Electronics

How do voltage and current share out?

Almost every electronic circuit has, somewhere on the board, a few resistors that set voltages and currents for the rest of it. The tools that solve these pieces come back in every later step, so it is worth going over them carefully.

In series, the same current flows through every resistor, and the supply voltage is shared in proportion to the resistances. This is the potential divider. In parallel, all the resistors have the same voltage and the current is shared in inverse proportion to the resistances, in the current divider.

When the circuit does not reduce to series and parallel, we turn to Kirchhoff's laws. The current law says that the current arriving at a node equals the current leaving it, because charge does not pile up in the wire, and the voltage law says that the voltages around any closed path add up to zero, since we return to the same potential.

A divider seems to deliver a fixed voltage, yet that voltage drops when we connect a load in parallel with the lower resistor. As a rule of thumb, the drop tends to stay small when the load is about ten times larger than that resistor, which we can check on the bench.

The Wheatstone bridge compares two dividers connected to the same supply. When their ratios are equal, the two midpoints sit at the same potential and the meter between them reads zero, so we can find an unknown resistance from three known ones without relying on the accuracy of the meter.

What changes when the voltage goes back and forth?

A fresh cell keeps roughly the same voltage for hours, and we say it is direct, or DC. A mains socket reverses its polarity 60 times a second and supplies an alternating signal, or AC, that follows a sine wave; the 127 V and 220 V at 60 Hz used on this bench are Brazilian mains values, while UK mains is 230 V at 50 Hz. Many circuits mix the two, with an alternating part added to a steady level.

To describe a periodic signal, we use the period \(T\), the duration of one cycle, and the frequency \(f = 1/T\), the number of cycles per second. The amplitude can be given by the peak value \(V_p\), by the peak-to-peak value \(V_{pp}\), which is what the oscilloscope shows straight away, or by the RMS value.

The RMS value of an alternating voltage is the steady voltage that would heat the same resistor with the same mean power. For a sine wave it equals \(V_p/\sqrt{2}\), and the 127 V and 220 V of the mains are RMS values, with peaks of about 180 V and 311 V. In a square wave the RMS is the peak itself, and in a triangle wave it is \(V_p/\sqrt{3}\), so signals with the same peak can heat in quite different ways.

On the oscilloscope, DC coupling shows the full voltage, and AC coupling removes the steady part so that we can look closely at what varies. The multimeter on V AC measures the RMS of the alternating part, and the models sold as true RMS get it right for any waveform, while the simpler ones are only right for a sine wave.

How long does a capacitor take to charge?

A camera flash takes a few seconds to be ready, and the LED on a charger stays lit for a moment after we pull it out of the socket. In both cases a capacitor gains or loses charge through a resistance. The pace of this process is set by a single number, the time constant \(\tau = RC\).

At the instant we switch on the supply, the uncharged capacitor behaves like a wire, and the whole voltage appears across the resistor. As charge builds up, the capacitor voltage rises and the current falls, because less and less voltage is left to push it. The rise is exponential. After one \(\tau\) the capacitor reaches 63% of its final voltage, and after \(5\tau\) more than 99%, which in practice we take as fully charged.

Discharging is the mirror image of charging. In one \(\tau\), the voltage falls to 37% of its initial value.

If the applied voltage changes all the time, we can picture the capacitor trying to follow it with this same lag. With a slow sine wave, whose period is much longer than \(\tau\), it follows with almost no loss. With a fast one it has no time to charge, and we see the voltage across it stay small. Taking the output across the capacitor gives a low-pass filter; swapping the resistor and the capacitor, with the output across the resistor, gives a high-pass filter.

The boundary between the two behaviours is the cut-off frequency \(f_c\), at which the capacitor's reactance, \(X_C = 1/(2\pi f C)\), equals the resistance. There the output falls to \(1/\sqrt{2}\) of the input, about 70.7%, with a phase shift of 45°. It is a lag in the low-pass filter and a lead in the high-pass one. On the oscilloscope, we measure the phase shift from the interval \(\Delta t\) between the zero crossings of the two traces, as a fraction of the period.

A square wave brings charging and filtering together, because each edge is a step. In the low-pass filter, the output rises and falls along charging curves, which only reach the top if half a period lasts a few \(\tau\). In the high-pass filter, each edge becomes a spike that decays with the same \(\tau\), and what is left between edges depends on how many time constants fit in half a period, something we can read straight off the screen. A square wave on the oscilloscope is therefore perhaps the quickest way to estimate the time constant of an unknown circuit.

How does a circuit pick out one frequency?

An old radio tunes in a station when we turn a knob that changes the capacitance of a capacitor connected to a coil. Of the thousands of signals that reach the aerial at the same time, the circuit lets through strongly only those close to one frequency, the resonance frequency. This preference comes from two oppositions to the current that change in opposite directions with frequency and that, at a single frequency, cancel each other.

With alternating current, the inductor and the capacitor oppose the current in opposite ways. The inductive reactance \(X_L\) grows with frequency, because the coil reacts to changes in the current. The capacitive reactance \(X_C\) falls, because the capacitor has barely begun to charge when the voltage already reverses. In the inductor the voltage leads the current by 90°, and in the capacitor it lags by 90°. For this reason, in a series circuit the two reactances subtract in the impedance \(Z\).

At the frequency where \(X_L = X_C\) they cancel, and in the series RLC the impedance reduces to \(R\) itself, with the current at its maximum and in phase with the generator. Equating the two reactances, we arrive at \(f_0 = 1/(2\pi\sqrt{LC})\). Something that may come as a surprise is that, at this frequency, the voltage across the capacitor and the voltage across the coil can be far larger than the generator voltage. They cancel each other at every instant.

The sharpness of the resonance is measured by the quality factor \(Q\), which in the series RLC equals \(X_L/R\) at \(f_0\). With a small \(R\), the curve of current against frequency is a narrow peak, and the bandwidth between the points where it falls to 70.7% of the maximum is \(f_0/Q\). In the parallel RLC, the tank circuit, the situation is reversed. At \(f_0\) the coil and the capacitor pass current back and forth between them, the pair presents a very high impedance and, when we measure the current coming from the generator, we see it fall almost to zero, even though a current that can be large circulates inside the tank.

The same physics appears when we apply a step, such as the edge of a square wave. With little damping, the capacitor voltage overshoots its final value and oscillates around it, at a frequency close to \(f_0\), until it settles. This is the underdamped response. With a large \(R\) it tends to rise slowly and without oscillating, in the overdamped response. Between the two lies critical damping, which reaches the final value in the shortest time without overshooting.

Why does a diode let current through in only one direction?

A diode works like a one-way valve for current. It is made of two regions of silicon doped in different ways. In the P region the lattice is short of electrons, and the carriers are holes, which behave like positive charges; in the N region there are spare free electrons. At the boundary, the PN junction, electrons and holes recombine and leave a strip with no carriers, the depletion layer, whose electric field blocks the passage of further charges.

When we connect the P side, the anode, to the positive terminal of the supply, in forward bias, we push the carriers towards the junction and narrow the barrier. From about 0.6 V in silicon, the current grows very rapidly. In reverse bias the barrier widens and only a tiny leakage current flows, until breakdown, which we leave for the Zener diode step.

The I×V curve sums up this behaviour. In practice we usually treat the diode as a switch that, when it conducts, has an almost fixed drop of about 0.7 V, an approximation that seems crude but works well in almost every circuit.

An LED is a diode in which the recombination of electrons and holes emits light, with a larger drop that depends on the colour, from about 1.8 V for red to 3 V for blue. Since the current rises exponentially with the voltage, we never connect an LED directly to a supply. It would burn out. A series resistor takes up the difference, and we choose \(R\) for the current we want, typically between 5 and 20 mA.

The one-way action of the diode turns an alternating voltage into a pulsating voltage in a single direction. In the half-wave rectifier, a diode in series with the load lets through only the positive half-cycles. The output has peaks 0.7 V below those of the generator and an average slightly below \((V_p - 0.7)/\pi\), since the diode only conducts once the input passes the knee. In the bridge of four diodes, both half-cycles reach the load in the same direction, each along a path through two diodes that we can follow on the diagram with a finger, and the output has twice the frequency and close to twice the half-wave average, with peaks 1.4 V below those of the generator.

So that we can see the input and the output of the bridge at the same time, the bench uses a 1:1 isolating transformer. The oscilloscope measures relative to earth, and connecting the generator's earth to one of the bridge's input terminals would short out a diode. It is perhaps the most common mistake on a real bench.

How do we get a steady voltage from a supply that varies?

A battery running down, a cheap charger or a supply with ripple delivers a voltage that rises and falls, whereas many of the circuits we build need a steady reference of 5 V or 3.3 V. The Zener diode is the simplest component that does this job.

In forward bias the Zener diode behaves like an ordinary diode, with about 0.7 V. In reverse bias it blocks up to the breakdown voltage \(V_Z\), and beyond it the reverse current grows very rapidly while the voltage hardly moves. It is made to work in breakdown without damage, as long as the power stays within its limit.

In the Zener regulator we connect the Zener diode with its cathode to the positive side, in parallel with the load, and a resistor \(R_S\) in series with the input. The voltage across \(R_S\) is the difference between the input and \(V_Z\), so the current \(I_S\) it lets through is set by the input; the load takes \(V_Z/R_L\), and the Zener diode absorbs the rest. When the input rises, almost all of the increase falls across \(R_S\), and it is the Zener diode that swallows the extra current.

Regulation has limits on both sides. If the load demands more current than \(R_S\) can deliver with the input at its minimum, the current in the Zener diode falls below about 1 mA, the diode leaves breakdown and the output starts to follow the divider formed by \(R_S\) and the load.

At the other extreme, with no load and the input at its maximum, all the current through \(R_S\) passes through the Zener diode, which dissipates \(V_Z\,I_Z\) and may exceed the power it can handle. When we design the regulator, we choose \(R_S\) between these two limits.

On the bench, the input carries a 120 Hz ripple, similar to that of a supply with rectification and smoothing, which we will build in step 9. With the two oscilloscope channels we can see how much of this ripple reaches the output, and perhaps surprisingly it drops to less than a hundredth.

How does a small current control a large one?

A microcontroller delivers a few milliamps per pin, while a relay, a motor or a row of LEDs needs tens or hundreds. The bipolar junction transistor (BJT) bridges the gap, because a small current into the base controls a much larger current between the collector and the emitter.

It has three layers of silicon, in the order NPN or PNP. In the NPN, once the base-emitter junction conducts, with \(V_{BE}\) close to 0.7 V, almost all the electrons the emitter injects into the very thin base cross over to the collector, and the collector current becomes proportional to the base current, \(I_C = \beta\,I_B\). The gain \(\beta\) typically lies between 50 and 300, and the PNP works the same way with every sign reversed, its emitter connected to the positive side.

We distinguish three regions of operation. In cut-off, \(V_{BE}\) stays below about 0.6 V, hardly any current flows and the transistor is an open switch; in the active region, \(I_C = \beta\,I_B\) holds with \(V_{CE}\) well above zero.

In saturation, the base receives more current than the load lets the collector use, and \(V_{CE}\) drops to 0.1 or 0.2 V, as in a closed switch. Four measurements tell the whole story, \(I_B\), \(I_C\), \(V_{BE}\) and \(V_{CE}\), and we can read all of them on the bench.

As a switch, we want only cut-off and saturation, because in both the power \(V_{CE}\,I_C\) is small. The rule of thumb is to give the base a few times the minimum current \(I_C/\beta\), which tends to guarantee saturation even in a transistor with a lower \(\beta\).

The coil of a relay stores energy in its magnetic field, and the flyback diode in parallel with it gives the current a path when the transistor turns off. Without it, the collector voltage would leap to hundreds of volts at switch-off and could destroy the transistor.

As an amplifier, we keep the transistor in the active region. In the common-emitter stage with potential-divider bias, \(R_1\) and \(R_2\) set the base voltage and \(R_E\) sets the quiescent current, almost independently of \(\beta\); a small signal added to the base through the capacitor changes the emitter current by \(v_{\text{in}}/R_E\), and this change, flowing through \(R_C\), produces a larger, inverted output, with a gain close to \(-R_C/R_E\).

How do we control a current without spending current on the control?

The bipolar transistor needs base current for as long as it conducts, and for a 2 A motor with \(\beta = 100\) that would mean 20 mA taken from the control circuit. With the MOSFET, we control the current in another way, through the electric field that the gate voltage creates.

In the enhancement-mode n-channel MOSFET, the gate is a conducting plate separated from the silicon by an extremely thin layer of oxide, which is an insulator. With a positive voltage between the gate and the source, the gate attracts electrons to the surface beneath it, and above the threshold voltage \(V_{th}\) they form a channel joining the drain to the source. Since the gate is insulated, we can treat it as a capacitor, and in steady state no current flows through it.

Below \(V_{th}\), we say the MOSFET is in cut-off. Above it, with a large \(V_{DS}\), it is in saturation, and the current depends only on \(V_{GS}\), through the square law \(I_D = k(V_{GS} - V_{th})^2\); with a small \(V_{DS}\), in the triode region, it behaves as a resistor controlled by the gate.

The names tend to confuse anyone coming from the bipolar transistor, because saturation in a MOSFET corresponds to the active region of a BJT. The MOSFET's closed switch is the triode region, with \(V_{GS}\) well above \(V_{th}\) and an on-resistance \(R_{DS(\text{on})}\) that is typically a few tenths or hundredths of an ohm.

To control the speed of a motor or the brightness of an LED, we switch on and off thousands of times per second with pulse-width modulation, or PWM. On average, the load receives the supply voltage times the fraction \(D\) of the time the switch is closed, the duty cycle, and the transistor dissipates little because it spends almost all its time either cut off or fully on.

The gate oxide is only a few nanometres thick and breaks down at a few tens of volts. The body's static electricity, at thousands of volts, can puncture it with a single touch, which is why we handle MOSFETs wearing an earthed wrist strap and store them in antistatic packaging. In the circuit, a resistor between gate and source stops the gate from floating, with stored charge and an undefined state.

How does the mains socket become a steady 5 V?

Most electronic circuits run on a few volts DC, whereas the socket supplies the 127 V AC mains. A linear power supply makes this conversion in four stages, which we can follow one by one with the oscilloscope, from the transformer to the regulator.

The transformer has two windings on the same iron core, and the voltage across each is proportional to its number of turns. In a 127 V to 9 V transformer, we find a turns ratio of about 14 to 1, and the peak on the secondary is \(9\sqrt{2} \approx 12.7\ \text{V}\). The two windings have no electrical connection between them, so the secondary is isolated from the mains.

In the bridge rectifier, two diodes conduct in each half-cycle and steer the current to the same side of the load in both halves of the cycle. Output P loses two drops of about 0.7 V and repeats the half-waves at 120 Hz, twice the mains frequency. Because the oscilloscope measures relative to earth, which here is the negative terminal of the bridge, S1 and S2 each show only half of the sine wave, and to see the whole secondary we measure between the two with the multimeter on V AC.

The smoothing capacitor charges at the top of each half-wave and carries the load on its own until the next peak, discharging along a ramp. This ripple grows with the current and shrinks with the capacitance. The estimate \(\Delta V \approx I/(f\,C)\) assumes that the discharge lasts until the next peak, without subtracting the recharge time, and so it tends to give a little more than we measure. This ripple is what we see on CH1 with AC coupling.

The 7805 voltage regulator delivers 5 V as long as its input stays about 2 V above that, the so-called dropout voltage. We can think of it as a resistance that adjusts itself, and the voltage difference times the current turns into heat inside it. When the trough of the ripple dips below about 7 V, regulation is lost and the ripple passes through to the output.

Today's mobile phone chargers generally use switch-mode power supplies, which are lighter and more efficient, but they too rectify and smooth the mains voltage right at the input.

What sets the gain of an operational amplifier?

The operational amplifier, or op-amp, is an integrated circuit with two inputs, the non-inverting input (+) and the inverting input (−), and one output. It amplifies the difference between its inputs with an enormous gain, typically of the order of a hundred thousand, and for that reason it rarely works on its own. What the circuit actually does is decided by the resistors and capacitors we connect around it.

When part of the output is fed back to the inverting input, this negative feedback makes the op-amp adjust its output until the two inputs sit at almost the same potential. From this we get the two golden rules, valid as long as the output stays away from its limits, which say that the voltage difference between the inputs is practically zero and that no current flows into them.

In the inverting amplifier, the + input is at earth and the − input sits at a virtual earth, and since the current arriving through \(R_1\) has nowhere to go but \(R_f\), the gain is \(-R_f/R_1\). If the signal goes straight into the + input, we have the non-inverting amplifier, with the output in phase and a gain of \(1 + R_f/R_g\), and the case with gain 1 is the voltage follower, which draws almost no current from whatever drives it and supplies the load current by itself.

With several inputs, each with its own resistor, the currents meet at the virtual earth and the inverting amplifier becomes a summing amplifier, typically the heart of an analogue audio mixer. Replacing \(R_f\) with a capacitor, we arrive at the integrator, in which a constant input charges the capacitor with a fixed current and the output falls in a ramp.

If we remove the feedback, the enormous gain throws the output to one of its extremes as soon as one input passes the other, and the circuit becomes a comparator. These extremes, the supply rails, lie a little below the supply voltage, and the output cannot go beyond them, however large the gain we ask for. When we ask for more, the wave comes out clipped and we say that the op-amp has saturated.

How does an eight-pin chip keep time?

The 555 is an integrated circuit from the 1970s that is probably still inside some flashing light, alarm or timer near you. On the outside it has eight pins, and on the inside it has only a few parts, which the 'inside the 555' panel on the bench shows together with the circuit voltages at every instant.

Three equal 5 kΩ resistors divide the supply and create two references, at 2/3 and at 1/3 of VCC. The threshold comparator watches pin 6 and signals when it goes above 2/3 of VCC, and the trigger comparator watches pin 2 and signals when it drops below 1/3. Both control a flip-flop, a one-bit memory. The trigger switches the output on, at pin 3, the threshold switches it off, and while the output is low a transistor connects pin 7, the discharge, to earth.

In astable mode, the capacitor charges through \(R_1 + R_2\) up to 2/3 of VCC, when the threshold switches the output off and the discharge pin starts to empty it through \(R_2\) alone. When it reaches 1/3 of VCC, the trigger switches the output on again. The cycle then repeats without stopping. Each stretch is an RC charge or discharge between 1/3 and 2/3 of the voltage, which takes \(\ln 2 \approx 0.693\) time constants, and the formula for the frequency follows from that.

Since the high time uses \(R_1 + R_2\) and the low time only \(R_2\), the output stays high for longer than it stays low, and the duty cycle of this astable circuit is above 50%. It gets close to 50% when \(R_2\) is much larger than \(R_1\).

In monostable mode, the output stays low until a tap on pin 2, which sets the flip-flop and makes the discharge pin release the capacitor, which then charges through R from zero up to 2/3 of VCC, when the threshold ends the pulse. This time, which we can measure on the oscilloscope, is \(R\,C\ln 3 \approx 1.1\,R\,C\) and does not depend on how long the tap lasts, provided the tap is shorter than the pulse.

With an LED at the output, we can see the 555 at work. At low frequencies the LED blinks, and above roughly twenty blinks per second the eye tends to merge them, so the LED seems to stay on all the time, at its average brightness.