Originally Posted by rotation
I just built this
psu and i have some questions; there are two caps before and after 5W resistor near gretz. Author says they can have even 400uF. This would make psu quite big as there are two more 450V caps.
I would like to know if i can replace the second cap near first 5W resistor with something much smaller, 10u or 47u? What would the effect of, say, 47u cap be? I guess i have enough filtering with nice big cap at the input and output.
What about the cap near zener string, does it really need to be that big? I've seen some other schematics with much much smaller caps at that position. Would that make big difference? Why?
You can't know how much capacitance you need without knowing your expected range of current draw.
The size of the capacitors determines how much filtering you get. If the caps are too small, you'll have some ripple (hum) left on the supply, which could find its way into your audio. Smaller capacitors will also increase the extent to which a variable current draw can modulate the DC supply voltage - something alluded to earlier in this thread ("sag" has become a part of the sound of tube guitar amps, but something you generally don't want in line-level circuits). On the other hand, if the capacitors are too big, your transformer and rectifier may have trouble charging them at start-up and you'll get nuisance fuse blows. So how much capacitance do you need? Enough to avoid audible hum in the audio circuit, and avoid any unwanted sag in the power supply (if you want some sag, then the amount you want will still determine how much capacity you need).
In addition to the capacitance, the stiffness of the supply will be determined by the current draw, the series resistance, the "regulation" spec of the power transformer, and the mains frequency. More current draw will discharge the caps more quickly, leaving you with more ripple for a given capacitance. More series resistance will improve the filtering of the capacitors, but will produce a current-dependent voltage drop across that resistance. Likewise, the transformer represents a non-zero impedance that will drop volts proportional to current draw. And 60Hz mains power charges the caps more frequently than 50Hz mains power (and a full-wave rectifier charges them twice as often as a half-wave rectifier). You'll want to choose the value of the series resistor and the filter capacitors in the context of your current needs. It looks like this drawing was made as part of a power amp design. If you're using it to power line-level circuitry, your current draw will probably be substantially lower, and so will your capacitance requirement.
But the biggest factor in your particular circuit is that TIP50 transistor. It's behaving simultaneously as a voltage limiter and a filter known as a "capacitance multiplier". The output of the transistor at its emitter likes to stay about two-thirds of a volt below the voltage on its base. This allows the zener diode on the base to limit the output voltage without requiring the zener (and the resistor feeding it) to handle ALL of the excess power. At the same time, the voltage on the base is filtered by an additional RC filter stage that is not loaded by the output of the supply. The effect is that the capacitance on the base of the transistor is multiplied by the Hfe (DC current gain) of the transistor. This has the general effect of requiring less total capacitance in the supply, but you have to make sure the ripple coming into the transistor doesn't allow the voltage to drop, even momentarily, below the desired output.
I like this circuit a lot, and I've used it many times in a number of projects and products. It can be thought of as a "Voltage Limiter" rather than a regulator. It works best when you choose the Zener diode to set some safe upper limit to the output voltage. If the input voltage is lower than the Zener voltage, the Zener stops conducting and the Capacitance Multiplier aspect of the circuit continues filtering happily. In other words, this circuit doesn't really have a "dropout voltage" like most voltage regulators do. Feed less than 26V into a fixed 24V regulator, for example, and you get no regulation and no filtering. When this circuit gets less voltage than it's expecting, it still works just fine and still filters very effectively.
The base resistor needs to be selected carefully to produce optimum output voltage and current capacity without overheating. The base current (which is output current divided by the Hfe of the transistor) and the Zener current will both flow through the resistor, causing a voltage drop across it. In order to determine the best value, you need to know the maximum and minimum input voltages the circuit will see, the maximum and minimum current draw of the circuit being powered, and the desired output voltage (and Zener voltage). It's a fairly involved set of calculations, but relatively straightforward and only requires some basic algebra applied to Ohm's Law.
The one thing you should keep in mind about this circuit as it's drawn is that there's no short-circuit protection. In theory, if you size the base resistor just right, you get some degree of current limiting. But if you short the output, you'll dissipate some watts in the transistor and may destroy it. As long as the rated Vce of the transistor is higher than the input voltage, destruction won't be automatic.