rozważania o CNC

There are many circuits for low voltage regulators. For higher voltages, such as supplies for valve circuits, the situation is different. That?s why we decided to design this simple regulator that can cope with these voltages. This circuit is obviously well suited for use in combination with the quad power supply for the hybrid amp, published elsewhere in this issue. The actual regulator consists of just three transistors. A fourth has been added for the current limiting function.

The circuit is a positive series regulator, using a pnp transistor (T2) to keep the voltage drop as low as possible. The operation of the circuit is very straightforward. When the output voltage drops, T4 pulls the emitter of T3 lower. This drives T2 harder, which causes the output voltage to rise again. R4 restricts the base current of T2. C1 and C2 have been added to improve the stability of the circuit.

These are connected in series so that the voltage across each capacitor at switch-on or during a short circuit doesn?t become too large. You should use capacitors rated for at least 100 V for C1-C3. D1 protects T2 against negative voltages that may occur when the input is short-circuited or when large capacitors are connected to the output. We use two zener diodes of 39 V connected in series for the reference voltage, giving 78 V to the base of T3.

Because R6 is equal to R7 the output voltage will be twice as large, which is about 155 V. T4 acts as a buffer for potential divider R6/R7, which means we can use higher values for these resistors and that the voltage is not affected by the base current of T2 (this current is about the same as the emitter current of T3). This is obviously not a temperature compensated circuit, but for this purpose it is good enough.

Circuit diagram:
(http://lh6.ggpht.com/_FdGFE8NBDgc/S_u8IB9iD_I/AAAAAAAAC6M/Kh5PTa0YaGU/high-voltage-regulator-with-short-circuit-protection-circuit-diagram.GIF)

The current limiting section built around T1 couldn?t be simpler. When the output current rises above 30 mA the voltage across R1 causes T1 to conduct. T1 then limits the base-emitter voltage of T2. R2 is required to protect T1 against extremely fast peak voltages across R1. R3 is needed to start the regulator. Without R3 there wouldn?t be a voltage at the output and hence there wouldn?t be a base current in T2. R3 lets T2 conduct a little bit, which is sufficient for the regulator to reach its intended state.

During normal operation with a voltage drop of 15 V across T2 and a current of about 30 mA there is no need for extra cooling of T2. The junction temperature is then 70 °C, which means you can burn your fingers if you?re not careful! The lower the input voltage is, the more current can be supplied by this regulator. This current is determined by the SOAR (Safe Operation ARea) of T2. During a short circuit and an input voltage of 140 V the current is about 30 mA and T2 certainly requires a heatsink of at least 10 K/W in those conditions.

To increase the output voltage you should use a larger value for R6. If you want to use a higher reference voltage, you should replace T4 with a MJE350. If you only ever need to draw a few milli-amps there is no need to include T4 and R4. The potential divider (R6/R7) can then be connected directly to the emitter of T3. The ripple suppression of the circuit is about 50 dB. The quiescent current is 2.5 mA and for small currents the dropout voltage is only 1.5 V.
Author: Ton Giesberts

Valves are enjoying increasing popularity in audio systems. With the European ?E? series of valves, such as the ECC83 (12AX7) and EL84 (6BQ5), the filament voltage is 6.3 V. Depending on how the circuit is wired, the ECC 81?83 series of twin triodes can also be used with a filament voltage of 12.6 V. In earlier times, the filament voltage was usually taken directly from a separate transformer winding, which (in part) was responsible for the well known ?valve hum?. With regard to the signal path, current valve circuits have hardly experienced any fundamental changes. In high-quality valve equipment, though, it is relatively common to find a stabilised anode supply.

Mains hum can have a measurable and audible effect on input stages whose filaments are heated by an ac voltage. The remedy described here is a stabilised and precisely regulated dc filament voltage. The slow rise of the filament voltage after switching on is also beneficial. The exact setting of the voltage level and the soft start have a positive effect on the useful life of the valves. Diagram shows a voltage regulator meeting these requirements that is built from discrete components. The two sets of component values are for a voltage of 6.3 V (upper) and 12.6 V (lower).

Circuit diagram:
(http://1.bp.blogspot.com/_FdGFE8NBDgc/TCH3ZXDE-1I/AAAAAAAADdc/-_GxeJ2MTq8/s1600/stable-filament-supply-circuit-diagram.GIF)

Thanks to the fact that the supply works with a constant load, it can do without special protective circuits and the additional complexity of optimum regulation characteristics for dynamic loads. The circuit in Figure 1 consists of a power MOSFET configured as a series-pass regulator and a conventional control amplifier. Zener diode (D5) sets the reference potential. A constant voltage is thus present at the emitter of the BC547 control amplifier (T3). The current through D5 is set to approximately 4?5 mA by series resistor R5. The output voltage UO (the controlled variable) acts on the base of the control amplifier (T3) via voltage divider R6/R7. If the output voltage drops, the collector current of T3 also decreases, and with it the voltage drop across load resistors R1 and R2.

The voltage on the gate of the MOSFET thus increases. This closes the control loop. The values of the resistors forming the voltage divider are chosen for the usual tolerances of Zener diodes, but they must be adjusted if the diode is out of spec (which can happen). The load resistance of the control amplifier is divided between R1 and R2. The current through the load resistance and the collector current of T3 are practically the same, since the MOSFET draws almost no gate current. Filter capacitor C2 is connected to the junction of R1 and R2 to reduce residual hum. Electrolytic capacitor C4 and power supply filter capacitor C1 serve the same purpose. The hum voltage also depends on the magnitude of the load current.

(http://2.bp.blogspot.com/_FdGFE8NBDgc/TCH3Y7vvTII/AAAAAAAADdU/JJ6sTN7XUk4/s1600/stable-filament-power-supply-circuit-diagram.GIF)

The voltage drop over the series-pass regulator is nearly the same for an output voltage of 6.3 V or 12.6 V. With a BUZ11 and a load of 1 A at 6.3 V, for instance, the average voltage across the source?drain channel is approximately 7V. The power dissipation of 7 W requires a corresponding heat sink. The slow rise of the output voltage is due to the presence of timing network R3/C3 and T1. When power is switched on, T1 holds the gate of the MOSFET at nearly ground level. As C3 charges, T1 conducts increasingly less current, so ultimately only the control transistor affects the gate voltage.

The mains transformer must be selected according to the required load current. The required value of the input voltage can be read from the chart. The transformer should have a power rating at least 30 % greater than what is necessary based on the calculated load dissipation. Where possible, preference should be given to a filament voltage of 12.6 V. When twin triodes in the ECC81?83 series are used, for example, the power dissipation in the series pass transistor is lower with a voltage of 12.6 V.
Author: Dr Alexander Voigt

This circuit was designed to allow a laptop computer to be powered from a solar power setup. The computer requires 12V at 3.3A. The circuit is a linear regulator with Mosfet Q4 as the series pass device. A 100kO resistor provides Q4 with a positive gate-source voltage. Any tendency for the output voltage to exceed ZD1's voltage causes Q2 to turn on. This turns on Q3 which reduces Q4's gate voltage and thus reduces the output voltage. Note that Q2's base-emitter voltage stabilizes at about 0.35V. This combined with the zener voltage gives an output of 12.4V. If a more precise output is required, first select ZD1 so that its voltage rating is at least 0.4V less than the required output voltage.

Circuit diagram:

(http://2.bp.blogspot.com/_FdGFE8NBDgc/TBOXJ2D1JwI/AAAAAAAADEo/1PIh_hDE0ls/s1600/high-current-low-dropout+-regulator.jpg)

You can then "trim" to the required output voltage by installing a resistor in series with ZD1. Q2's base-emitter voltage and the 680W base resistor set the current through ZD1 to 0.5mA. This means that the output voltage will be boosted by 0.1V for each 200O of resistance in series with ZD1. Zener diode ZD2 ensures that Q4's maximum rated gate-source voltage is not exceeded. Mosfet Q1 provides reverse polarity protection. Note that Q4 requires a heatsink since it will dissipate about 10W under worst-case conditions. No heatsink is required for Q1. At 3.3A, the regulator reduces the output voltage by just 0.2V. This can be further reduced by paralleling Q1 & Q4 with additional Mosfets.
Author: Andrew Partridge

The losses in a bridge rectifier can easily become significant when low voltages are being rectified. The voltage drop across the bridge is a good 1.5 V, which is a hefty 25% with an input voltage of 6V. The loss can be reduced by around 50% by using Schottky diodes, but it would naturally be even nicer to reduce it to practically zero. That?s possible with a synchronous rectifier. What that means is using an active switching system instead of a ?passive? bridge rectifier.

The principle is simple: whenever the instantaneous value of the input AC voltage is greater than the rectified output voltage, a MOSFET is switched on to allow current to flow from the input to the output. As we want to have a full-wave rectifier, we need four FETs instead of four diodes, just as in a bridge rectifier. R1?R4 form a voltage divider for the rectified voltage, and R5?R8 do the same for the AC input voltage. As soon as the input voltage is a bit higher than the rectified voltage, IC1d switches on MOSFET T3.

Just as in a normal bridge rectifier, the MOSFET diagonally opposite T3 must also be switched on at the same time. That?s taken care of by IC1b. The polarity of the AC voltage is reversed during the next half-wave, so IC1c and IC1a switch on T4 and T1, respectively. As you can see, the voltage dividers are not fully symmetrical. The input voltage is reduced slightly to cause a slight delay in switching on the FETs. That is better than switching them on too soon, which would increase the losses.

Circuit diagram:

(http://lh6.ggpht.com/_FdGFE8NBDgc/S_uLOGdGvLI/AAAAAAAAC5I/CwOll-9L4sI/s640/power-mosfet-bridge-rectifier-circuit-diagram.GIF)

Be sure to use 1% resistors for the dividers, or (if you can get them) even 0.1% resistors. The control circuit around the TL084 is powered from the rectified voltage, so an auxiliary supply is not necessary. Naturally, that raises the question of how that can work. At the beginning, there won?t be any voltage, so the rectifier won?t work and there never will be any voltage... Fortunately, we have a bit of luck here. Due to their internal structures, all FETs have internal diodes, which are shown in dashed outline here for clarity.

They allow the circuit to start up (with losses). There?s not much that has to be said about the choice of FETs ? it?s not critical. You can use whatever you can put your hands on, but bear in mind that the loss depends on the internal resistance. Nowadays, a value of 20 to 50 mW is quite common. Such FETs can handle currents on the order of 50 A. That sounds like a lot, but an average current of 5 A can easily result in peak currents of 50 A in the FETs.

The IRFZ48N (55 V @ 64 A, 16 mW) specified by the author is no longer made, but you might still be able to buy it, or you can use a different type. For instance, the IRF4905 can handle 55 V @ 74 A and has an internal resistance of 20 mR. At voltages above 6 V, it is recommended to increase the value of the 8.2-kR resistors, for example to 15 kR for 9V or 22 kR for 12 V.
Author: Wolfgang Schubert

When running 12V electronic devices from lead-acid battery banks, the voltage to the appliance can vary from below 11V with discharged batteries, to well above 14V during charging. Many appliances will not tolerate such a wide fluctuation and may perform poorly or be damaged.

This step-up inverter, combined with a 12V fixed regulator, is a good solution. Q1 & Q2, together with the ferrite pot-core transformer, comprise a DC-to-AC inverter. The turns ratio steps down the input voltage by about 30%. The square wave output voltage is rectified and added to the input DC voltage. The stepped up DC is then fed to a 7812 12V regulator (REG1).

Circuit diagram:

(http://lh6.ggpht.com/_FdGFE8NBDgc/S53VNMzgmhI/AAAAAAAACMY/HePZjYSxEsk/s640/12V-regulated-inverter-supply-circuit-diagram.jpg)

The specified regulator will supply 1.5A at 12V out, from any input into the inverter between 9V and 15V, with the inverter making up the shortfall. Current requirements are kept to a minimum by not having the inverter supplying the total current.

By substituting a higher rated linear regulator, up to 5A can easily be supplied by this simple circuit. The transistors can be almost any general-purpose power type while the twin diode (D4/D5) is a high-speed device commonly found in defunct computer power supplies. Normal rectifier diodes can be used with a slight decrease in efficiency. The same comment applies to D2/D3. D6 is a protection diode and any 3A type will be suitable.

By slightly modifying the turns ratio, and substituting the linear regulator, 24V devices can be operated from a 12V supply. Laptops requiring around 18V can be powered as well.

(http://lh4.ggpht.com/_FdGFE8NBDgc/S53VNNa-FfI/AAAAAAAACMc/fChT-chl-WI/s640/12V-regulated-inverter-supply-diagram.jpg)

This diagram shows how to wind the transformer. L1 & L2 are six turns bifilar wound using 1mm-diameter enameled copper wire, while L3 & L4 are four turns bifilar wound.
Source: Silicon Chip 01 May 2009

A series diode is often used as a means of protecting equipment from accidental power supply reversal, particularly in battery-powered equipment. Due to forward voltage losses, this is sometimes impractical. One solution is to use an enhancement mode P-channel power Mosfet (Q1) in series with the positive supply rail. A device with low drain-source "on" resistance can be selected to minimise voltage losses, which in turn extends battery life and reduces heat dissipation.

Circuit diagram:

(http://1.bp.blogspot.com/_FdGFE8NBDgc/TBttMcJqiAI/AAAAAAAADSg/-sqbYAq7toU/s1600/reverse-polarity-protector-circuit-diagram.jpg)

Zener diode ZD1 must be included to protect against excessive gate-source voltage, while a 100k? resistor limits zener fault current. A second 100k? resistor across the output ensures that the gate doesn?t float when the input is disconnected. A series fuse and bidirectional transient voltage suppressor (TVS1) could be included to provide over-voltage protection, if desired. If common input & output grounds are unimportant, then a version of this circuit employing an N-channel power Mosfet in series with the negative (0V) rail could also be employed.
Author: Bruce Griffiths

This circuit shows that dimmers intended for use at mains voltage do not always have to contain a triac. Here, a MOSFET (BUZ41A, 500 V/4.5A) in a diode bridge is used to control the voltage across an incandescent bulb with pulse-width modulation (PWM). A useful PWM controller can be found elsewhere in this issue. The power supply voltage for driving the gate is supplied by the voltage across the MOSFET. D6, R5 and C2 form a rectifier. R5 limits the current pulses through D6 to about 1.5 A (as a consequence it is no longer a pure peak rectifier). The voltage across C2 is regulated to a maximum value of 10 V by R3, R4, C1 and D1. An optocoupler and resistor (R2) are used for driving the gate.

R1 is intended as protection for the LED in the optocoupler. R1 also functions as a normal current limiting device so that a ?hard? voltage can be applied safely. The optocoupler is anold acquaintance, the CNY65, which provides class-II isolation. This ensures the safety of the regulator. The transistor in the optocoupler is connected to the positive power supply so that T1 can be brought into conduction as quickly as possible. In order to reduce switching spikes as a consequence of parasitic inductance, the value of R2 has been selected to be not too low: 22 k? is a compromise between inductive voltages and switching loss when going into and out of conduction.

Circuit diagram:

(http://2.bp.blogspot.com/_FdGFE8NBDgc/TCH_9dHdkAI/AAAAAAAADes/f1kW4JikjlU/s1600/dimmer-with-a-mosfet-circuit-diagram.GIF)

An additional effect is that T1 will conduct a little longer than what may be expected from the PWM signal only. When the voltage across T1 reduces, the voltage across D1 remains equal to 10 V up to a duty cycle of 88 %. A higher duty cycle results in a lower voltage. At 94 % the voltage of 4.8 V proved to be just enough to cause T1 to conduct sufficiently. This value may be considered the maximum duty cycle. At this value the transistor is just about 100 % in conduction. At 230 V mains voltage, the voltage across the lamp is only 2.5 V lower, measured with a 100-W lamp. Just to be clear, note that this circuit cannot be used to control inductive loads. T1 is switched asynchronously with the mains frequency and this can cause DC current to flow.

Electronic lamps, such as the PL types, cannot be dimmed with this circuit either. These lamps use a rectifier and internally they actually operate off DC.A few remarks about the size of R3 and R4. This is a compromise between the lowest possible current consumption (when the lamp is off) and the highest possible duty cycle that is allowed. When the duty cycle is zero, the voltage across the resistors is at maximum, around 128 V with a mains voltage of 230 V. Because (depending on the actual resistor) the voltage rating of the resistor may be less than 300 V, two resistors are connected in series. The power that each resistor dissipates amounts to a maximum of 0.5 W. With an eye on the life expectancy, it would be wise to use two 1-W rated resistors here.
Author: Ton Giesberts