Linear Power Supplies – Part 1
A linear power supply is a power supply that does not contain any switching or digital components. It has some outstanding characteristics compared to switching power supplies, such as very low noise and ripple, immunity from mains-borne noise, simplicity, robustness, ease of design, and repair. They can also generate very high voltages (thousands of volts) and very low voltages (less than 1V). They can easily generate multiple output voltages. On the flip side, they are large in size and heavy, and need more heat-sinking. Linear power supplies have been around for several decades, long before the advent of semiconductors.1
Designing A Regulated Power Supply 2
Regulated power supplies are a mystery. Almost every IC construction project includes a regulated supply and most solid-state equipment built for 117 V also has a regulated supply or supplies to power the low-voltage solid-state devices. But the mystery is that while most hobbyists have a good idea how to use transistors and integrated circuits in simple applications, few have the remotest idea of how the regulated supply works and fewer still could design one from scratch if required to.
We will deal with the operation of regulated power supplies for low-voltage applications and gives all the necessary information needed to design a regulated supply from scratch. Information is given so that the designer may select the proper components such as the transformer, diodes, capacitor, regulator, pass transistors, and heat sinks. Sources for all parts are given so that the designer won’t be stuck for some hard to find parts. The regulated power supply consists of an unregulated DC power supply feeding a regulating a regulating circuit. The unregulated DC supply may consist of a full wave rectifier feeding a filter capacitor as shown in Fig. 2.


The regulating circuit may be a circuit made up of discrete components, or it may be a regulating IC, such as the NE550. Components and design options are chosen according to the voltage and current requirements of the project needing the regulated supply. Integrated circuit voltage regulators are commonly used, rather than discrete components, because of their low cost and ease of use. The basic design comes from the Signetics Digital Linear, Mos manual and is based on the Signetics NE550 regulator IC. This basic design is simple and permits numerous output voltages and limiting currents by merely selecting readily available resistor combinations.
The DC Power Supply
The DC Power supply used for most low-voltage power supplies is a capacitive load circuit as shown in figure 2. Inductive filters are occasionally used Instead of capacitors, but high-value, high current inductors are more difficult to locate and more expensive that low-voltage high-value capacitors. Either a full-wave (Fig. 2) or a half-wave circuit (Fig. 3), may be used to supply the DC; however, a full-wave circuit is preferred, because it provides better basic regulation.

In order to determine the voltage and current ratings of the components to be used in the unregulated DC power supply, it is first necessary to determine the voltage and current requirements of the equipment or device to be powered. When determining these power requirements it is best to allow reasonable safety factors in order to prevent overheating and to insure that the equipment will operate correctly. Normally a current safety factor of 10% is allowed in cases where peak current is being drawn 50% of the time, or less. In all other cases allow a safety factor of one-third. To determine the required current rating for the transformer, use the formula I=1.3 X Ip, where Ip is the anticipated peak current requirements of the equipment.
The designer may design the basic unregulated DC power supply so that the DC output voltage is anywhere from 30% to 98% of the peak AC voltage of the transformer. If a large value filter capacitor is used, the 98% value may be achieved and very little ripple will appear on the output of the basic supply. Unfortunately, very high-capacitance capacitors are expensive and in some cases may be hard to find. Smaller value capacitors are less expensive and easy to locate but will give lower DC outputs and will produce appreciable ripple on the output. For given DC output voltage (under load), the AC output of the transformer will have to be higher for small filter capacitors as compared to large value filter capacitors. Note that in general it is less expensive to use a transformer of higher voltage with a low-capacitance capacitor for a given DC output than it is to use a large capacitor and a lower voltage transformer to produce the same DC voltage. This is logical since the cost of a transformer does not increase appreciably as the voltage goes up, while the cost of a capacitor increases significantly as the value of capacitance goes up. In order to minimize the cost of the supply, one of the design factors is to keep the filter capacitor to moderate size and low cost.
Keep in mind that with no load on the output, the DC output from a simple capacitive filter supply will be virtually ripple free. When a load is placed across the supply, ripple will be evident. Furthermore, the amount of AC ripple on the DC output will increase as the size of the filter capacitor decreases. The AC ripple can be significant and not affect the regulation of the regulator.
Determining the DC output voltage for a given transformer voltage can be a difficult task if exact values are required. For practical purposes, however, only minimum values, not exact values are needed. For example, if our computations show us that we will get 18V DC output from a DC supply but we really get more than 18V, then this is of no consequence. We only want to assure ourselves that we will get at least the minimum required under load. With this in mind, the following formulas can be used to determine the AC (RMS) value of the transformer required:
EPEAK = 1.4 X ERMS
EOUT = .71 X EPEAK
and then
EOUT = 0.71 X 1.4 X ERMS
Where EOUT is the minimum DC output voltage from the unregulated DC power supply, and ERMS is the secondary voltage of the transformer. The anticipated DC output voltage under load from a simple unregulated supply, a shown in Fig 1, will be equal to or greater than the AC voltage from the secondary of the transformer. This will only be true if the current ratings of the transformer are not exceeded. The above formula takes into consideration that a moderate size capacitor will be used and is based on the assumption that ripple on the DC output voltage can be 10% or less. The NE550 regulator IC, which is used for this design, can tolerate 10% ripple provided that the lowest input voltage (low part of the ripple) is at least 3V higher than the desired DC regulated voltage. We will have to consider the DC input to be the bottom of the ripple as shown in Fig. 4.

Note that the peak voltage cannot be higher than the maximum ratings of the NE550. As defined, EDC input = EREG +3, where EREG is the desired regulated voltage.
The DC input voltage is also 95% of EOUT (DC input is 5% lower than EOUT because of the ripple): thus EDC input = .95 EOUT = .95 ERMS. Solving the two equations gives ERMS = (EREG + 3)/0.95. We now have a very simple formula to use to determine the secondary (ERMS) value of the transformer given only what we want for a regulated voltage and assuming that we will not exceed the manufacturer’s current rating for the transformer chosen.
These formulas for I and ERMS will hold true for virtually all low-voltage, high-current supplies provided that good quality properly designed transformers are selected. The transformers recommended mentioned later fall into that class. If low grade transformers with high internal resistance are used, the ERMS may approach the value of (EREG + 3)/0.5. As an example, assume we want a power supply to deliver 5A at 12V regulated. The minimum ratings of the transformer would be determined as follows:
I = 1.3 X IP = 1.3 X 5 = 6.5A ERMS = (12 + 3) ÷ 0.95 = 16V.
Some Suggested Electronic & Hardware Sources of Supply
All Electronics 800 826 5432 https://www.allelectronics.com/category/285/integrated-circuits/1.html
DigiKey 800 344 4539 https://www.digikey.com/
Mouser Electronics 800 346 6873 https://www.mouser.com/
Jameco Electronics 800 831 4242 https://www.jameco.com/
Electronic Surplus 440 205 8388 https://www.electronicsurplus.com/
Circuit Specialists 480 464 2485 https://www.circuitspecialists.com/electronic-parts-and-general-supplies
NTE Electronics 800 631 1250 https://www.nteinc.com/
Surplus Electronic Sales Note: They do not take orders over the telephone https://www.surplus-electronics-sales.com/index.php?main_page=about_us
Skycraft 321 272 8357 https://skycraftsurplus.com/contact-us/
Note: The designer/builder may have some difficulty in locating the NE550 dip package. They are, however, easily found as a surface mount, flat pack. I have found notes, on the Internet, that indicate that the LM723 may be a suitable substitute. Or at least, close enough to work, with some minor circuit modifications |
Looking through the various catalogs you probably won’t find a transformer that has a secondary exactly matching our requirements, but you would find one that exceeds the requirements. It is a good idea to check the peak DC output voltage obtainable under any circumstances to see that this voltage does not exceed the voltage ratings of the NE550 regulator. The maximum is given by EMa = 1.4 X ERMS. Thus in our case EMAX = 1.4 X 18.9 = 26.5V. The maximum voltage rating of the NE550 is 40V. We are within the limits in this case. In a case where EREG is 37V, the maximum allowable for the NE550 or any case where EMAX exceeds 40V, then an overvoltage protection circuit (Fig 5), seen below, will be required to be used to provide the DC input voltage to the regulator.

In our example, diodes in a bridge would have a minimum PIV rating of 75.6V, while diodes in the half-wave configurations should have a rating of 113 volts. These are oddball values, so would use diodes of the next higher rating. A bridge with a 100 PIV rating could be used, or two diodes in a center tapped configuration with a rating of 150 V would do.. Note that the current rating for a complete diode bridge (as compared to individual diodes in a bridge) is NOT divided in half. In this example 6.5A is the requirement, so a 10A bridge would be required. Diode bridges are preferred since they are usually epoxy encapsulated and may be mounted directly to a heat sink without having to worry about mica insulators and special means to provide insulation.Selection of diodes can be made in a fashion similar to the transformer. Diodes in a full-wave configuration pass only one half the total current, therefore, ID = 0.51, where ID represents the current requirements of the diodes. In our example the maximum current is 6.5A, so the diodes would need to handle 3.25A each. Since this is an oddball value, the next higher current rating would be used such as diodes with a 6A rating. To be conservative for low-voltage supplies, the voltage ratings of the diodes should be greater that the maximum peak voltage that can be encountered. For a bridge rectifier configuration, each diode should have a PIV (peak inverse voltage) rating of four times the ERMS value of the transformer secondary, where for a center-tapped rectifier configuration the PIV should be six times the ERMS value of the transformer secondary.
The Filter Capacitor
The filter capacitor smooths the pulsating DC and gives steady state DC with some percentage of ripple on top. One of the design criteria is that we can tolerate 10% ripple. If the wrong capacitor is chosen, the ripple may exceed 10% (if the capacitor is too small) and the output voltage may be too low, causing the regulator to regulate poorly for heavy loads. If the capacitor is too large the ripple will be smaller and the output voltage from the unregulated supply will be higher, but this is of little consequence. We need to determine the minimum size of the capacitor. Note that excessively large filter capacitors cause enormously large surge currents through diodes during turn on. Most silicone diodes can handle large surges for a few seconds, so this shouldn’t be too much of a problem. If the designer sticks close (50% to 100% larger) to the value of capacitance determined during calculations, problems should not be encountered with popping diodes.
To determine the proper capacitor, it is necessary to first determine the load resistance. This load resistance is determined by the formula RL = EOUT ÷ I, where EOUT is the output voltage we need from the unregulated DC supply in order to supply EREG. I is the maximum current to be drawn. Note that load includes power dissipated in the regulating circuitry. In our example EOUT = ERMS = 18.9v (theoretical ERMS was 16V, but we used three transformers (the secondaries wired in parallel), to give us 18.9V) and I was 6.5A. Thus RL = 18.9V ÷ 6.5 = 2.9?. The basic formula to be used for the value of the capacitor is 2πfRLC = 5, where π = 3.14 and f = the line frequency. C is the desired capacitance in farads. Solving the equation for C we get C = 5 ÷ (6.28 f RL). In our example the line frequency is 60Hz and RL is 2.9?. Therefore C = 5 ÷ (6.28 X 60 X 2.9) = 0.0046 or 4600 μF. Since 4,600 μF is not a stock value, the next highest value would be used. The voltage rating of the capacitor should be at least double the ERMS voltage, which in our case is 38V.
You may question the 10% tolerable ripple figure previously given. For tube type power supplies no ripple was tolerable. While this 105 figure may seem like a lot, remember that the only thing that the 550 regulator wants to see is the lowest DC input voltage is a least 3V above the regulated output voltage. The amount of ripple on top of this minimum DC input voltage is insignificant as long as maximum ratings are not exceeded.
The Regulator

The NE550 regulator I an operational amplifier with an internal reference voltage and current limiting capabilities. The operational amplifier compares and internal standard reference voltage (internal zener) fed into the noninverting input with a sample of the desired regulated voltage. The difference between the standard voltage and a sample of the regulated output is amplified and inverted, producing a control voltage. This control voltage controls a pass transistor which is in series with the regulated output. As the regulated output drops, the control voltage increases, which in turn causes the regulated output to increase. A stable point is eventually reached where the output voltage remains constant. The stable point depends on the ratio of two resistors (RA and RB in Fig 1-7) connected as a voltage divider to deliver a sample of the regulated output to the inverting input of the operational amplifier. By changing the ratio of the values of the two resistors, the output of the voltage divider changes, which in turn produces a change in the regulated output. The value of the regulated output, can be simply changes by altering the ratio of the values of the two resistors. The following tables give the various values of RA and RB for selected values of regulated voltage.
1% Resistors 5% Resistors Trimming
Ereg RA k Ohms RB k Ohms RA k Ohms RB k Ohms Resistor
5 | 6.13 | 2.97 | 5.6 | 5.6 | 1k |
10 | 12.3 | 2.39 | 11.0 | 11.0 | 1k |
12 | 14.7 | 2.31 | 13.0 | 13.0 | 2k |
13 | 16.0 | 2.29 | 15.0 | 15.0 | 2k |
15 | 18.4 | 2.24 | 16.0 | 16.0 | 5k |
20 | 24.5 | 2.18 | 22.0 | 22.0 | 5k |
30 | 36.8 | 2.11 | 23.0 | 33.0 | 5k |
Fig. 1-8 Listings of values of resistors RA and RB at 1% and 5%
Current limiting is provided by the circuit (Fig. 6) shown below:
The basic idea is that as current through RE1 increases, the resulting increased voltage drop across RE2 will increase current in RB and increase VB , thereby limiting IE1 . This simple 5-component circuit is not trivial to analyze because of the tight interaction of the two BJTs.

Analysis is simplified by the observation that the emitter current of Q 1 consists of two components: the currents of RE1 and IB2 . The base current of Q 1 has both components, and IB1 from IB2 is IB2 /(β1 + 1) or IC2 /β2 x (β1 + 1). This current and IC2 flow through RB to contribute to VB and they are proportional, through the β values. Yet the IB1 component is very small compared with IC2 , by a ratio of 1/β2 x (β1 + 1). The total current in RB caused by Q 2 is

For a typical β value of 150, the β factor is 44.15 x 10 – 6 or 44 ppm. Analysis is simplified by making the approximation that this current is negligible, and omitting it. This is equivalent to having β2 → ∞. Then IB2 = 0 A. Once we know about this 1/β2 x (β1 + 1) factor, it is not hard to put it back into the subsequent equations to make them exact, as we shall see. Again, the same design goals hold as for the previous circuits.
Heat Sinking
Semiconductors dissipate heat and in some cases get VERY HOT to the touch. But just how hot can semiconductors get without being destroyed? The data sheets for both diodes and transistors give power derating factors for reducing the allowable dissipation or allowable current for a given temperature rise. These factors must be used to prevent overheating and destruction of the device. Note that a power derating factor is a multiplier, based on temperature, which serves to lower the power rating as the temperature of the device increases. Fig. 1-13 gives derating factors for selected transistors.
Max Minimum Max Thermal Transistor Motorola
Wattage Current DC current Voltage Derating Number HEP
Gain (hFE) Factor Equivalent
20 | 2A dc | 25 | 60 | .133 W/° C | MJ2249 | HEP241 |
40 | 2 | 25 | 125 | .266 | 2N5050 | HEP241 |
87.5 | 4 | 30 | 40 | .5 | MJ480 | HEP247 |
85 | 4 | 750 | 60 | .343 | MJ4200 | - |
90 | 10 | 20 | 60 | .718 | MJE3055 | S5001 |
115 | 15 | 20 | 60 | .657 | 2N3055 | HEP704 |
150 | 20 | 500 | 40 | 1.2 | 2N6355 | - |
Fig. 1-13 Listing for transistor selection
Let’s assume that in this power supply we used 100V, 6A diodes, such as the Motorola 1N3880. To determine the power dissipation of the diodes use the empirically derived formula PDIODE = 1 X 1.5, thus for our example PDIODE = 3.25 X 1.5 = 5 W (approximately). Furthermore from Fig. 1-14 we find that the maximum usable temperature without derating is 100° C.
Maximum Current Maximum Voltage Max.Temperaure Diode Number
in Amps Without Derating
1 | 200 | - | HEP 156 | Heat Sink Not Needed |
2.5 | 1000 | - | HEP 170 | Heat Sink Not Needed |
6 | 100 | 100° C | 1N3880 | HEP 153 |
6 | 200 | 100 ° C | 1N3881 | HEP 153 |
Fig. 1-14 Listing for diode selection. Temperatures shown are in degrees Celsius
At this point we must determine the cooling capacity of the heat sink needed. Heat sinks are rated in degrees Celsius per watt. For every watt of dissipation, the temperature of the heat sink will rise so many degrees Celsius above ambient room temperature. If a heat sink is rated as 10° C ÷ W, then the temperature of the heat sink will rise 10° for every watt dissipated. The formula for determining the cooling capacity of the heat sink needed is:
Cooling Capacity (° C ÷ W) = max ambient temp/watts dissipation
From figure 1-14, the maximum temperature without derating is 100° C. While our actual dissipation will be 5W, a 25% safety factor is applied to 5 watts and gives 6.25W or 7 watts, rounded up. .Using the formula above, Cooling capacity + (100 -23) ÷ 7=11° C÷W. (Fig. 1-15) gives various heat sinks selected according to various cooling capacities. The Thermalloy # 6111B might be used. This heat sink will provide a temperature reduction of 10° C ÷ W. When selecting a heat sink keep in mind that the specifications of the heat sink must give less heat rise than the device – the degree rise per watt will be smaller or equal to the permissible heat rise of the device. Note when mounting the diodes to the heat sinks that silicone conducting grease should be used on the diodes as well as the mica insulating washers as shown in Fig. 1-16.


As previously determined the transistor chosen will dissipate 61W under the conditions defined. In order to determine the amount of heat sinking required it is necessary to determine the maximum permissible device temperature rise for a dissipation of 61W, the maximum dissipation that we previously calculated. If you have access to charts, this temperature rise is easy to find. In the absence of charts, the derating factor given in Fig. 1-13 is used. The derating factor is a factor that tells us the reduction in wattage which must be applied against the example, for the 2N6355 the derating factor is 1.2 W ÷ °C.
Thus for every degree Celsius above room temperature, the wattage must be reduced 1.2W. As the temperature rises, the power dissipation decreases. This factor can be used to determine the maximum permissible device temperature rise. This temperature rise is determined by the formula: Temp rise = maximum wattage – required wattage ÷ derating factor. In our example, the maximum wattage is 150W, the required wattage dissipation is 61W, and the derating factor is 1.2 W ÷ °C, and:
Temp rise = 150 – 61 ÷ ½ = 74°C
In this case, as long as the temperature of the device does not increase by more than 74°C, we will be operating within the safe bounds. Note that we are talking about a temperature rise in this case, so in order to use the previous formula we would have to add the ambient room temperature to get the maximum room temperature, but then we turn right around and subtract the room temperature. Thus, the previous formula can be modified for cases where we are talking about temperature rise as follows:
Cooling Capacity = temperature rise ÷ dissipation (watts)
In our example we have: Cooling capacity = 74 ÷ 61 = 1.2° C ÷ W. The heat sink required must have a thermal resistance of less than 1.2° C ÷ W. A Thermalloy 6421B with a thermal resistance of 1° C ÷ W would be a good selection in this application. It is important to use a liberal coating of heat conducting silicone grease on the transistor and the associated insulating mica washers.
Summary
It is difficult to go into ever detail required to produce a correct and exact power design to fit a set of requirements, so the approach that has been taken here is to include substantial safety factors so that the analysis and arithmetic could be simplified. It is recognized that lower rating components could be used, but then exact calculations and extensive analysis would be required. If the experimenter follows the design steps given here, the result will be a reliable, moderately priced power supply which most experimenters can easily build.
ESTIMATING POWER TRANSFORMER RATINGS
Junk box transformers sometimes have a way losing their identity with time. It isn’t too difficult to determine the voltages of the various windings, but current ratings are another matter. The best approach is to first get an idea of the overall power rating of the transformer. This is determined, primarily, by the amount of copper and iron in the transformer, so as an approximation of the weight of the transformer will tell us, roughly, how many watts a transformer will handle.
With this idea in mind, several transformers of known ratings were weighed and their watts determined by dividing the wattage rating by the weight. Here are some typical results:
Transformer Type | Watts | Wt/lb | Watts/lb |
TV power | 300 | 13.5 | 22.2 |
Old Radio power | 132 | 7.5 | 17.6 |
Battery Charger | 600 | 24.0 | 25.0 |
Small radio power | 40 | 4.25 | 9.4 |
Instrument power | 20 | 1.6 | 12.5 |
Generally speaking, the largest transformer, the more watts per pound you get. This is to be expected since a larger transformer is more efficient that a smaller one. The table may be used as a rough guide in determining the wattage of your transformer. Weigh the transformer and multiply the weight by the estimated watts per pound for that weight. Next estimate the current requirements for your application to see if the total number of watts is within the transformer ratings.
A word of caution when checking transformer windings on the AC line. Always use a test lamp in series with the winding until you are sure the winding connected to the AC line is indeed the primary. An unloaded transformer will show little, or no primary current (lamp very dim) with the primary connected to the line
SEVEN WAYS TO PROTECT AGAINST SMOKE
Monolithic IC regulators such as the LM309, and others are here to stay, as are the more sophisticated dual-tracking regulators like Raytheon’s RM4149, they include a number of fail-safe features – most notably, short-proof circuitry and thermal overload shutdown, but in many cases, that’s not enough protection. Presented here are some other possible failure modes, and how to avoid them
Reverse Bias Across the Regulator
This accounts for a number of otherwise unexplainable regulator failures. Too see why this problem occurs, examine the typical three-terminal regulator supply in Fig. 1-17. If the input capacitor should go rapidly to ground (through a short, for example) the output becomes more positive that the input, setting up a reverse bias across the regulator’s series-pass transistor, which can destroy it. Adding a diode in series with the DC input, as shown in Fig. 1-18, can eliminate the problem; but the diode adds a series resistance and consequent series resistance and a voltage drop. Fig 1-19 normally biased off because VIN is greater than VOUT; but should VOUT become more positive, the diode conducts, dumping the current back to the input without going through the regulator, itself.


Improper Polarity Transients at the Output
If a large negative transient hits the output of a negative regulator, or if a positive transient hits the output of a negative regulator, all kinds of troubles can occur. Unfortunately, transients riding the power supply line be a fairly common occurrence; once again a diode solves the problem. Fig 1-20 shows a simplified diagram of a typical dual-tracking regulator. By connecting two diodes as shows, opposite polarity transients can do no damage. Any positive transients on the negative line shunt to ground through diode; D2 performs a similar function for negative transients.D1
Excessive Input Voltage to the Regulator
The popular LM309 and several similar regulators are rated at a maximum 35V input. Anything over 35V can easily zap the regulator. Even if you are running around, say, 33V, a good voltage spike or upward change in the line voltage can cause the 35V figure to be exceeded. The best way to deal with this is to use a 35V zener across the input of the regulator, catching any possible overvoltage problems. See Fig. 1-21
Sources Cited
Circuit Basics 1
https://www.circuitbasics.com/linear-power-supplies/
TAB Books © 1979 The Power Supply Handbook by the Editors of 73 Magazine2 Unless otherwise noted, all the figures,& images in this document, were obtained from this book.
Electronic Tutorials https://www.electronics-tutorials.ws/diode/diode_6.html3 Fig. 2
Circuits Today . https://www.circuitstoday.com/half-wave-rectifiers. Fig. 3
Wavelength Electronics https://www.teamwavelength.com/power-supply-basics/
Fig. 4
Circuit Finder – (Extreme Circuits) https://www.learningelectronics.net/circuits/protection-for-voltage-regulators.html. Fig. 5
Planet Analog
https://www.planetanalog.com/seemingly-simple-circuits-user-proof-external-supplies-circuit-3-the-two-bjt-current-limited-supply/ Fig 6
| KI4VEO | 2022-05-04 | |
|---|---|---|
| Re: Linear Power Supplies – Part 1 | ||
| If the polarity is incorrect, that is because I didn't double check it. The schematic was taken directly, from one of my references, in my listings of works cited. Thanks for pointing that out, Reply to a comment by : NA3CW on 2022-04-27 Thanks for such an informative article. One nit, though, on your first schematic: the filter caps on the negative supply are connected backwards which, of course, can be exciting to watch. :-) 73! | ||
| NA3CW | 2022-04-27 | |
|---|---|---|
| Linear Power Supplies – Part 1 | ||
| Thanks for such an informative article. One nit, though, on your first schematic: the filter caps on the negative supply are connected backwards which, of course, can be exciting to watch. :-) 73! | ||
| G4AON | 2022-04-23 | |
|---|---|---|
| Linear Power Supplies – Part 1 | ||
| The critical item to include with any power supply, is over Voltage protection. The older solution was to use an SCR and Zener diode in a crowbar circuit, which blows a fuse if the output rises above a preset limit. More recently Mosfet switches driven by a control I/C are often used to turn off the output, rather than shorting the output to blow a fuse. There was a project to build one of these in a recent RSGB RadCom magazine (see: https://gm4wzg.co.uk/wp/home/projects/bob/) I saved a transceiver from damage with a crowbar circuit added to a bench supply that had a pass transistor fail. I also had a small audio amp blow when it’s wall wart supply went over Voltage, fortunately it was supplied with the amp and both were replaced under warranty. | ||