Question regarding basic transistor circuit
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 11th September 2009, 09:56 AM #1 300b   diyAudio Member   Join Date: Aug 2008 Question regarding basic transistor circuit Hello Guys, I know its not really about amplifiers, but it is SS, so I guess its the right forum. I have some experience in building a couple of basic amps, but I've decided to learn electronics better since I like it. I'm reading "The Art of Electronics", and have a problem understanding one of the circuits, here it is: Can someone please explain to me how exactly it works (keep in mind my knowledge is very limited, so a more detailed explanation will be highly appreciated)? What difference does the 180ohm resistor make? Thank you.
 11th September 2009, 12:43 PM #2 Bigun   diyAudio Member     Join Date: Jan 2009 Location: Waterloo, ON or Herefordshire UK I'm a bit of newbie myself, so this is just a guess The 2nd device is the active device set up to behave as an amplifier in 'common emitter' mode (the emitter is common to both the input and output signals) and the output is from the collector which means it will be inverted with respect to the input. It is a Class A amplifier - the output device is always 'on' and needs to be held 'on' so that there's enough current flowing through it without an input signal. It is marked up to indicate that the output should be held at a potential mid-way between the positive rail and ground - so that with a signal applied the output voltage has plenty of room to swing up (towards the positive rail) and down (towards ground). [this isn't the optimal bias point for maximum output, the half-way rule applies when the output device is loaded by a constant current source - but that's a whole other story] To correctly bias this transistor means holding it's base at the right potential with respect to it's emitter (about 0.65V will start to turn on the transistor). The first transistor is set up so that it provides this bias. It's pretty much wired to act as a diode with it's base connected to the collector to turn on the transistor via a current limiting resistor and hence current will flow from collector to emitter. This first device together with it's collector resistor forms a potential divider to generate a bias voltage which can be used to hold the base of the 2nd device at a suitable operating point. The advantage of doing it this way is temperature compensation. The collector-emitter current is a function of the base-emitter potential in a bipolar transistor, but this behaviour changes with temperature. Roughly, a 1 deg. C change produces the same effect as change of 2mV in the base-emitter voltage. But here, since both transistors will track roughly the same the bias point generated by the 1st device will move with temperature just the right amount to keep the 2nd device stable over a reasonable range in temperature. This is important because without it, the 2nd device will heat up due to the current flow through it and this can result in thermal runaway. Both devices will need to be on the same heatsink so that their temperatures track together. The 180 Ohm emitter resistors reduce the gain of the devices (the gain is roughly collector resistance divided by emitter resistance). The benefit of this is that the gain is now more controlled, depending on fixed resistors rather than the intrinsic gain and internal resistances of the transistor that may vary with normal manufacturing tolerances. The reduced gain will also result in improved linearity (look up 'emitter degeneration'). Hope my guess is not too far off the mark __________________ "The test of the machine is the satisfaction it gives you. There isn't any other test. If the machine produces tranquility it's right. If it disturbs you it's wrong until either the machine or your mind is changed." Robert M Pirsig. Last edited by Bigun; 11th September 2009 at 12:55 PM.
 11th September 2009, 01:34 PM #3 300b   diyAudio Member   Join Date: Aug 2008 Hi Bigun, Thank you for the explanation. Can you please tell me why the 10K is needed for the Q1 base? And I'm not sure I understand, how does the change of temperature makes Q1 compensate for Q2?
 11th September 2009, 02:14 PM #4 chrisbx1975x   diyAudio Member   Join Date: Sep 2009 Location: new hampshire, usa the 180 ohm resistor improves the stability of the circuit. limits current and allows for setting the voltage gain. otherwise the gain would be 10000 divided by the internal transresistance of the transistor approximately, which is extremely high, very dependent on the actual beta of the transistor and extremely susceptible to changes in temperature. an emitter resistor is common place for setting and limiting the gain and increasing the input impedance. q2 is the only actual amplifier here, common emitter. stabiized by q1. 180 ohm on each emitter will improve the stability of both transistors.
 11th September 2009, 02:28 PM #5 chrisbx1975x   diyAudio Member   Join Date: Sep 2009 Location: new hampshire, usa also, by stability i mean improving the stability of the 10v point. if is unstable it drifts around (10, 10.8, 9.4, etc)which is very undesireable and affects all parts of the circuit. this makes it hold more steady, although it will still drift with a circuit like this.
 11th September 2009, 02:37 PM #6 chrisbx1975x   diyAudio Member   Join Date: Sep 2009 Location: new hampshire, usa the 10k resistor turns on q1. otherwise, the transistor is completely off.
 11th September 2009, 03:03 PM #7 Bigun   diyAudio Member     Join Date: Jan 2009 Location: Waterloo, ON or Herefordshire UK There are a couple of 10k resistors in this circuit. They are both there to limit the current flow into the bases of the two transistors and in doing so they ensure that the input to these bases is essentially a voltage since it is the base-emitter voltage difference that controls the current flow through the transistors. When the temperature changes, the current flow through Q1 will change even though the base-emitter voltage applied through the 10k resistor is virtually the same. This change in current flow results in a change to the voltage at the collector of Q1 which in turn changes the voltage presented to the base of Q2. This change nullifies the affect that the temperature change in Q2 has on the current flow through Q2. You can replace Q1 with a single diode to achieve the same effect. The benefit of using a transistor of the same type of Q2 is potentially a much better match of their thermal behaviour. __________________ "The test of the machine is the satisfaction it gives you. There isn't any other test. If the machine produces tranquility it's right. If it disturbs you it's wrong until either the machine or your mind is changed." Robert M Pirsig. Last edited by Bigun; 11th September 2009 at 03:12 PM.
 11th September 2009, 05:38 PM #8 Bigun   diyAudio Member     Join Date: Jan 2009 Location: Waterloo, ON or Herefordshire UK I will add another note as it may well come up during your learning. Something that came up during my learning were statements that bipolar transistors are 'current devices' and field effect transistors are 'voltage devices'. Then there were statements to the contrary and some wise words about the fact that current and voltage go hand in hand. Many of the comments are correct in their context and are intended to help by providing a simple concept to understand circuit behaviour without needing to know the physics of how the devices work. But unfortunately there's no free lunch and it's worth knowing how the devices work if you want to understand the circuits better. This issue is important for designing circuits since it affects the basic issues of proper biass/operating point, impedance matching, linearity etc. Here's my take (so far, on my journey of learning) on the bipolar transistor FWIW... it is a voltage-controlled current device (i.e. it is neither a 'voltage device' nor a 'current device'). The voltage across the base-emitter junction is the primary determinant of the current that flows between collector and emitter. The other factor is the voltage across the collector-emitter, if there's no voltage there to drive a current flow then no matter the base-emitter voltage you won't get any current. So there must be a voltage across the collector-emitter and this voltage also has some impact on the amount of current that will flow when the base-emitter junction is suitably biassed. The base input of a bipolar transistor does not have an infinitely high impedance, in other words it has some (non-linear) resistance to the other terminals of the transistor. So if you apply a voltage to the base, a current will flow. If the voltage source applied to the base has the ability to source lots of current (i.e. a low output impedance) then it will easily source this current and maintain the proper voltage at the base of the transistor. If not, the voltage will change. So whilst I would describe the bipolar as a voltage controlled device it does require that the voltage be able to supply a suitable current to maintain this voltage. This has many consequences in circuit design. [FETs have a gate capacitance to charge/discharge and so require ac current at their controlling input too] __________________ "The test of the machine is the satisfaction it gives you. There isn't any other test. If the machine produces tranquility it's right. If it disturbs you it's wrong until either the machine or your mind is changed." Robert M Pirsig. Last edited by Bigun; 11th September 2009 at 05:42 PM.
jan.didden
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Join Date: May 2002
Location: The great city of Turnhout, BE
Quote:
 Originally Posted by 300b Hi Bigun, Thank you for the explanation. Can you please tell me why the 10K is needed for the Q1 base? And I'm not sure I understand, how does the change of temperature makes Q1 compensate for Q2?
There are two 10k resistors. The base current for the first transistor runs through the one 10k. The base current for the second transistor runs through the other 10k. We can assume matched transistors so the two base current are the same. Since the two 10k's are sourced from the same point (collector of 1st transistoor) the voltage drop across the two 10k's is the same. The voltage at the lower end of the 10k's is therefore equal. Therefore, the two base voltages are equal. Neat huh

Edit: The actual voltage across the 10k is very low (base current is very low) so the 1st transistor collector voltage is almost the same as the base voltage. The supply voltage minus this (very low) collector voltage is across the 20k collector resistor. Now the same Ic runs through Q2 as through Q1 (because the base voltage/currents are the same). If we make the Q2 collector resistance half the Q1 collector resistance, it drops also only half the voltage, only half Vsupply. Therefore, Q2 Vcollector is 10V. Neat again, huh

jd
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Last edited by jan.didden; 11th September 2009 at 05:55 PM.

 11th September 2009, 06:51 PM #10 Wavebourn   Designer & Technologist diyAudio Member     Join Date: Sep 2006 Location: Pleasant Hill, CA Looking from a different perspective, the thingy is a current mirror. That means, if transistors are equal, their collector currents will be equal. Degeneration resistors in emitter allow mismatch of transistors, including mismatch due to different temperatures, so if resistors are equal currents are again equal and more stable than without them. The current through both collectors is defined by 20K resistor, so a voltage drop on the 10K resistor in collector of the right transistor will be half of what drops on the 20k resistor since it's resistance is 2 times smaller, i.e. nearly half of a power source voltage.

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