 1mA) = (5v / 1mA) = 5K ohms, closest value for my collection is 5.1K ohms.
NOTE: after I built and tested the circuit I found that with 220 ohms in the emitter of Q3 raised the voltage at the base of Q1 from 1.2V to around 1.4v, that increased the voltage across Q1, (which I don't want), so I lowered RE2 to 200 ohms and the VBQ1 went to around 1.17V, so now the stages are working together.
The base voltage of Q1, (VBQ1) needs to be as low as possible, to be used in switching mode.
This demonstrates how you need to design and build and test each stage added, if you can keep the voltages at the testpoint close to the same values through out the entire design then the circuit will be a success, if the voltages change at the test points, than going back and redesigning values for the stage that caused it is necessary, as I did in this case.
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Ok from here I will just show the math as I go to the next 3 stages of design.
When I get to stage Q7, I'll be coupling it by a capacitor, to the previous stages and I'll give my design explanations about that part.
Stage Q4 design:
VCEQ4 = 4V chosen value.
VRE4 = (VBQ3 - VCEQ4) = (5v - 4v) = 1V
RE4 = (VRE4 / 1mA) = 1K ohms.
VBQ4 = (VRE4 + 0.7Vbe) = 1.7V= (VBQ4)
Divider current for Q4 base is 1/10th of collector current (1mA / 10 ) = 100uA.
RBA4 = (VBQ4 / 100uA) = 17K (use 16K standard for me)
RBB4 = {(VCC - VBQ4) / 100uA}= 63K (I'll use 62K standard) ohms.
After testing I found that 62K ohms was too high it changed the testpoint voltages, so I lowered it to 56K ohms, and got the test voltage back to normal, NOTE: (Test voltage is the voltage at the base of Q1)
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Stage Q5 design:
I wanted to use another NPN transistor stage, so that leaves now a CC amp to be designed, this is to get the input imprdance up high, at the final stages.
VCEQ5 = 4V.
ICQ5 = 100uA
divider current for Q5 is now 1/10th of that = 10uA.
(VBQ4 + VbeQ5) = (1.7v + 0.7v) = 2.4V > (VBQ5)
RBA5 = (VBQ5 / 10uA) = (2.4V / 10uA)= 240K ohms.
RBB5 = [(VCC - VBQ5) / 10uA] =[(8v - 2.4v) / 10uA]= 560K ohms.
VCQ5 = voltage at the collector of Q5 with respect to ground (VBQ4 + VCEQ5) = ( 1.7v + 4v) = 5.7V = VCQ5
RE5 = {(VCC - VCQ5) / IC5} = [(8v - 5.7v) / 100uA]= 23K ohms. (24K closest standard)
Again after testing at the testpoint (VBQ1) I found I had to raise the RBA5 value up to 270K ohms to keep the test point value close to original value of 1.2V.
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Q6 stage design.
To increase the input resistance further another CC stage with a NPN transistor is used.
VBQ6 = (VBQ5 + 0.7v) = (2.4v +0.7v) = 3.1V VBQ6.
divider current = (10uA /10) = 1uA.
RBA6 = (VBQ6 / 1uA) = (3.1v / 1uA) = 3.1Meg ohms. standard 3.3Meg can be used.
RBB6 = {(VCC - VBQ6) / 1uA}= [(8v - 3.1v) / 1uA]= 4.9Meg ohms.
NOTE: when you get this small of collector and base divider currents, you are now ready to amplify very miniscule signals, meaning that, resistors in the divider network cannot be calculated anymore, the amplification of noise and such comes into play, its at this stage in the design, where you incorporate trimmer resistors into the bias network, here is where you need to really watch the testpoint value (VBQ1), you put a voltmeter reading at the base of Q1, then begin tweaking the base resistors of Q6, until you attain close to value of your voltage of 1.2V or lower, for Q1 base. Once this is worked out your LED should be ON, this ensures your circuit is working properly. Now to make it have an actual use, you need to turn off the LED so that an input signal can trigger it into conduction. So adjust the trimmer pot on Q6 stage until the LED goes off and remains off, at this point the VBQ1 will be well above 1.2V most likely in cutoff, however because of the amplification that is designed into this it won't take much of a voltage input to trigger it into conduction again.
Once this is attained, then the DC portion of this amp is completed, and it should pick up and amplify the noise coming from around the circuits environment, the LED will light up brighly when you touch any leads on Q6 stage, when it does you know your ready for the input stage Q7.
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Stage Q7 design.
Now that the input impedance to this amp is way up in the megohm range, it is nothing to design a capacitive couple CE amp stage to it, and know that the output of this stage will not be bogged down as it is feeding into this Dc amp we just built.
I arbitrarily chose to use 3.9K ohm resistor for the collector load of Q7.
Now using the equation voltage gain = (RC7 / RE7) I decided to use a stage gain of around 10.
So that would make RE7 10 times less than RC7. or 390 ohms.
Now you want to make the voltage drop across the collector resistor be close to half the supply voltage.
So VRC7 = 4V
now the collector current for Q7 is solved as follows (VRC7 / RC7) = (4v / 3.9K) = 1mA
now volt drop across RE7 is (1mA x 390 ohms) = 399mV. >(VE7)
now VBQ7 is (VE7 + 0.7Vbe) = (0.39v + 0.7v)=~1.1V
Now divider current 1/10th of collector current makes it = 100uA.
and RBA7 = (VBQ7 / 100uA) = 11k ohms.
and RBB7 = [(VCC - VBQ7) / 100uA] = 69K (standard 68K can be used)
You can use any capacitor for (CC7). from 0.1uf to 47uf.
Experiment with different size capacitors for (BP7) to get more sensitivity to the circuit performance.
Build and test this stage, however because it has no influence on the other stages due to capacitor isolation, than the test point for this stage would be the collector volrtage VCQ7 of Q7, tweak the values of RBA7 or RBB7 until the VCQ7 is around 4V. or so.
Now again you may need to trim the Q6 syage to get the LED to turn off and remain off until a valid input signal is applied to Q7 stage.
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To test it hook up a long wire through a capacitor to the base of Q7 and see if it will light up the LED, it may take some tweaking but it should pick up noise when electrical appliances are close to the wire antenna.
I hooked it up to a small inductor and hooked up the other end of the inductor to ground, and used it to detect electrical currents flowing through my cell phone and computer peripherals.
It is up to you to design and build circuits to add on to this to turn on different loads, or different input circuits, ect...
This circuit is one building block, for more involved circuit designs.
Enjoy your sucess in designing a complex small signal amplifier.
