Published in Practical Wireless, December 2002.
These pages contains additional information for anyone who intends to build any of the power supply modules featured in my article in the December 2002 issue of Practical Wireless. The information may also be of some help to anyone who wishes to understand the circuits or would like to modify them for their own purposes. Links to relevant data sheets can be found at the end of this page.
If you want the circuits then you'll need a copy of the magazine. Back issues are available. Don't ask me for copies of either the original article or the circuits because they're copyright Practical Wireless. Sorry.
Goof count so far: One
The Schottky diodes in Figure 3 (D4) and Figure 6 (D5) have been drawn as Zener diodes. Yes, I do know the difference but I didn't notice the mistake when the drawings were checked. One to me.
Most of the components specified are readily obtainable.
However, the switched-mode integrated circuits and the 470uH
inductor may be more of a problem.
Fortunately, Electrovalue Ltd. have available two low-cost inductors which
can be used in place of the 470uH inductor shown in the circuit diagrams.
These are: 680uH at 1A - suitable for the MC34063A design,
and 330uH at 1.4A - which works well with the LM2575T.
Either can be used in the optional filter shown in Figure 3.
Both inductors are radial types and so need mounting clear
of other components and any metalwork.
Electrovalue can also supply the integrated circuits.
Contact Electrovalue Ltd. on 01784 433604. E-mail sales@electrovalue.co.uk. Or see the Electrovalue advertisement in Practical wireless.
The modules are primarily intended to power portable valved radio sets which use miniature (B7G-based) 1.4V-filament battery valves. In most cases, they can also be used with older sets having either 1.4V-filament or 2.0V-filament valves.
There are two series of miniature valves; one series has 50mA filaments, while the other, later series, has 25mA filaments. In both series, output valves have twin filaments which may be wired in series or in parallel. Connecting the filaments in series doubles the filament voltage (to 2.8V) but keeps the current the same (at 50mA or 25mA). The parallel connection doubles the required current: 100mA for a 50mA-series valve and 50mA for a 25mA-series valve.
Most portable sets have a mixer/oscillator stage, an i.f. amplifier stage, an audio/detector stage and an audio output stage. The total l.t. current required is therefore 250mA (for 50mA-series valves) or 125mA (for 25mA-series valves). The valve manufacturers state that the filament voltage should be 1.3V (not 1.4V) if a regulated filament supply is available.
The h.t. voltage was sometimes 67.5V but in the main, 90V was favourite. The current required seldom exceeded 10mA, which is the design goal here.
Assuming a total filament current of 250mA, and assuming a 1.3V l.t. supply, the l.t. power required is 325mW. The h.t. power required is simply 90V x 10mA = 900mW. Total power is therefore 1.225W. Power converters are never 100% efficient; I guessed 70% averaged over both l.t. and h.t. converters. Taking into account the conversion efficiency, the power source needs to supply 100/70 x 1.225W = 1.75W.
As some losses in the converter will be proportional to the square of the current - 'I-squared-R' (resistive) losses - it makes sense to keep the input current low. Therefore, a very low input voltage - which means a high input current - is not desirable. Neither is too high a voltage (too many cells in the battery). I settled on 12V, as the current drawn is then quite sensible; about 145mA.
While this current is a little high for AA-size NiCd cells, high-capacity (1400mAh plus) AA-size NiMH cells should prove quite satisfactory. With luck, the cells will last close to ten hours between recharges. If there's sufficient room inside the set, a 12V sealed lead-acid battery is a good choice.
In most cases, the h.t. and l.t. supplies do not share a common ground. However, the l.t. negative is very often connected to the set's chassis. The switched-mode regulators described in the article are not isolated, so the input to the power converter is connected through to the set's chassis. This is an advantage as it's safer when an grounded power supply is used or the set is powered from a vehicle battery; the set's chassis will then be at earth potential too.
Normally, positive voltage linear regulators use an n.p.n.
transistor as the series-pass element.
However, with a p.n.p. transistor, the minimum input-output
voltage differential can be as low as the collector-emitter
voltage of the series-pass transistor, maybe 200mV or less.
A conventional regulator might need 1.5V or more.
Diode D3 aids symmetry by causing a similar voltage drop in the collector circuit of Tr2 as that caused by the base-emitter junction in the collector circuit of Tr1. Notice that the two transistors share a common current source: R4.
The circuit has limitations, particularly due to the use of ordinary silicon diodes as the voltage reference. As the input voltage varies, so the current through the diodes will vary. And that will cause the voltage across the diodes to vary in sympathy. Not by much, but it is noticeable. For example, when testing my prototype, varying the input voltage over the range 1.8V to 2.8V resulted in a change in output voltage of 0.1V.
Normally, low-value emitter resistors are fitted to help transistors wired in parallel share the load current. In this design they're not strictly necessary as long as the two ZTX550 transistors are reasonably matched. Even if the transistors aren't matched, there's no great consequence. One will simply work harder than the other. If you do have several ZTX550 transistors available and can measure their gain, pick the two with the highest gain. By the way, a single ZTX789A or ZTX790A will easily replace a pair of ZTX550s in this application.
This design can be also be used to power 2V-filament valves from three NiCd or NiMH cells. It's much better than just wiring a fixed resistor in series with the valve filaments.
Performance isn't wonderful, but it's more than adequate. At a load current of 50mA, the minimum input-output differential is less than 100mV (single ZTX550). And it's still below 200mV with a load current of 250mA (two ZTX550s). Even when the NiCd/NiMH cells are down to 0.9V each, the regulator will still be well within its operating range.
To measure the power dissipated in the ZTX550 transistors, simply multiply the voltage across them by the load current. With fully-charged NiCd cells (1.4V per cell - 2.8V the pair) and a load current of 250mA, the dissipation is 1.5V (2.8V minus 1.3V) times 0.25A, or 0.375W. Which is well within the dissipation of even a single ZTX550.
With feedback circuits there is always the possibility of oscillation. And this circuit is no exception. Positive voltage regulators that use n.p.n. transistors as series-pass elements can be made unconditionally stable quite easily. But p.n.p. regulators are highly dependant on the regulator's output capacitor for stability. It is essential that the output capacitor's equivalent series resistance (e.s.r.), lies between certain limits.
You can make C1 a little larger in value or even substitute a similar aluminium electrolytic. But if you do, check for stability with a dummy load. Use either an oscilloscope or an a.c. millivolt-meter if you can. Actually, it's a good idea to check the circuit for stability regardless.
In the case of a linear 12V to 1.3V regulator, almost 90% of the input power would be wasted in the form of heat. Hardly good news for a battery-powered radio. That's why I'm suggesting a switched-mode regulator.
While there are fast alternatives to Schottky diodes, here they're false economy. The 1N5819 is efficient and inexpensive. Naturally, a Schottky diode with an equivalent or better specification can be used in place of the 1N5819.
A hint on laying-out the circuit of Figure 3:
If you sketch the circuit with the input on the right
and the output on the left, and with the i.c. connections
as viewed from above (as is normal),
then you'll automatically end up with the Veroboard layout
shown on the right.
(Look at Figure 3 in a mirror - you'll see what I mean.)
Fast switching speeds generate noise. The MC34063 can be quite bad in this respect as the output pulse-train can contain frequency components within audible range. To explain why, I need to mention the second switched-mode regulator i.c. shown in Figure 6; one of National Semiconductor's Simple Switcher regulators. The internal switch in this i.c. produces a regular sequence of pulses (at its output), at a fixed frequency of 52kHz. To achieve the desired output voltage, the i.c. adjusts the duty cycle of these pulses. This type of operation (basically) produces noise (mainly harmonics) at frequencies of 52kHz and above.
The MC34063 is a rather old design which, if you're really lucky, will produce the same regular sequence of output pulses as the National device. But far more often, the i.c. will produce a seemingly random sequence of pulses of both variable frequency and duty cycle. Yet averaged over time, the duty cycle will produce the desired output voltage. If the noise is at 52kHz (and above), we won't be able to hear it, even if the audio stages of the radio fail (unlikely) to filter it out. But when the noise is within audible range, as can happen very often with the MC34063, the set will amplify it.
The circuit can be used for other output voltages and load currents. Principally, the output voltage can be altered by changing R7. However, you ought to follow the design guide in the data sheet (see links below) if you intend to use the regulator to supply more than 6V, or if the load current is less than 100mA.
The measured performance (at 12V input, 1.35V output)
of my prototype was:
quiescent current = 2.8mA
efficiency (125mA load) = 58%
efficiency (250mA load) = 59%
The efficiency was substantially constant over the
design input voltage range of 10V to 15V.
The output voltage did not change more than a few millivolts
when either the input voltage was varied between 10V and 15V,
or when the output current was varied between 50mA and 250mA.
At just 58%, the efficiency is somewhat low for a switched-mode regulator. But it has to be remembered that the output voltage is very low. Both the voltage drop across the transistor switch inside the i.c. (almost 1V), and the voltage drop across the Schottky diode (about 0.5V) are significant fractions of the output voltage. And both will impact efficiency. Even the small voltage drop across inductor L1 will have a proportionately greater effect on efficiency at 1.3V than at an output voltage of, say, 5V.
Designing switched-mode regulators with National's Simple Switchers is very easy. There's even a computer program which will do the design work for you!
Actually, the circuit (Figure 6) is very similar to the one shown in Figure 3, but there's no timing capacitor (C3 in Figure 3). Neither is there a current limiting resistor (R6 in Figure 3). Both the switching frequency and the current limit are internally pre-set.
The efficiency of my prototype was 55% and remained substantially constant over an input voltage range of 10V to 15V. Regulation was excellent except at very low output currents.
If you want to use the LM2575 for anything other than powering a radio, then I suggest you read the device's data sheet (see links below). And if you need more than one amp, try the three-amp LM2576T-ADJ. As with most switched-mode regulators, the inductor and output capacitor are critical components. While a professional designer will (probably) use exactly what's required for a design, amateurs will (probably) use whatever they have to hand and hope it works. Following the design procedure in the data sheet will, at least, give you an idea of what components not to use.
Now the fun really starts....
Some h.t. inverters use capacitor multipliers, while others use inductors or transformers. As a d.c.-isolated output is needed, a transformer is the obvious choice. The next consideration is whether to use switched-mode techniques - which need special transformers - or stick to low frequencies and use an off-the-shelf mains transformer. Having to find a suitable core and winding a load of wire around it can be off-putting, so I opted for a low-frequency inverter using an off-the-shelf transformer.
I'm using power MOSFETs because they suit the HEF4013, they're inexpensive and very robust. You'll find that most power MOSFETs are specified at a gate-source voltage of +10V. There is no point in driving the gates of these MOSFETs with more than 10V, yet our allowable supply voltage is 10V to 15V. The potential dividers in the outputs of the HEF4013 reduce the drive voltage so that it never exceeds +10V. At a supply voltage of 10V, our working minimum, the gate voltage will be around 6.5V. At this gate voltage, a MOSFET will not pass as much current, and it's 'on' resistance will be higher than the specified minimum. Here the drain current is comparatively low (about 100mA) so the lack of gate voltage will not be a problem.
If you use the ICM7555/HCF4013/MOSFET combination in another application, you may have to change the values of R16-R19 to ensure sufficient gate voltage. Do consult all the relevant device data sheets.
While MOSFETs do not need any 'd.c.' drive, they have considerable gate-to-source and gate-to-drain capacitance. They also suffer from that bane of thermionic triodes: the Miller effect. That is, their gate-to-drain capacitance appears magnified by the device's voltage gain. Fortunately, capacitance is not a problem in this application, but it can be very important when driving MOSFETs at radio frequencies.
To help minimise switching noise, C13 should be located close to the MOSFETs' source connections and to the point at which the wire from the transformer's centre-tap connects with the positive supply. For comments concerning the snubbers and problems caused by the transformer's leakage inductance, see the note about leakage inductance here.
Although nominally 12V, the battery voltage may be as high as 14V, or as low as 10V. Portable valved sets were designed to operate effectively even when their h.t. battery was almost exhausted, so the low voltage condition isn't a concern. But a transformer producing 90V across C14 when the battery is at 12V, will produce 105V across C14 when the battery is fully-charged and up around 14V. Such a high voltage may damage the set. It's for this reason I added the output regulator. If the inverter's output voltage does not exceed 95V (across C14) with a fully-charged battery, then the output regulator isn't needed.
Zener diodes are somewhat noisy, hence the 220nF bypass capacitor (C16).
A metallised polyester film capacitor rated at 160V or more, is ideal.
Ideally, Tr6 needs to have a reasonably high gain,
so if you have a 'bad-un', simply make it into a darlington
by adding a MPSA42 as shown.
If you do use a darlington connection,
it's worthwhile increasing R20 to 47Kohm.
Transistor Tr6 doesn't need a heatsink, but do check the dissipation
if the regulator circuit is used in another application.
The output regulator part of Figure 8 can be very useful when powering battery-valved sets from the mains. It's possible to use a 24-0-24V, 6VA mains transformer followed by a voltage doubler to replace T1, BR1 and C14. To add further versatility, a string of eight 15V zener diodes can substituted for the three 30V diodes shown. By tapping the base of Tr6/C16 up and down the zeners, you will have a regulated supply from 15V to 120V in 15V steps. Just keep an eye on the current delivered and the dissipation in Tr6. Of course, there's nothing stopping you doing a similar thing when running off batteries. Just make sure the input voltage is high enough.
Remember, as the transformer receives a squarewave drive, there's no '1.414'-factor anywhere. The peak, average and r.m.s. values of a squarewave are all the same.
Ratio wise, we need 90V/12V = 7.5:1. And we only need about 1W of power so you'd think a 3VA transformer ought to be adequate. But things are not that simple.
Readily available mains transformers usually have either a single 240V primary, or two separate 120V primaries (which are wired in series for 240V mains, or in parallel for 120V mains). A primary voltage of 120V is reasonably close to what we want as our secondary (output) voltage, so the ideal (manufacturers) secondary we're looking for is: 120V/7.5 = 16V. All we have to do now is wire our 16-0-16V transformer into circuit, and away we go. Unfortunately, a 120V to 16-0-16V, 3VA transformer will probably produce no more than 70V across C14. So what's going on?
Like every other 'real' component, transformers are not perfect. Fortunately, only two of their imperfections really concern us: iron losses in the core and the resistance of the primary and secondary windings. There is also the matter of primary inductance; it's not in itself an imperfection, but it does have an impact on the efficiency of our inverter.
It's easier to describe the effect these imperfections have on a transformer by using an example as seen from a transformer manufacturer's point of view. Let's assume a mains transformer with a 240V primary and a 3VA, 12V secondary is required (the numbers make the maths easier, that's all). Through experience, the manufacturer will know what core size to use to give an acceptable primary inductance and core loss.
A high primary inductance minimises the current drawn from the mains when there's no load on the secondary. This is good; we would like our primary to have a high inductance too, for the same reason. But a high inductance means lots of turns of wire around the core, and that means lots of copper wire. To keep core losses as low as possible, a low-loss core material is needed, and the flux density in the core needs to be kept low too. That means lots of high quality iron. Again, we want to keep losses in the core to a minimum too. After all, we want the battery to power our radio, not keep the transformer warm.
For a manufacturer, lots of copper and lots of iron means lots of cost, so compromises have to be made. A little 3VA transformer will have a small core, probably run at a high flux density. Consequently, core losses will be proportionately high. But so what if 2W (this is power, not VA) is lost in the core. What's an extra 2W drawn from the mains supply. Well, it's a lot when you're running from batteries! (Two watts is just an example; it hasn't 'come' from anywhere.)
A small core needs a small bobbin; and very thin wire to get all the required turns on the bobbin. The primary resistance could easily be as high as 1600ohm, and the 12V secondary, 12ohm. Theoretically, the ratio needed is 240/12 = 20:1, but see what happens when we (fully) load the transformer with a 48ohm resistor.
Assuming the magnetising current in the primary is negligible (for the sake of this example), the open circuit output voltage will be 12V. But when we connect the 48ohm resistor the output voltage will fall to 9V. What's happened? Well, we've forgotten the resistance of the windings, that's what.
The primary resistance can be 'moved' to the secondary by dividing by the square of the ratio. In this case 1600ohm/(20 * 20) = 4ohm. Add the actual secondary resistance of 12ohm and we get a total (secondary plus reflected primary resistance) of 16ohm. So the total load is really 48ohm + 16ohm = 64ohm. Doing the maths, the current is actually 12V/64ohm = 0.1875A, and the voltage across the (external) 48ohm resistor is 0.1875A * 48ohm = 9V. That accounts for the missing 3V; it's lost in the windings of the transformer.
Any potential customer for this transformer would not be happy. The specification calls for 12V at 3VA (0.25A load) and that's what the transformer should deliver. The transformer manufacturer gets around the problem by winding the secondary with more turns to make up the lost voltage. Of course, the extra turns will have resistance themselves, so they need extra turns to compensate for them. And the extra, extra turns will need compensation too....
Once sufficient extra secondary turns have been added, probably by trial and error, the transformer will be able to deliver 12V at the rated load current. Unfortunately, these extra secondary turns will have changed the ratio, and the open-circuit secondary voltage will now be significantly higher than 12V.
This drop in secondary voltage between no load and full load happens in all transformers. The percentage by which the voltage changes is called the regulation. In the example above, the open-circuit secondary voltage will be at least 15V, giving a regulation of 25% (15V minus 12V, divided by 12V, times 100%). Note: in the USA, regulation is defined slightly differently, but the numbers turn out about the same. While a figure of 25% might seem high, it's not uncommon for very low power transformers. In general, the larger the transformer, the smaller the regulation (meaning less change in secondary voltage).
All this has thrown our calculations concerning the ratio we need into confusion. We can't calculate the ratio of a small mains transformer with any accuracy unless we know the regulation. What's even worse, the winding resistance now works against us. So we need to make allowances for the resistance too. It's like the regulation is working against us twice over.
I managed to get a 15VA toroidal transformer with a 230V primary and a 25-0-25V secondary. This proved ideal. With a 10V supply the output voltage was 72V, rising to 100V with a 14V supply. With ten NiMH cells as the supply, Tr6 and its associated components could probably be removed. Even more encouraging was the overall efficiency: 84% with a 10V supply, 83% with a 12V supply, and 80% with a 14V supply.
Whatever transformer you choose, make sure it has a low resistance secondary. That's the winding we'll be using as our primary and its resistance has a direct bearing on the efficiency of the inverter. Say it were 12ohm like in the example. If the circuit of Figure 8 takes 100mA at 12V, the voltage drop in our primary winding would be 1.2V (0.1A times 12ohm). So with a 12V supply, our primary will only 'see' 10.8V! That's 10% of the supply voltage gone, and with it, 10% of the power drawn from the battery. Even if there were no core or any other losses, the efficiency of the inverter could never exceed 90% simply due to the resistance of our primary.
We need to keep losses in the core to a minimum (low flux density) and we'd also like our primary to have a high inductance. Both requirements can be helped by using a transformer with a 240V primary, and with a much higher VA rating than we actually need. The higher VA rating will ensure a large core and a lower regulation figure (hence lower resistance windings and less voltage drop), and the extra turns on the transformer (because the primary is wound for 240V) mean a higher inductance.
I substituted a similar 15VA toroidal transformer with a 12-0-12V secondary, and twin 115V primaries wired in parallel. Efficiency was lower (due to increased core losses because of the higher flux density) but still respectable at 75% with a 12V supply. Although the efficiency as measured at C14, was maintained at over 70% with a 15V supply, the efficiency when measured at the output terminals was less than 50%. This can be explained by the loss across Tr6. With a 15V supply there was 115V across C14 and 25V across Tr6 (giving 90V output). The loss in Tr6 effectively reduces the efficiency.
It can often prove difficult to get information about a transformer sufficient to decide whether it will work effectively in an inverter such as this. Simply through trial and error, I've found two types of transformers - which need to be rated at 10VA or more - satisfactory. Firstly, those with 12-0-12V secondaries and twin 115V/120V primaries (wired in parallel). Secondly, and to be preferred, those with 24-0-24V or 25-0-25V secondaries which have 230V/240V primaries. Of course, a transformer with twin 115V/120V primaries is also suitable as the two windings can be wired in series.
My prototype inverter worked well, at least in terms of its electrical characteristics and efficiency. However, radiated noise, sounding rather like ignition interference, was a big problem. If only strong stations are of interest then only a little shielding will be necessary. But to receive weak stations, and those on Long Wave, complete shielding of all the electronics may be essential. Very much a case of try it and see.
One final point about the Veroboard layout: please ensure no high-voltage track comes within 2.5mm (0.1inch) of a low-voltage track.
Any rechargeable battery suitable for our purposes will have a low internal resistance, and a large current will flow if the battery is accidentally short circuited. This can pose a fire hazard. Always use a fuse.
I'll be very pleased to receive feedback from anyone who builds any of these modules. Particularly with regard to how they perform.
An informative note on low dropout regulators can be found at: http://www.national.com/appinfo/power/files/f10.pdf
Data on the ON-Semiconductor MC34063A can be found at: http://www.onsemi.com/pub/Collateral/MC34063A-D.PDF with applications information at: http://www.onsemi.com/pub/Collateral/AN920-D.PDF
Data on the National Semiconductor LM2575T-ADJ and LM2576T-ADJ can be found at: http://www.national.com/ds/LM/LM1575.pdf and http://www.national.com/ds/LM/LM2576.pdf respectively.
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