# Need Help Understanding a Circuit (Atari 2600 XTAL Oscillator) Above is the image, it's the clock from an Atari 2600 Jr.

I'm new to electronics and all the books I've gone through so far don't cover XTALs. I have no idea why, seems pretty important to electronics. My searching over google/youtube hasn't gotten me any usefully direct answers either.

This particular circuit has a 3.579575Mhz XTAL, 2 PNPs, 7 Resistors, and 2 Capacitors. I'd like to know what's happening at each node and what's happening at CLOCK OUT in terms of the wave form (what's the end oscillating frequency at CLOCK OUT?).

Is there a circuit simulator (open source) that can allow you to input XTALs in a circuit? So far I haven't come across one.

• For the record, "XTAL" is crystal. Looking up "Crystal Oscillator" may have more results than looking up "XTAL". – R Drast Oct 27 '16 at 6:30
• Crystals can be modelled by putting an inductor and a capacitator in a row. At the resonant frequency, the impedance of such an arrangement drops to a minimum. The difference to a real-world cyrstal is the latter has a much sharper resonant peak than any real-world inductor/capacitator arragement. But in simulation, you can make inductors/capacitators ideal elements. – Janka Oct 27 '16 at 8:50
• @Janka: In this case what would the right values for an inductor/capacitor in sequence to substitute for a crystal in a simulation? – John Ernest Oct 27 '16 at 23:33
• @Janka: Here's a link to a Falstad simulation I'm working on: tinyurl.com/hbdkght – John Ernest Oct 27 '16 at 23:35
• In simulation, you can choose any combination. f0=1/(2pi*sqrt(LC)) – Janka Oct 28 '16 at 0:24

It may be easier to understand if you just start with a long tailed pair -- a differential amplifier. Let me lay it out that way to see if that helps: simulate this circuit – Schematic created using CircuitLab

I've shown a current and a few voltage values on the schematic. These are roughly true once it starts running. But $C_1$ starts out discharged, so those values take a while to reach. At first, $C_1$ pulls hard on $Q_1$'s base and saturates it, adding current from $R_e$ to current from $R_{b1}$, in order to charge up $C_1$. At some point in this process, and I would guess that it would take about $\Delta t= \frac{\Delta V\cdot I}{C_1}=\frac{3.4\:\textrm{V}\cdot 15\:\textrm{mA}}{220\:\textrm{nF}}\approx 50\:\mu\textrm{s}$ before things started rolling, $Q_1$ will come out of saturation and things will start to operate more normally. (There's also some diffusion charge storage that was built up into $Q_1$, but that's probably only another few added $\mu$s. Nothing big.)

As $Q_1$ comes out of this initial saturation, it's $\beta$ starts to take hold, its base ceases to contribute to charging $C_1$, and the voltage drop across $R_1$ declines by at least a few tenths of a volt as $Q_1$'s $V_{CE}$ finally exceeds $1\:\textrm{V}$. This drop is passed through the crystal and appears as a slight pull-down at the base of $Q_2$, which diverts current through its collector and away from the collector of $Q_1$, which causes the drop across $R_1$ to decline still further and helps to emphasize this diversion for a moment. However, $R_3$ rapidly charges the crystal back up and drives $Q_2$'s base back upwards causing it to divert less and allowing $Q_1$ to begin diverting more current through its collector, causing the drop across $R_1$ to increase again and drive $Q_2$'s base further up, as well.

($R_{b1}$ and $R_{b2}$ form a Thevenin resistance of about $680\:\Omega$. I seem to recall these crystals were about $100\:\textrm{pF}$ or so, so working around the loop I see $680\:\Omega+220\:\Omega+220\:\Omega$ in series with about $100\:\textrm{pF}$ giving about $110\:\textrm{ns}$ per half-cycle, which works out to about $4.5\:\textrm{MHz}$. So I must be a little wrong about the capacitance of these crystals, I think. The idea is that the natural frequency should be close to the crystal and that the crystal would then "pull it in" tightly.)

Once equilibrium is reached, $Q_1$'s base will be very tightly held to its nominal value by $C_1$. I think $C_1$ is mostly there to hold the base voltage pretty solid and, perhaps indirectly I suppose, to set the start-up timing. It probably could be larger still, with the price being that it would take longer to start up. Most of the variation is taking place at $Q_2$'s base. Remember that the collector current will vary by a factor of 10 for a $60\:\textrm{mV}$ change in $V_{BE}$. $Q_2$ will be entirely turned off when its base reaches the $4.05\:\textrm{V}$ of its emitter. So I expect that to be the highest voltage seen at the base of $Q_2$. The lowest voltage there will be approximately that much, less the voltage drop of $R_1$ caused by half the current through $R_e$, so I think about $1.2\:\textrm{V}$ less, or $2.85\:\textrm{V}$ at a guess. In short, I expect $V_{PP}=1.2\:\textrm{V}$ at $Q_2$'s base, centered relatively close to the $3.4\:\textrm{V}$ experienced at $Q_1$'s base (with perhaps a very tiny relative offset to it.)

I don't think it would work with either $R_1$ or $R_2$ changed. Much larger would force the BJT collectors back up against their bases and it wouldn't oscillate. That's one reason they are where they are at. And so you'd have to change $R_e$, if you changed them. Also, this would change the natural frequency to be too far away from the crystal and it would oscillate at some other frequency or just fail, even if you changed all three values. So all the pieces seem about right as they are, to me.

This is Series Resonant Oscillator with Series resonant Xtal with very low Rs. Non-inverting current gain via Q2-E and non-inverting voltage gain Q1-E to Q1-C.

Both Bases are biased around 4V with Q2 saturated until it starts up. Q2-C is inverted outside the loop and is probably pretty ugly looking square wave with lots of droop from 10pF with 240 on the collector and 90 on the emitter.

It would look much better with 1k on Q2-C with a 4.5V square wave with more V gain on Q2 or maybe change Re from 91 to 50 Ohms to get more gain.

These drove the Xtals with pretty high power and would be the huge HC-49 cans with big wafers, so modern microslice xtals might fail in this circuit.