mauifan wrote:So the good news is that I am getting amplification, but why isn't the output inverted? I double-checked my connections and Channel 2 on my scope (yellow trace) does indeed connect to OUT-.
firstname.lastname@example.org wrote:Long answer: When you design a transistor amplifier, you need to choose certain parameters so you can calculate the rest. One of the middle values that helps to calculate all the others is called 'g.m', where 'g' is the traditional symbol for conductance (the inverse of resistance).
g.m describes the transistor's 'transconductance'.. how much the collector current changes in response to a change in base voltage:
g.m = dI.c / dV.b
where 'd' means 'the change in'. (the actual equations use greek letters, subscripts, and other things I can't type, so 'g.m' means 'g-subscript-m')
There's also an equation that defines g.m in terms of 'thermal voltage'. When you get down to the electron level, the difference between 'heat' and 'voltage' gets kind of fuzzy. They're both essentially 'energy per electron'. Thermal voltage (V.t) allows us to include the energy from thermal effects in electrical calculations, and it's *really* important when you start working with semiconductors. The good news is that we can choose a temperature, calculate Vt, and treat it as a constant from then on. At room temperature (25C or 300 Kelvin), Vt =~ .0258v or about 26mV.
The thermal equation for g.m is:
g.m = I.c / V.t
where I.c is the quiescent current. For I.c = 1mA, g.m =~ .0385 siemens.
If g.m is the transistor's conductance, 1/g.m will be its effective resistance.. 1/.0385s =~ 26 ohms (assuming I.c = 1mA). That value is called 'r.e'.
For a common emitter amplifier, the gain (A) is inherently limited by the ratio of your collector resistor (r.c) to r.e:
A.max = r.c / r.e
Dividing by r.e is the same as multiplying by g.m though, so:
A.max = r.c * g.m
g.m is the ratio of collector current to thermal voltage, so:
A.max = r.c * I.c / V.t
but 'r.c * I.c' is just the voltage across r.c when the amp sits at its operating point. That gives us:
A.max = V.rc / V.t
which tells us how much headroom we need in order to get a certain amount of gain. The interesting bit is that A.max is completely independent of the supply voltage. It's just a ratio of resistance to current that has to exist before you can get a certain amount of gain.
So.. if you know V.rc you can calculate the maximum possible gain easily (divide by .026v or multiply by 38.5s). If you know how much gain you want, you can get V.rc by running the equation the other way (V.rc = A*.026v). Either way, you end up with a voltage and need to choose a resistor. Choosing I.c = 1mA makes that simple.
Engineers like those equations so much that lots of common transistors are specifically designed to perform best near I.c = 1mA. The 2N3904 (my usual choice for such jobs) has a typical current gain of 300 when I.c = 1mA.
mauifan wrote:Are you talking about Ebbers-Moll here, mstone?
mauifan wrote:I started reviewing the section about Ebbers-Moll in AoE and... well... let's just say that I think I need to read that section a few more times.
mauifan wrote:That said, I tried an experiment varying the frequency on my amp circuit and saw that gain diminished with higher frequency.
mauifan wrote:Thinking ahead, I imagine that this will be fine when I add a 32kHz crystal to my circuit, but it is going to start becoming problematic when I get over 1MHz or so. The gain was definitely <1 at 4MHz, which means that it isn't going to oscillate unless I up the gain or something.
email@example.com wrote:That's normal. There's a bit of lag between "something going into the transistor" and "something coming out".. usually a few nanoseconds. The difference between the input and output during that lag is stored energy that the output hasn't caught up to yet.
Oscillating signals change direction though, so at some point the input will start going down while the output is still trying to go up. They'll meet somewhere in the middle of the stored energy, then the input will start pulling the output down rather than up. That means some of the stored energy will never make it to the output, and instead will be cancelled by the input.
For slow moving signals, the difference between input and output is practically zero, so there's practically no stored energy to lose. As the signal moves faster, the input can get farther ahead of the output during the lag. That means there's more stored energy to lose. If the input makes a full cycle during the lag, practically all the input turns into stored energy, gets cancelled by the input, and is lost.
mauifan wrote:So a real-world BJT transistor is an "ideal" transistor with tiny capacitors (perhaps 1-2pF?) across the junctions?
mauifan wrote:If so, it makes perfect sense to me why gain falls off at higher frequencies: Higher frequencies would bypass the ideal transistor through the caps.
mauifan wrote:Thanks for your response, mstone. I will have to read through it a couple of times.
In the meantime -- and perhaps while you were typing your post -- I did a quick experiment. Obviously I did something wrong, because it didn't work as I expected.
I verified that my transistor amp circuit was still working. Indeed I still get amplification at OUT- that is 180 degrees out of phase with the input. I then connected OUT- on the amp to INx in the circuit below. Likewise, I connected the amp's IN to OUTx. I fired up my scope, turned on the power to my circuit... and NOTHING. The junction at Vb was at .7V. OUT- varied with Rc (OUT- was @3V with Rc=4.8k).
For this test, I used a 32kHz crystal with C1 and C2 as shown.
As I recall from the op amp thread, a 32kHz crystal has an ESR of about 32k. The reactance of a 22pF cap at 32kHz is about 225k. Therefore, the capacitance reactance should "dominate" the ESR and cause a near 90 degree phase shift. Given that there are two caps in this "pi network," shouldn't there be a phase shift close to 180 degrees -- and thus invert that output so that the input is [almost] in phase?
mauifan wrote:Based on my understanding, the phase shift is the "inverse tangent" of Xc/R
The crystal in combination with C1 and C2 forms a pi network band-pass filter, which provides a 180 degree phase shift and a voltage gain from the output to input at approximately the resonant frequency of the crystal. To understand the operation, note that at the frequency of oscillation, the crystal appears inductive. Thus, it can be considered a large, high Q inductor. The combination of the 180 degree phase shift (i.e. inverting gain) from the pi network, and the negative gain from the inverter, results in a positive loop gain (positive feedback), making the bias point set by R1 unstable and leading to oscillation.
mauifan wrote:I am somewhat of a visual/intuitive learner, and... well... if the crystal is acting as an inductor (which in itself makes sense to me), won't it basically act to cancel out the impedance of the capacitors rather than add another 180 degrees of phase shift?
mauifan wrote:What I am less clear about is how you got the "phaseshift = cos(theta)" part. It has been a while since I last had a need to worry about trigonometery, though I do remember tan(theta) = sin(theta) / cos(theta)... or something like that.
mauifan wrote:In my mind, I am aware of the following "rules" --
- In an inductor, voltage leads current by up to 90 degrees
- In a capacitor, voltage lags current by up to 90 degrees
- In a resistor, voltage and current are in phase
mauifan wrote:Why didn't I see a phase shift around 32kHz as Wikipedia promised?
mauifan wrote:A second experiment I tried was... well... I realized that C1 and C2 were effectively in series from the perspective of X1. With C1=C2=22pF, the total capacitance across X1 was about 11pF (perhaps a little more due to the breadboard wiring). I replaced the crystal with a 180uH inductor and calculated that the resonant frequency of this tank circuit was about 3.5MHz. I fed a sine wave signal into one side of the tank and monitored the output signal. I saw a phase shift approaching 180 degrees, but it occurred at a much higher frequency than 3.5MHz.
Users browsing this forum: Majestic-12 [Bot], MSN [Bot], Zener and 10 guests