Pentode Constant Current Sinks [8]

Pentodes are even better as CCSs because of their high μ, and are particularly useful when the allowable voltage drop across the sink is quite low.

If we needed a CCS of 10 mA, but were only allowed a 100 V drop across it, an E88CC could achieve an output resistance of only ≈100 kΩ, which is still a 10-fold improvement over a 10 kΩ resistor, but a pentode can do rather better.

If we leave even a 2 kΩ cathode resistor unbypassed, a pentode can increase its output resistance to >10 MΩ. This is a stunningly good CCS, but it should be remembered that pentodes tend to be noisy, so this would not be a good choice in the first stage of a sensitive pre-amplifier (see Figure 2.35).

When using a pentode as a CCS, it is vital to remember that the cathode resistor passes not only the desired constant current, but also the g₂ current. Note also that the g₂ decoupling capacitor must be taken to the cathode, and not to ground. This is because we want cathode feedback to increase r_a, but we do not want the voltage between g₂ and the cathode to vary, as this would cause positive feedback that would reduce r_a.

Some pentodes make better CCSs than others because their anode characteristics are flatter, giving a higher output resistance, or because the flat part of their anode characteristics swings closer to 0 V. Table 2.1 shows single pentodes that are particularly suitable in CCSs.

Table 2.1 shows optimum currents much lower than I_a(max), partly because at higher currents the anode curves tilt away from the horizontal, indicating reduced r_a, but mostly because the major contribution to output resistance actually comes from the unbypassed R_k, whose value is multiplied by a factor of g_m·r_a(μ), which is typically 1,000 or more for a pentode. Higher currents require less bias and a reduced value of R_k, thus reducing output resistance. For maximum output resistance, it is better to use a valve with an oversized P_a, requiring a large R_k, than that with a perfectly rated P_a, requiring a smaller R_k. Unfortunately, the disadvantage of a CCS operated at very low I_a is that $I_{{\mathrm{g}}_2}$becomes a significant proportion of I_k, making the circuit inefficient.

As an example, we might require an 8 mA CCS. However, this is the lower end of efficient operation for an EF184, and the curves show that it is likely to require $I_{{\mathrm{g}}_2}\approx 3\text{mA}$, which means that total HT current has been increased by ≈38%. If there is only one sink, then this isn’t a problem, but if there are many such sinks, this can greatly increase the cost of the HT supply. If we were willing to reduce the sink current to 6 mA, this would bring it within range of the EF91, which only requires $I_{{\mathrm{g}}_2}\approx 1.55\text{mA}$ in this instance, reducing HT current from 11 mA to 7.55 mA. Although the EF91 does not have quite such attractive curves as the EF184, it is far cheaper to use, and if HT current is limited, it can be worth bending a design to allow its use.

Optimally biassed, most small-signal pentodes split I_k between I_a and $I_{{\mathrm{g}}_2}$ by ≈4:1 ratio. Thus, 8 mA of I_a typically requires 2 mA of $I_{{\mathrm{g}}_2}$. It is very important to check that the pentode’s I_a, P_a, and especially $P_{{\mathrm{g}}_2}$ are not exceeded. Successful design of pentode CCSs requires full datasheets with curves or a valve tester/rig to allow voltages to be imposed and currents to be determined experimentally (which is more reliable).

If the CCS is to be used in a stage with a low signal level, it is worth considering hum and screening. The EF184 has an integral metal screen, the EF91 has a conductive paint screen on the inside of the envelope, but the EL83 and EL822 power valves are completely unscreened.

Analysis of the Self-Contained White Cathode Follower

The lower valve is fed with a signal from the upper valve, which, in turn, it feeds back into the cathode/grid circuit of the upper valve. At the input to the lower valve, the circuit may be considered to be a cascode amplifier (see Figure 2.37).

Provided that μ is reasonably large and the cathode is bypassed:

$A_v\approx g_{\mathrm{m}}·R$

And the resistance looking into the cathode of the upper valve is:

$r_{\mathrm{k}}=\frac{R+r_{\mathrm{a}}}{\mu +1}$

But the gain of the lower valve is devoted to reducing the output resistance at the cathode of the upper valve, so combining the two equations gives:

${r΄}_{\mathrm{k}}\approx \frac{R+r_{\mathrm{a}}}{(\mu +1)g_{\mathrm{m}}·R}$

μ is usually rather greater than 1, even for power triodes, so if we substitute μ=g_m·r_a:

${r΄}_{\mathrm{k}}\approx \frac{R+r_{\mathrm{a}}}{g_{\mathrm{m}}^2·R·r_{\mathrm{a}}}\text{,}\mathrm{or}\frac{1}{g_{\mathrm{m}}^2}·\frac{R+r_{\mathrm{a}}}{R·r_{\mathrm{a}}}$

We can now recognise the R·r_a term as the inverted parallel combination of R and r_a. This is significant because it indicates that there is a point beyond which increasing R has no effect and that final output resistance is limited by r_a:

$r_{\text{out}}={r΄}_{\mathrm{k}}\approx \frac{1}{g_{\mathrm{m}}^2·R΄}$

where

$R΄=\frac{R·r_{\mathrm{a}}}{R+r_{\mathrm{a}}}$

It should be noted that two rather dubious approximations were made to derive this result, both of which relied on high μ. The example in Figure 2.36 was optimised for low output resistance, and R≈10r_a, beyond which limit no practical improvement is possible.

Although superficially completely different, the self-contained White cathode follower and the shunt-regulated push–pull (SRPP) amplifier described later in this chapter are both shunt-regulated amplifiers because both valves contribute to the AC load current. Rigorous equations for gain and output resistance derived by Amos and Birkinshaw [10] are as follows:

$A_v=\frac{{\mu }_1({\mu }_{2}R+r_{{\mathrm{a}}_2})}{r_{{\mathrm{a}}_2}({\mu }_1+1)+r_{{\mathrm{a}}_1}+R[{\mu }_2({\mu }_1+1)+1]}$

$r_{\text{out}}=\frac{r_{{\mathrm{a}}_2}(R+r_{{\mathrm{a}}_1})}{r_{{\mathrm{a}}_2}({\mu }_1+1)+r_{{\mathrm{a}}_1}+R[{\mu }_2({\mu }_1+1)+1]}$

where V₁ is the upper (amplifying) valve and V₂ is the lower (regulating) valve.

Using an E88CC as an example, with g_m≈5 mA/V and μ≈28, the rigorous equation predicts r_out=6.6 Ω, whereas the approximate equation is only 4% high at 6.9 Ω despite the two dubious approximations. Experimentation with a spreadsheet reveals that this version of the White cathode follower is unsuited to low-μ valves, since a 6080 (μ=2) predicts r_out≈35 Ω, which is worse than when used as a standard cathode follower (r_out≈15 Ω). However, a triode-strapped E55L (μ=30) predicts r_out<2 Ω, and a triode-strapped D3a (μ=80) should easily achieve r_out<1 Ω. It is easy to become excited about predicted low output resistances, but we must always remember that all such equations contain the implicit assumption that the output resistance of the HT supply is 0 Ω, which is normally only approximated by a regulated supply.

Since the White patent suggested that the circuit was particularly suitable for driving analogue video cables (which were typically 75 Ω transmission lines), it is not surprising that the stage makes an excellent output cable driver for a pre-amplifier.

Note that because the feedback that causes the low output resistance is AC coupled, output resistance rises at low frequencies not to 1/g_m, but to:

$r_{\text{out}(\text{LF})}=\frac{R+r_{\mathrm{a}}}{\mu +1}\Vert r_{\mathrm{a}}=\frac{r_{\mathrm{a}}(R+r_{\mathrm{a}})}{r_{\mathrm{a}}(\mu +1)+R}$

In this instance, r_out rises to 1.5 kΩ, rather than 200 Ω, which is what a normal cathode follower would achieve. The practical implication is that the stage will not short circuit low-frequency noise (such as mains hum) induced into the output cable as effectively as a stage with a true 6 Ω output resistance from DC to light.

Usually, we do not need to calculate the gain A_v precisely, and the general cathode follower approximation of A_v=μ/(μ+1) is perfectly adequate, but if the stage was to be used as the basis of a Sallen & Key filter, the rigorous calculation of gain might be needed.

The White Cathode Follower as an Output Stage

The primary use of the White cathode follower is as an output stage for OTL amplifiers. A series resistor in either of the HT rails is a serious waste of power, so we must use the version preceded by a phase splitter (see Figure 2.38).

Assuming that neither valve switches off under any signal conditions (Class A), the gain of the lower valve is:

$A_{v(\text{lower})}=\frac{\mu ·R_{\mathrm{L}}}{R_{\mathrm{L}}+r_{\mathrm{a}}}$

The upper valve no longer has a resistor in its anode circuit, so r_k=1/g_m, and this is the anode load of the lower valve. Substituting:

$A_{v(\text{lower})}=\frac{\mu ·(1/g_{\mathrm{m}})}{(1/g_{\mathrm{m}})+r_{\mathrm{a}}}$

Multiplying by g_m and simplifying:

$A_{v(\text{lower})}=\frac{\mu }{\mu +1}$

A cathode follower with R_k=∞ would have the same gain, and because the lower valve strives to produce exactly the same signal as a standard cathode follower if it saw R_k=∞, there is no voltage difference between the two valves, so the upper valve does see R_k=∞. However, the input to the lower valve must be inverted, requiring an external phase splitter.

The lower valve no longer reduces the output resistance of the upper valve, since with a gain of 1 it cannot apply feedback to the upper valve, which is why OTL amplifiers need considerable global feedback to bring their output resistance down to a suitable value for damping moving coil loudspeakers.

The Importance of the AC Loadline

Up until now we have tacitly assumed that the input resistance of the following stage had little, or no, effect on the performance of the preceding stage. This would not be true if we used the anode output of the μ-follower because the value of the following grid-leak (typically ≈1 MΩ) is not merely comparable with the value of R_L, it is actually less than R_L and, therefore, lowers the effective value of R_L from ≈2 MΩ to 670 kΩ. This has a negligible effect on gain, but trebles the distortion, so using the anode output is not recommended.

Whilst R_g≥10R_L, it is legitimate to ignore its effect on the preceding stage, but once it becomes smaller than this, we should consider drawing an AC loadline to investigate whether it will cause a problem. Stages with active loads must take into account the input resistance of the following stage.

An accurate AC loadline is easily drawn. First we find the AC load, which is usually just the anode load and the following grid-leak in parallel. We know that the AC loadline must pass through the operating point, so all we need is a second point. The simplest way to do this is to move a convenient number of squares horizontally (change the voltage by 100 V or so), and calculate the increase, or decrease, in current through the AC load to give our second point. The line through these points is then the AC loadline, so inspection of this line gives the gain and linearity of the stage including the effects of the following load resistor.

Upper Valve Choice in the μ-Follower

There is no reason why the upper valve should be the same as the lower valve in a μ-follower. As a rule of thumb, the AC load resistance seen by the lower valve can be found by:

$r_{\mathrm{L}}\simeq {\mu }_{\text{upper}}·R_{\mathrm{L}}$

Maximising r_L minimises distortion in the lower valve, but experiment shows that once r_L≥50r_a, there is no further benefit to be gained and it is more profitable to investigate the distortion produced by the upper valve. Because a cathode follower operates with 100% feedback, increasing μ increases feedback and reduces distortion. However, high-μ valves need a higher value of V_a to avoid grid current, reducing available V_a for the lower valve and lowering maximum voltage swing.

High g_m is also useful in the upper valve if the stage is to feed a passive equalisation network because the resultant low (but changeable) r_out is a smaller proportion of the network’s series resistance.

The 6545P single triode has μ=52 and g_m≈20 mA/V at sensible anode currents, but its main advantage as the upper valve is that it can swing V_gk close to 0 V without distortion, allowing a high output voltage swing from a given HT voltage.

When triode strapped (g₂, g₃ to a), the D3a pentode is also a good choice as μ=80, and g_m≈20 mA/V is easily achievable even at quite low currents, but significant grid current begins at V_gk≈−1.1 V. The D3a has gold-plated pins and was produced in the era when gold plating genuinely meant special quality. Not only does it meet its published specification, but it is very consistent from one sample to another. By contrast, the (Soviet era) 6545P generally only just meets the lower limits of its specification and is rather variable, although its anode curves are extremely linear.

Limitations of the μ-Follower

Although the μ-follower is an excellent gain stage, it does have limitations. We have seen that it has a low output resistance and low distortion, and it is, therefore, tempting to use it as a line stage to drive long cables or low input resistance transistor amplifiers.

However, a low-impedance load steepens the AC load line of the cathode follower upper valve. Although this valve has 100% feedback, the steeper load line slightly reduces gain and the cathode follower can no longer bootstrap the lower stage’s R_L as effectively, so the lower valve sees a reduced load resistance, and distortion in the lower valve rises; connection of an external load always increases distortion within a μ-follower. As an extreme example, the r_out of a 6J5/6J5 μ-follower was tested at 0 dBu. At +28 dBu, the stage produced 0.29% THD+N, so it was predicted to produce ≈0.01% THD at 0 dBu. However, an output resistance measurement that dropped the output from 0 dBu to −6 dBu by loading it with a 720 Ω resistance increased THD+N to 0.85%.

If a low-load resistance must be driven with minimum distortion, the μ-follower can be buffered by a cathode follower. To drive the load effectively, the cathode follower should pass ≥10 mA, and the valve should be a frame-grid type with high g_m and high μ; 6545P and triode-strapped D3a are ideal. The cathode follower is a high-impedance load, so it can be direct coupled to the lower output of the μ-follower to avoid the inevitable distortion of the upper valve in the μ-follower. For lowest possible distortion, the cathode follower should have a CCS as its load.

A well-designed μ-follower enters overload very suddenly. A 6J5/6J5 μ-follower driven from a 51 kΩ source was driven into grid current, resulting in an output of +38.1 dBu (61.6 V_RMS) at 0.87% distortion. A high source resistance causes hard clipping at the onset of grid current, so we should expect a decaying series of odd harmonics. However, because grid current only clips one half-cycle and asymmetry causes even harmonics, we can expect all possible harmonics (see Figure 2.41).

Dropping the level by 1 dB to +37.1 dBu reduced the distortion to 0.54%, and the higher harmonics entirely disappeared (see Figure 2.42).

The μ-follower is a gain stage and cathode follower connected in such a way that it provides a CCS load to the gain stage, making it much more linear. The downside is that the cathode follower is in series with the gain stage and consumes excess HT voltage because the (normally triode) cathode follower cannot swing to V_a=0, and also because of the voltage dropped across R_L. If we don’t combine the two valves in this way, but instead use a semiconductor CCS load for the gain stage and another for the (DC-coupled) cathode follower, we can obtain the same distortion as the μ-follower but a higher maximum output swing for a given HT voltage. We will investigate semiconductor CCSs at the end of this chapter.

Gain of the Differential Pair

When driven by a signal connected between the two grids, the gain of the differential pair is identical to that of a standard common cathode stage, but the output voltage is found between the anodes of the stage. Therefore, if we look between one anode and ground, we only see half of the output voltage, and the gain appears to be halved.

If we use the differential pair as a phase splitter, and apply the same input voltage as before between one grid and ground, instead of each grid seeing half the input voltage, one sees the entire input voltage and the other none. Because the voltage difference between the two grids is the same, the gain remains the same.

Output Resistance of the Differential Pair

Provided that the output of the differential pair is not unbalanced in any way, r_out at each terminal is identical to that of a simple common cathode amplifier (r_a‖R_L).

However, if only one output is loaded, the output resistance rises considerably. Working backwards from the path to ground (HT supply) via the first R_L, we see:

$r_{\mathrm{k}}=\frac{R_{\mathrm{L}}+r_{\mathrm{a}}}{\mu +1}$

But we now also see a path R_k to ground (0 V), which is in parallel with r_k:

${r΄}_{\mathrm{k}}=\frac{R_{\mathrm{k}}·((R_{\mathrm{L}}+r_{\mathrm{a}})/(\mu +1))}{R_{\mathrm{k}}+((R_{\mathrm{L}}+r_{\mathrm{a}})/(\mu +1))}$

Multiplying through by (μ+1):

${r΄}_{\mathrm{k}}=\frac{R_{\mathrm{k}}(R_{\mathrm{L}}+r_{\mathrm{a}})}{R_{\mathrm{k}}(\mu +1)+R_{\mathrm{L}}+r_{\mathrm{a}}}$

Looking down through the second anode, we see r_a in series with this multiplied by a factor of (μ+1):

${r΄}_{\mathrm{a}}=r_{\mathrm{a}}+\frac{R_{\mathrm{k}}(\mu +1)(R_{\mathrm{L}}+r_{\mathrm{a}})}{R_{\mathrm{k}}(\mu +1)+(R_{\mathrm{L}}+r_{\mathrm{a}})}$

If we divide by R_k(μ+1), we obtain:

${r΄}_{\mathrm{a}}=r_{\mathrm{a}}+\frac{R_{\mathrm{L}}+r_{\mathrm{a}}}{1+((R_{\mathrm{L}}+r_{\mathrm{a}})/R_{\mathrm{k}}(\mu +1))}$

As R_k tends to ∞, the right-hand term on the bottom line reduces to zero, resulting in a maximum value of ${r΄}_a$:

${r΄}_{\mathrm{a}}\approx R_{\mathrm{L}}+2r_{\mathrm{a}}$

This high value of r_a will become significant when we investigate the PSRR of the differential pair.

If R_L>>r_a, then the output resistance (only one terminal loaded) is:

$r_{\text{out}}\approx \frac{R_{\mathrm{L}}(R_{\mathrm{L}}+2r_{\mathrm{a}})}{2(R_{\mathrm{L}}+r_{\mathrm{a}})}\approx \frac{R_{\mathrm{L}}}{2}$

AC Balance of the Differential Pair and Signal at the Cathode Junction

Provided that the anode load resistors are perfectly matched, the only way that there can be an AC imbalance at the anodes is if some signal current is lost to ground by a finite tail resistance. Thus, an infinite tail resistance would allow two entirely different valves to achieve perfect balance provided their anode loads were matched.

If the two valves are perfectly matched, then they must have identical gain to their identical anode loads and there can be no signal at the cathode junction (grids driven by signals of equal and opposite polarity). If they are not perfectly matched, their gains must differ, so there must be a signal at the cathode junction.

If the two valves are not driven by signals of equal and opposite polarity, there must be a signal at the cathode junction. This statement is easily understood by considering the extreme case of a phase splitter (signal at only one grid); the only way the valve with no signal on its grid can have its V_gk modulated is by moving its cathode, requiring a signal at the cathode junction.

The two previous statements show that any imperfection of drive or valve matching results in a signal at the cathode junction – there should ideally be no signal. Worse, if there is a signal, it implies that the valves are being forced to amplify that signal in order to produce their output signal, and any interference signal at the cathode junction will also be amplified.

Common-Mode Rejection Ratio (CMRR)

Rather than assuming signals of equal and opposite polarity on each grid, we will now consider the response to identical signals on each grid. If we apply +1 V to both grids, the cathode voltage simply rises by 1 V, the cathode current remains constant, and the anode voltages do not change, because we have not modulated V_gk. The amplifier only responds to differences between the inputs or differential signals. Applying the same signal to both grids is known as a common-mode signal.

This property of rejecting common-mode signals is significant, since it implies that the circuit can reject hum on power supplies, or common-mode hum on the input signal, so we will investigate it further.

The signal at each output can be expressed in terms of currents using Ohm’s law:

$V_{\text{out}(1)}=i_1·R_{\mathrm{L}(1)}$

$V_{\text{out}(2)}=i_2·R_{\mathrm{L}(2)}$

Each output will be an exact inverted replica of the other if i₁=i₂, provided that the two load resistors are equal. There are two main ways in which this nirvana may be eroded:

• If signal current is lost through an additional path to ground. A signal current i₁ flowing down V₁ out of its cathode must split, with some current being lost down R_k and the remainder flowing up into the cathode of V₂, to become i₂. However, if R_k=∞, no current can flow down R_k, and i₁=i₂. If μ₁=μ₂, and R_L(1)=R_L(2), then:

$\text{CMRR}\approx \frac{\mu R_{\mathrm{k}}}{R_{\mathrm{L}}+r_{\mathrm{a}}}$

This indicates that we should use high-μ valves, and maximise the ratio of R_k to R_L. As an example, an EF184 constant current source (r_sink=R_k≈1 MΩ) and E88CC differential pair (μ=32) and R_L=47 kΩ would have CMRR≈57 dB.

• Results from the above equation will be degraded if μ₁≠μ₂, or if R_L(1)≠R_L(2), or a combination of the two. With the easy availability of low-cost, accurate digital multimeters, mismatching of load resistors is avoidable, but matching the valves is harder. If μ₁≠μ₂, then:

$\text{CMRR}\propto \frac{{\mu }_1·{\mu }_2}{{\mu }_1-{\mu }_2}$

which indicates that high-μ valves are still desirable, but that matching is important.

Because the simple equation for CMRR ignores mismatched valves, mismatched load resistors and stray capacitances, any predictions of CMRR>60 dB should be treated with caution. Nevertheless, it is handy for checking that your tail resistance R_k is sufficiently high to ensure that predicted CMRR>40 dB, since 40 dB is an easily achievable CMRR in practice.

Power Supply Rejection Ratio (PSRR)

Since hum and noise on the power supply line are a common-mode signal, they must also be attenuated by the CMRR. We might also expect the potential divider formed by r_a and R_L to give significant additional attenuation. However, at either terminal, the only path to ground is via the other anode and R_L up to the HT supply, which is exactly the scenario we investigated when determining r_out with only one output loaded, therefore:

${r΄}_{\mathrm{a}}\approx R_{\mathrm{L}}+2r_{\mathrm{a}}$

And the attenuation of power supply noise (solely due to potential divider action) is thus:

$\text{Attenuation}=\frac{R_{\mathrm{L}}+2r_{\mathrm{a}}}{2(R_{\mathrm{L}}+r_{\mathrm{a}})}\le 6\text{dB}$

If R_L>>r_a, we achieve the maximum attenuation of 6 dB! Our previous example (R_L=47 kΩ and r_a=4.95 kΩ) attenuates by 5.2 dB, and together with the 57 dB due to CMRR, PSRR=62 dB.

It is now well worth comparing the PSRR of the common cathode stage, μ-follower and differential pair (same DC conditions for the amplifying valve) (Table 2.2).

The differential pair is the best, and will remain the best, since an improved constant current source for the μ-follower could be adapted to become an improved CCS for the differential pair.

Knowing the PSRR enables us to design power supplies correctly because it gives an indication of the allowable hum on the HT supply.

For example, the second (differential pair) stage of a balanced pre-amplifier might need the 100 Hz power supply hum to be 100 dB quieter than the maximum expected audio signal. At this point, the signal has not received RIAA 3,180 μs/318 μs correction, so the level at 100 Hz is 13 dB lower than at 1 kHz. However, peak levels from LP are +12 dB compared to the 5 cm/s line-up level, so the maximum audio signal at 100 Hz is 1 dB lower than the 1 kHz calculated signal level at the anode (2.2 V_RMS)=2 V. We want 100 dB signal/hum, but because 62 dB of this will be provided by the differential pair’s PSRR, we only need the hum on the power supply to be 38 dB quieter than 2 V, so we could tolerate 25 mV of hum on the power supply – which is easily achievable.

Using Transistors as Active Loads for Valves

All the previous sink circuits can be mirrored about 0 V and PNP transistors substituted for NPN. If the circuit is then connected to the HT supply, all the previous sink circuits become constant current sources, allowing a triode to achieve A_v=μ. More significantly, they permit a valve to achieve low distortion from a low HT voltage.

As an example, in common with all high-μ valves, the ECC83 needs considerable V_a before it can be biassed out of grid current; 150 V is typical. As a general rule of thumb, R_L>2r_a, and as r_a≈75 kΩ for the ECC83, we might use R_L=150 kΩ. If I_a=0.7 mA, we would drop 105 V across R_L, so we would need 255 V of HT. But we might only require the stage to produce an output swing of 5 V_pk–pk, so most of the HT is wasted. If we replace the 150 kΩ resistor with a constant current source, the valve sees a much higher value of R_L, and we can set the HT voltage independently to accommodate the maximum required output swing (see Figure 2.54).

In Figure 2.54, the concept of operating a high-μ valve from a low HT was taken to the extreme because the author needed a high gain differential pair stage (ECC83: μ=100), but only had 150 V of positive HT available. Note that high-voltage transistors are required to withstand either anode swinging towards 0 V.

At fist sight, the circuit has two constant current sources, both trying to define the current in the same wire. The trick is that the cathode CCS is deliberately superior to the anode constant current sources, enabling it to enforce its behaviour upon them.

The anode constant current sources have an output resistance of ≈600 kΩ (h_fe=40 and R_E=15 k), but the cathode CCS has an output resistance of ≈130 MΩ (h_fe1=420, h_fe2=40 and R_E=7k5). Thus, fine adjustment of the cathode CCS simply changes the voltage drop across the 600 kΩ output resistance of the anode constant current sources. The output resistance of the anode constant current sources doesn’t change appreciably with temperature, so the stability of the final circuit is down to the behaviour of the cathode CCS. The cathode CCS enforces 0.82 mA, ideally split equally between the anode CCSs so that each passes 0.41 mA. By Ohm’s law, a 1 μA change in current through the 600 kΩ output resistance of an anode CCS would cause a change in voltage drop of 0.6 V. The LED reference voltage and ordinary resistor in the cathode CCS circuit should be stable to <1% drift in current, resulting in <2.5 V drift in each anode voltage, which is perfectly acceptable.

Although Zener diodes are normally bypassed to reduce noise, the noise generated by both diodes is a common mode and is, therefore, rejected by the next (differential) stage. On test, the circuit achieved the required differential swing of 7V_pk–pk at 1 kHz with only 0.04% distortion.

A cascode greatly increases r_out, flattening the loadline and reducing distortion in the valve. If we wanted to maximise output swing and minimise distortion, we might operate a 7N7 (loctal equivalent to 6SN7) at I_a=8 mA because μ becomes more nearly constant when I_a>6 mA. We assume that our cascode will provide a horizontal loadline, so we plot this at 8 mA (see Figure 2.55).

Normally, as V_a rises, we have to consider the cramping of curves as I_a falls and cut-off is approached, but I_a is now a constant, and the only limit to positive swing is that the cascode requires sufficient voltage to operate correctly. 15 V is quite adequate for the cascode, so a 400 V HT would allow V_a to swing to 385 V. Looking in the opposite direction along the 8 mA loadline, grid current is likely to begin at ≈100 V. The maximum possible swing is, therefore, 385−100 V=285 V_pk–pk≈100 V_RMS.

Although the cascode forces I_a=8 mA, we must adjust valve bias to set V_a. At just over maximum swing, we should clip positive and negative half-cycles equally, so the operating point should be halfway between the maximum and minimum permissible anode voltages – which is their average:

$V_{\mathrm{a}}=\frac{V_{\max }+V_{\min }}{2}=\frac{385+100}{2}=242.5\mathrm{V}$

Looking at the curves, we see that V_gk≈8 V is required to set V_a correctly, and this can be provided by an 8.2 V Zener diode (see Figure 2.56).

Since the stage is intended to swing large voltages, noise is not a problem, so it is not essential to bypass the Zener diode with a capacitor.

If there is 242.5 V across the valve, then there is 147.5 V across the lower transistor, so it must dissipate 1.18 W when quiescent. When V_a swings to 100 V, the transistor must withstand 285 V at 8 mA, so it momentarily dissipates 2.28 W, and it might seem that this is the required rating of the transistor. However, the flat loadline has reduced distortion almost to zero, so the positive and negative swings are equal, and the average power dissipated in the transistor over one cycle of audio is equal to the quiescent power.

As before, it is only the transistor nearest the valve that must be able to withstand high voltages and dissipate significant power, so the additional transistor can be as fast and fragile as we like.

Optimising r_out by Choice of Transistor Type

Table 2.3 compares transistors that are useful in support circuitry for valves.

Output resistance at low frequencies is partly determined by 1/h_oe, but is dominated by h_fe, since any resistance in the emitter circuit is multiplied by h_fe. Impedance at high frequencies is shunted by the capacitance seen at the collector of the transistor, which will partly be determined by strays, but also by the transistor itself. In general, high-voltage/high-power transistors have a larger silicon die area, and greater capacitance, which is reflected in their lower f_T. Additionally, f_T varies significantly with I_C, and operating a transistor below its optimum I_C could reduce f_T by a factor of five. If in doubt, download the transistor’s datasheet from the Internet – all the semiconductor manufacturers have excellent websites. As a consequence of these considerations, a small cascode CCS would ideally use two BC549Cs, or if low output capacitance (≈0.5 pF excluding strays) was essential and sufficient voltage was available, a triple cascode composed of a BFR90 with two BC549Cs beneath it. It is only critical for the outer transistor to have low output capacitance because the capacitances of the other transistors are at lower impedance points.

Any bipolar transistor needs a minimum V_CE for it to operate linearly. For a low-voltage transistor at currents ≤30 mA, ≈1 V is sufficient, but higher currents may require 2 V (see Figure 2.57).

High-voltage transistors such as MPSA42 or MJE340 may require V_CE>2 V. A CCS in a differential pair operating as a phase splitter has half the input signal across it, so this point can become significant. In a cascode sink, the lower transistor has hardly any AC signal across it, so it can operate with only 2–3 V_DC, leaving the remaining DC for the upper transistor, which supports the bulk of the AC signal.

Field-Effect Transistors (FETs) as Constant Current Sinks

FETs are described as being depletion or enhancement mode. An enhancement mode device needs bias similar to a bipolar transistor to turn it on, whereas a depletion mode device needs bias similar to a valve to turn it off. Both can be used for making CCSs, but a depletion mode device can be self-biassed, whereas the enhancement mode device requires an external voltage, making it more cumbersome to use. ‘Constant current’ diodes are actually depletion mode FETs with their gate and source tied together. They tend to have a low maximum voltage, and disappointingly low slope resistance for their price.

As stated earlier, the output capacitance of any constant current circuit must be minimised if it is to maintain its performance at high frequencies, and that means choosing transistors with low output capacitance. Sadly, many high-voltage FETs simply aren’t suitable for valve electronics because of excessive output capacitance. As an example, the 700 mW power rating of the ZVN0545A suggests that it ought to be useful but it actually has higher output capacitance than the 15 W (TO220) DN2540N5. The reason for this is that semiconductor manufacturers make far fewer different devices than you think – they just package the dies differently. Thus, the DN2540N3 is the same silicon die, but in a TO92 package, reducing its power rating to 1 W.

Happily, the 400 V 15 W Supertex DN2540N5 depletion mode JFET is very nearly ideal for supporting valve audio. Unfortunately, the phrase ‘device variation’ was invented for FETs and never more so than for the voltage V_GS(ID=0) required to turn the device on (or off); Supertex specify a spread of V_GS(ID=0) from −1.5 V to −3.5 V. Fortunately, it’s easy to measure V_GS for an individual device under any operating conditions, so devices can be matched or other circuit values can be determined for a specific device (see Figure 2.58).

The drain of the device is connected to a positive supply having the same voltage expected in the final circuit. The gate is connected to ground via a 1 kΩ gate-stopper resistor to prevent RF oscillation. The source is connected to a convenient negative supply via a programmable CCS adjusted to the required current. A Digital Volt Meter (DVM) connected between ground and source measures V_GS. If we had more than one device, we would quickly run them all through the test jig to find each individual V_GS. Devices with similar V_GS would be deemed to be pairs.

Designing Constant Current Sinks Using the DN2540N5

The very simplest CCS using the DN2540N5 is little more than the contents of the constant current diode vilified earlier (see Figure 2.59).

The carbon gate-stopper resistor prevents Very High Frequency (VHF) oscillation, and the source resistor programmes the current. Unfortunately, the value of the source resistor has to be found empirically for each DN2540N5, using the test jig of Figure 2.58. Alternatively, a variable resistor could be used and Adjusted On Test (AOT), but we still need to know the approximate value required in order to buy the correct value of variable resistor and if accidentally set to 0 Ω at switch-on, the saturation current might cause damage. It really is easier (and cheaper) to use the jig and fit a fixed resistor.

As an example, we might want to use a DN2540N5 as a CCS shared by the cathodes of a pair of EL84s. We know that when the EL84s are working, their cathodes will be at ≈11 V and will pass a total of 80 mA. The programmable CCS probably needs a minimum of 6 V across it to operate correctly, so we connect it to a negative supply of perhaps −9 V. We then apply +11 V to the drain of our Device Under Test (DUT), and using an ammeter adjust the programmable CCS to draw 80 mA through the DN2540N5. Having set the required current, we measure the voltage between 0 V and the FET’s source. (We don’t attempt to genuinely measure V_GS on the FET because connecting a DVM to it will almost certainly cause oscillation, so we assume zero gate current and measure from 0 V.) Perhaps, we find that for our particular sample, V_GS=−1.873 V at I_ds=80 mA. All we need to do to turn the JFET into a 80 mA CCS for our EL84 is to add a source resistor that will drop 1.873 V when 80 mA passes through it (just like a triode), 1.873 V/0.08 A=23.4 Ω. The nearest standard value of 24 Ω will almost certainly do.

Like the pentode CCSs we investigated earlier, the FET multiplies the value of its programming resistor by μ. Unsurprisingly, μ is not given by the manufacturer, but can be found from the relationship:

$\mu =g_{\mathrm{m}}·r_{\text{ds}}$

The drain slope resistance r_ds can be found from the device curves by measuring output currents on one V_GS curve at two widely spaced voltages (see Figure 2.60):

$r_{\text{ds}}=\frac{V_1-V_2}{I_1-I_2}=\frac{100-20}{32.6-30}\ge 31\mathrm{k}\mathrm{\Omega }$

The mutual conductance g_m at V_ds=100 V can also be found from the device curves:

$g_{\mathrm{m}}=\frac{I_1-I_2}{V_{{\text{GS}}_1}-V_{{\text{GS}}_2}}=\frac{33.2-18.4}{-1.375\mathrm{V}-(-1.475V\mathrm{)}}=148\mathrm{mA}\mathrm{/}V$

Note that although this value of g_m is much lower than given on the datasheet, it is far higher than a valve could achieve at the same current. Combining the two values, we obtain μ≈4,500, which is a little higher than we might expect from a pentode. More significantly, a 39 Ω programming resistor in this FET would produce a ≈40 mA CCS having an output resistance of 175 kΩ. To put this into context, a genuine 175 kΩ resistor would drop 7 kV rather than 100 V and dissipate 282 W rather than 4 W.

The previous worked example was given not to suggest that such a process is needed, but to show just how good a CCS can be made with the DN2540N5.

Nevertheless, considerable improvement is possible, and like the simple bipolar transistor, the cascode is the way to go. Remembering how the single BJT CCS evolved into a cascode, we could add a voltage reference and second FET (see Figure 2.61a and b).

Unfortunately, by adding the voltage reference, we lose the very desirable property of the circuit being a two-terminal device. Because the DN2540N5 is a depletion mode device, the voltage reference is not strictly necessary, and we could use one DN2540N5 as the reference resistance for the other (see Figure 2.61c).

However, there are disadvantages because the lower device in the cascode is forced to operate with a very low V_ds, but they can be ameliorated by returning the upper device’s gate stopper not to 0 V but to the source of the lower device (see Figure 2.61d).

This simple circuit is deservedly popular because (being two-terminal) it can be used equally well as an anode load or as the cathode tail resistance in a differential pair. The programming resistance is multiplied by both amplification factors, and even though the lower device’s performance suffers due to the low V_ds, our previous example is likely to improve from ≈175 kΩ to ≈30 MΩ.

However, things are not quite as rosy as they might seem. The high resistance applies at DC, but not at AC because of output capacitance C_OSS [21] (see Figure 2.62).

As can be seen from the graph, C_OSS is alarmingly high at low voltages and it is not until V_ds>15 V that it falls to its asymptotic value of ≈12 pF. Fortunately, we would always operate at V_ds>15 V (preferably 40 V, or more) because a BJT cascode is a better choice at lower voltages. An EF184 pentode CCS could achieve an output capacitance of 3 pF, but once we add typical strays of 3 pF to both circuits, the valve advantage degrades to 6 pF against 15 pF. This somewhat poorer output capacitance is the price we pay for the undoubted convenience of a two-terminal CCS.

The device has a maximum continuous current rating of 500 mA, so it should come as no surprise to learn that performance degrades at low currents. If you need a current <10 mA, then you owe it to yourself to see if there’s an alternative solution because the device is much better >10 mA, and at 25 mA it really comes alive (r_out of a cascode CCS at 25 mA was four times that at 10 mA, all other factors kept constant).