Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
This resistance is to represent the load connected at the output. If this your supply is suppose to deliver 15mA at 15V, you will simulate this by adding 1K(R= V/I =15/15m) resistance.
A bleeder resistor is a resistor placed in parallel with a high-voltage supply for the purposes of discharging the energy stored in the power source's filter capacitors or other components that store electrical energy when the equipment is turned off - wiki.
Nominally, it's a safety device to discharge the capacitor when your source turns off. In a RF detector circuit, you can vary the size of C and R to set up the rise/fall time characteristics of an envelope detector.
For a rectifier like the circuit you gave, make the R a few kOhms. You'd calculate the value to make sure 1) that you don't burn up the part when you have DC voltage present, and 2) to determine how quickly your storage capacitor will discharge after the source power is disconnected. For matters of user safety, consider the following scenario... the system's power input is unplugged, but there is still energy in the capacitor... stick a screwdriver in the wrong place, and BANG!!! sparks fly, unless you use a bleed off resistor, as shown, to safely discharge the energy in the cap.
but i also have seen a capacitor in series with this resistor than why the capacitor and resistance in series? is capacitor and resisotr in series is some type of dummy load before voltage regulator >
Its a load resistor.......without it the current does not discharge and there is no ripple, think of it as an open-circuit without the resistor, the current has no where to go. You should find a couple good analog books on DC/DC AC/DC converters such as buck, boost, buck-boost, flyback etc. The picture you posted is a simple rectifier circuit which is used to turn AC to DC with a drawback of some "ripple" obtained because the current going through the capacitor cannot change instantaneously, it requires some finite discharge time to discharge through the resistor. Think of the current as water, the capacitor as a bucket and the resistor as a well......the bucket has to constantly bring water to the well. If you go more advanced then these basic rectifier circuits are later tuned with snubbers and used in DC to DC converters. Hope that helps alot
---------- Post added at 09:21 ---------- Previous post was at 09:14 ----------
The cap and resistor are never in series at the load.....they are in shunt for RC time constant. You can think of a load as anything connected to your AC plug at home like your labtop. If you see one before the load, its acting as a filter to remove some noises caused by the cheap diodes. You see more snubber circuits or an inductor in series with the shunt res + cap for the noise issue too.
All of you are correct because the resistor does both jobs, while the power supply is on it works as a load and when it turn of it works as a discharge path for the capacitor.
We have no idea if the designer of the circuit had the intension to use this resistor as a load or if it was only added to discharge the capacitor when the circuit is turned off.
A series RC from Vout to ground is a simple de-Q'ed filter element. The resistor gives the filtering element some real resistance to keep multiple reactive elements in a circuit from ringing/resonating with each other.
A series RC from Vout to ground is a simple de-Q'ed filter element. The resistor gives the filtering element some real resistance to keep multiple reactive elements in a circuit from ringing/resonating with each other.
but i also have seen a capacitor in series with this resistor than why the capacitor and resistance in series? is capacitor and resisotr in series is some type of dummy load before voltage regulator >
Close, but not quite. If you had just a capacitor on the output, it's voltage would eventually charge up to the peak voltage of the incoming rectified signal (AC peak voltage minus a few tenths of a volt across the diodes). When you add a resistor in shunt to the cap, it will begin to discharge the capacitor as soon as the incoming AC voltage drops enough that the rectifier diodes are no longer forward biased (turned "on"). This small current drain causes the capacitor voltage to droop, until the rectified AC voltage can once again forward bias the diodes and "top off" the capacitor's energy supply. This causes a ripple voltage on your DC... small, but it's there.
The ripple is very small in the case of a big C and big R, but if you want to make a fast-tracking RF envelope detector (a diode, cap and resistor), you'd make the bleed off resistor smaller, so that the cap's DC voltage will track the envelope of the incoming RF signal, but not track the faster moving carrier frequency. Fundamentally, that's what you're doing with an AC rectifier... the mains voltage remains at a constant 110V or 220V, so the output of the rectifier is a constant voltage, because the incoming AC signal is a constant amplitude.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
