A WIZARD'S ELECTRONICS COMPANION
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A WIZARD'S ELECTRONICS COMPANION |
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In order to understand the basic concepts I use when I design circuits (functional blocks, source & load), I will demonstrate how they are built from the simplest of circuits.
In the Basic Circuits figure In A, an ideal battery is in series with an ideal resistor comprising perhaps the most basic practical circuit. The simple DC voltage of the battery is not loaded down by the current drawn from it. The current is calculated from I = E / R = 6V / 2kohms = .003A = 3mA. The power dissipated by the resistor (that is the work done on the resistor) is P = I * E = 3mA * 6V = 18mW.
In B, the resistor does not connect to ground but is available to whatever you wish to connect to it. This is a typical model for a DC voltage source or a practical (as opposed to an ideal) model for a battery showing it has an internal resistance which wastes a significant portion of the power if too much current is drawn from it. Since there is no current path, no current is flowing and no power is being dissipated in the battery's internal resistance and the full battery voltage is seen on both sides of the resistor. Note no voltage drop occurs across the resistor under these conditions. When reasonable amounts of current are drawn from this battery model, a reasonable voltage is available to power our load. Usually, you've decided to use a specific battery, so you design your circuit within the reasonable current limit that battery can provide. The battery's internal resistance can then be ignored for all practical purposes.
In C, there are two resistors in series across an ideal battery or voltage source. The battery combined with R1 can be viewed as a practical battery. R2, then, can be considered the external load resistance which is using the battery's power.
The total resistance is R[total] = R[1] + R[2] = 300 ohms + 600 ohms = 900 ohms.
The current throughout the circuit is I = E / R = 9V / 900 ohms = 10mA.
The voltage dropped across R1 is E[R1] = I * R[1] = 10mA * 300ohms = 3V.
The voltage dropped across R2 is E[R2] = I * R[2] = 10mA * 600ohms = 6V.
The voltage of the node between the resistors with respect to ground is 6V.
The power dissipated by R1 is P[R1] = I * E[R1] = 10mA * 3V = 30mW.
The power dissipated by R2 is P[R2] = I * E[R2] = 10mA * 6V = 60mW.
The total power dissipated in the resistors is P[total] = P[R1] + P[R2] = 30mW + 60mW = 90mW.
If a carbon-zinc 9V battery is used, the rating is 9mA@50hrs (9 * 50 = 450mA-hrs); 450mA-hrs / 10mA = 45 hours of operation.
If an alkaline 9V battery is used, the rating is 18mA@33hrs (18 * 33 = 594mA-hrs); 594mA-hrs / 10mA = 59 hours of operation.
The maximum capacity of a real battery can only be achieved if the current demand is small. The higher the current, the lower the apparent capacity of the battery, so the model in this example does not account for this effect.
In D, a load resistor is placed across our practical battery source. The current is the same throughout the circuit, but different voltage drops occur across the resistors so our load resistor never sees the full voltage of the battery (unlike our ideal circuit in A). In C, consider the addition of a AC signal source component added to the DC and fed through the resistor. These three taken together comprise a useful model for a generic source. In D, a load is added to circuit B.
In E, two practical batteries are connected together with their negative terminals common to one another. We choose this point as being our ground. The direction of current flowing through one or more resistors depend on the relative voltage. In this case our load is at 9V DC and our source is at 3V DC, so the source is sinking current and the load is sourcing current. Since this is again a series circuit, the current is the same throughout.
R[total] = R[source] + R[load] = 4 + 14 = 18 ohms.
The total DC voltage across the resistors is V[source] - V[load] = 3V - 9V = -6V.
The current is I = E / R = -6V / 18ohms = -0.333A = -333mA.
The voltage across the source resistor is E[source] = I * R[source] = -0.333A * 4ohms = -1.33V
The voltage across the load resistor is E[load] = I * R[load] = -0.333A * 14ohms = -4.67V
At their nominal full-charge battery voltages, the voltage at the connection node between the batteries is 3V + 1.33V or 9V - 4.67V = 4.33V.
The 9V battery is discharging while it is charging the 3V battery (rather inefficiently). Two-thirds of the voltage is lost in the transfer, and given the equal currents, two-thirds of the power.
The power lost as heat in the source resistance is P[source resistance] = I * E[source resistance] = 333mA * 1.33V = 444mW
The power lost as heat in the load resistance is P[load resistance] = I * E[load resistance] = 333mA * 4.67V = 1555mW = 1.555W
The power taken from or put into the ideal part of the batteries are assumed to be conserved as chemical energy storage and not lost as heat:
The charging power at nominal voltage in the source is P[source] = I * E[source] = 0.333A * 3V = 1W
The discharging power at nominal voltage in the load is P[load] = I * E[load] = 0.333A * 9V = 3W
This is, in practice, no way to charge batteries as they get damaged by boiling their electrolytes, or even exploding from rapid heating. The current needs to be reduced through the addition of a third resistance or some other means. The only difference from the foregoing calculations is that the total resistance would be:
R[total] = R[source] + R[limiter] + R[load]
The total current then can be calculated, the voltages across each resistor, the power dissipated by each resistor, and the charging/discharging powers.
Power is being converted into a different form--electricity to sound (speaker), sound to electricity (microphone), heat, light, sound, movement--whatever is needed. One generally needs to match impedences for efficient power transfer if strict power limitations exist. Plugging small appliances into a power strip is an example of having more than enough power to do the job. The voltages from the source match the voltage requirements of the load and each pull a tiny fraction of the total available current.
Decide which direction your energy is flowing--which is the source and which is the load. Characterize your load: how much is R, L, and C? What are the requirements of your load--the voltage and current needed, and whether AC or DC? How about AC frequency? What voltage and current (and frequency, if AC) can your source provide?
If you need less voltage, the simplest fix is a divider. Categorize your transducer as R, L, or C, choose the series componet you want, and calculate the voltage drop needed as shown below. Line power (120V, 60Hz AC) can be altered using a power transformer. In the past, audio circuits also took advantage of audio transformers. RF circuits can and do use RF transformers. Transformers, while the most expensive of the R, L, C trio, can step voltage up as well as down. They also provide DC isolation. Useful, if the DC bias voltage is different from one stage to the next.
A signal containing pulses is one of the bugaboos of any technician. What is the power in the signal? It may contain high-currents around the time of the pulse, but out-of-phase with the voltage pulse. Regular meters are generally terrible, having been optimized for 60Hz AC for meters costing two-figures, perhaps a few kilohertz in meters costing three figures, and ones that might measure it successfully accurately costing four figures. A better way to measure these is with a Tektronix oscilloscope with a current clamp. This will give you a current waveform plotted along with your voltage waveform. The newer 'scopes will even calculate the power at each instant in time for you and plot that. Very useful. If you have time to burn, one trick from the calibration field is to use heating effects to measure power. A small heating element (whose temperature changes will occur quickly enough to be useful) is used as a transfer standard. The unknown pulses are dumped into the heating element, knowing that pretty much all the energy is being honestly dissipated as heat. Then, a nice, steady DC voltage and current is applied to the heating element, instead. When the heating element reaches the same temperature from each source, then you know the same power is being produced by each source, and you can look at your DC source for the RMS voltage and current. You might try this trick next time you've built you're next "free energy" device and you don't have access to a good 'scope. If the frequency being generated is particularly high (or if a lot of the energy is in high-frequency harmonics) be sure to design your heater so it can absorb those frequencies. Few humans perform real investigative science well. If the answer is truly unknown, you sometimes have to question the very premises your experiments are based on.
The load impedence must be the same as the source impedence for maximum transfer of power. Less current and the power goes unused, more and the voltage drops too much--remember that power is the product (voltage) * (current). So, to best use the 1V, we must know what the effective resistance or impedence of our "black box" is. One way is to measure the voltage without a load, then again under a known load current. The internal resistance of our "black box" is then easily computed from Ohm's Law:
I = E / R ==> R = E / I = (change in voltage) / (change in current)
Having calculated 1k ohms in this example, we can best use the power with the addition of an external 1k ohm load.
Any two components in series will divide a voltage being fed to them, and if reactive components are present, those voltage will probably change with frequency. Any two components in parallel are going to provide multiple paths for current to get to the other end, and once again, reactive components will likely have current flows that do not simply follow the input voltage waveform. As the power flows through your circuit, the signal nodes, usually isolated from the power supply rails by components, are being biased by only a couple of components at that stage. This section will detail how to find the voltages and currents in any pair of components, be they resistor, capacitor, or inductor. These may be the most important and most used of all analyses.
The first rule to remember is deciding how to categorize the parts in question: R, L, or C? Of course, hybrids may cause some confusion, but if we have discrete components, or at least ones that act like resistors and draw power cleanly, it's easy enough to answer.
A 10% reduction in voltage will double the life, 20% will allow it to run indefinitely (if I recall correctly.) So you choose 20%. (6V)(1.0-0.2) = 4.8V. Now, say the lamp had been rated for 1W. The resistance of the filament of an incadescent lamp changes radically with resistance, from a few ohms when cold to dozens or hundreds when hot. Assuming the lamp is fully powered on, what is its resistance?
I = P / E = (1W)/(6V) = 0.17A
R = E / I = (6V) / (0.17A) = 36 ohms
We will now assume a simple linear ratio for running the lamp at a different voltage (not true, but close enough.) We will also assume 36 ohms for all voltages (even though that's not true at all, but the filament is mostly pretty hot.) The resistance is going to be higher, and the current is going to be lower by the same propertion (just inverted.)
(36 ohms)(6V / 4.8V) = 45 ohms.
The current could be calculated by the same ratio:
(0.17A)(4.8V / 6V) = 0.14A
Now we have both current and resistance (approximate, of course) of our cooler-running lamp. How much voltage do we have to lose?
12 - 4.8 = 7.2V
The cheapest solution (though granted not the most efficient) is a simple series resistor. We now know it will have about 7.2V across it and 0.14A through it:
R = E / I = (7.2) / (0.14) = 51 ohms
This is a heat-dissipating device, so how much power is it dissipating?
P = I x E = (0.14A)(7.2V) = 1W
Resistors are rated by their power dissipation in free air just shy of burning themselves up, so give yourself plenty of wattage-headroom -- a factor of two or more is a good idea unless you like burning your fingers, so I would use at least a 2W resistor.
Let's look at a simple circuit with a 9V battery, a switch, and a 500 Ohm resistor in a simple series circuit. The dotted line shows a simple equivalent circuit of a chemical battery--current is not unlimited, it is as if a power source had a resistor inside the case in series with it, R(B). When the switch is closed, current will flow through the resistor. If the battery measures 9.1V with the switch open, but only 8.2V with the switch closed and current flowing, what's with that? First, the current in the 500 Ohm resistor with the switch closed is I = E / R = 8.2V / 0.5k ~ 16 mA. "In a series circuit the current is everywhere the same," quoth my high school electronics instructor like a litany until never forgot it. Since it is in series with our external load, the current in our internal resistance is also 16 mA. The voltage across this internal resistance would be the difference between the open-circuit voltage (no voltage drops occur in a circuit where there is no current flow, so we are reading the actual battery voltage as though there were no internal resistance) and the closed-circuit voltage, because the sum of these two voltage would have to equal the battery voltage, because that's where all the voltage comes from. So we have:
V[int] = V[open] - V[closed] = 9.1V - 8.2V = 0.9V
Another way to look at any circuit loop is with Kirchoff's Voltage Law (KVL), which says that if you add up all the voltages across each component around ANY series circuit loop (keeping the signs consistent -- ie: sources "+" and loads "-"), they will always sum to zero. In this case the series loop is the entire circuit, but no matter how many paths there are, you can meander all you want, even looping back through the same components going the other way (and reversing the sign) until you finish an entire twisty loop. In this case, KVL would say:
(+V[batt]) + (-V[int]) + (-V[loadR]) = 0
(+9.1V) + (-V[int]) + (-8.2V) = 0 V
V[int] = 0.9V
Now we have the voltage across the internal resistance as well as the current through it. The internal resistance can then be calculated from Ohm's Law: R = E / I = (0.9V) / (0.016A) ~ 56 Ohms
In analytical terms, the "upper" resistor (the internal resistor--the one closest to the positive side of the battery) can be viewed as a pull-up resistor, and the "lower" one as a pull-down. The node between the two is the point we are trying to get to, the voltage we want to create. The core of the power source can be likened to our input signal, that which must be massaged to achieve our goal, the energy source which we use for our purposes. From one end of our power source to the other, there will be voltage drops if there is current flow.
This fact is very useful troubleshooting a complex board with small traces, trying to find a short circuit. The parts on the board probably don't pull enough current in normal operation to allow convenient checking of voltage drops along the traces. One can take a separate power supply (if the circuit does not supply enough current) and, with the aid of a sensitive DC voltmeter, place one probe firmly at the "bottom" end of the stream of current (often the negative power supply connection) and probe from one end of the stream to the other for changing voltage drops. The copper traces on the board do have some small resistance, and act like a continuous string of very small resistors. If the short circuit can safely pull (say) three amperes, and your voltmeter is sensitive to tens of microvolts, chances are very good that you can find the short. If you are following a trace where the voltage is not changing at all, then there can be no current flowing. I once isolated a short to within a millimeter in this manner. It proved to be a hairline bit of metal beneath a capacitor that I had already removed and tested (then replaced with a new tested good one, just to be sure). In troubleshooting, sometimes the cause is hard to credit, especially if you know you already checked something thoroughly.
Two 2k resistors in parallel make 1k total (or three 3k resistors.) The water in a bunch of streams adds together to make the river:
I[t] = I1 + I2 + I3 + ... + I[n]. For those of you who don't mind a little algebra, another powerful little trick is to substitute the other side of a common equation (Ohm's Law, in this case) into the above equation for each term. Also, we know that the voltage across each branch is the same, so:
I[t] = (E / R1) + (E / R2) + (E / R3) + ... + (E / R[n]). Then, factoring out the voltage:
I[t] = E * ( (1 / R1) + (1 / R2) + (1 / R3) + ... + (1 / R[n]) ). One might see (murkily) that the whole equation is also Ohm's Law where R is:
R[t] = 1 / ( (1 / R1) + (1 / R2) + (1 / R3) + ... + (1 / R[n]) ). The equation for resistors in parallel.
Another example: We have three resistors in parallel, 1k, 2k, and 3k. From the equation, then:
R[t] = 1 / ( (1 / 1k) + (1 / 2k) + (1 / 3k) ) ~ 0.545k
If you just want to attenuate the voltage then a two-resistor voltage divider is all you need, but if both the source and load need to "see" a certain impedence and the source is capable of having plenty of drive voltage and current for the load then a T-network may serve. Note that R3 is common to both, so the impedence of R3 in parallel with the opposing side is the minimum either side can see. As R2 approaches zero, it drops into the simpler two-resistor voltage divider.
If the impedence one side must see is small compared to the other, the pi-network is used.
Now let's take the last few examples and take it to the next level--a network of three resistors. Calculations can quickly grow out-of-proportion to the size of the problem with networks.
Approach #1: I will leave the rigorous solution to this one to the brave few Algebra students with only the advice: Use the calculations for resistors in series and resistors and in parallel, along with Kirchoff's Laws, with the voltages fixed at their limits (that is, 5V and 7V for the high end, 0V and 6V at the low end, for the inputs and outputs needed, in order to produce enough equations to solve for the three resistor values.
Approach #2: The voltage divider composed of R1 and R2 is a biasing network which sets the average DC voltage of the system. If the input is averaging about 2.5VDC, then the problem becomes: (1) Signal attenuation: View R3 as the "upper" resistor and the combination of every path to the power supply "downstream" as a parallel network of resistances which can be reduced to a single "lower" resistor. This forms a voltage divider for the signal--its attenuation. In this case we want the signal to be five times smaller (as can be seen from the ranges of the input and output voltages), so R3 will be five times larger than the parallel combination of R1 and R2. (2) DC biasing: Choose R1 and R2 based on the average DC voltage shift needed. Make a momentary approximation ignoring R3 for the moment. It is just over half the supply, so R1=R2 plus a small R. Now give that roughly 20% (5-to-1) correction. Suddenly, we realize we haven't enough information. There is no current demanded by our output, so we can nail one of these values down to whatever we please. So I choose to make the R1/R2 divider total 1200 ohms for a convenient similarity to the 12V supply, and the fact that 1k to 10k ohms is a good choice for power-saving, yet supplies enough current for medium-speed electronics given the current state of the technology.
That means, uncorrected, the upper resistor is 550 and the lower 650 ohms, and with correction...one-fifth of 650 is roughly 130, plus 650 equals 780 for the lower resistor, and 1200-780=420 ohms for the upper resistor. We are correcting upwards because the source is lower than the output--we need to pull up. Now, R3 is five times bigger, so the in our calculations for R3 will have five times less effect. R3 is five times the parallel combination of R1&R2, very roughly around 300 ohms times five equals 1500 ohms. We have just made a first-order approximation (barely, but more than a "zeroth-order") of the solution. (The term "first order" comes from the common power series which is used to approximate curves and stuff in engineering: A+BX+CX^2+DX^3+EX^4... Each successive term increases the accuracy of the model, but adds complexity. A+BX is the general form of a linear equation whose highest exponent of X is one(1), thus the term "first-order".)
Approach #3: Check out values in a SPICE program. Hm. About 7V +/1 4V signal. Having gotten close (within 50% or so), you could tweak it closer with SPICE. The first approach is rigorous beyond the ability of the electronics to achieve, the third requires you have some idea where to start at least, and the second is quick and requires no special tools except your brain. You will have noted #2 is fraught with errors. As you choose to eliminate them one-by-one, the accuracy of this method goes up, plus you have a greater intuitive understanding of the dynamics in an electronic circuit.
Here's a general case for a biased node between a source and a load. We have a source impedence (R3), a load impedence (R4), and a voltage-divider biasing network consisting of pull-up (R1, source of positive charges) and pull-down (R2, sink for positive charges) resistors.
Rc Time-Constant: one time-constant is the time for the RC node to shift two-thirds of the way to the voltage you would see if you waited long enough -- in practice by about five time-constants. A second time-constant period will allow the voltage to cover two-thirds of the remaining one-third, and so on. Unlike the description of phasor relationships at a single AC frequency, the RC time-constant is an example of a transient condition. Essentially, all frequencies are represented in the charging of a capacitor, from the arbitrarily high down to the lowest being represented by however long you allow the capacitor to charge. The initial current into the capacitor would be the total voltage across the resistor divided by the resistance. The current would gradually decrease by stealing more of the available voltage from the resistor to put across the capacitor. The final voltage across the capacitor approaches the total voltage input (theoretically never quite reaching it).
If one were to charge the capacitor with a constant current, the RC time-constant curve would become a straight line at some angle because you would be filling the capacitor at a constant rate. To see the phase-shift itself put to use, check out the phase-shift oscillator.
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