Explaining DC return? - Page 1
Forum section: electronics / beginners

author: gotwood
author_rank: Contributor
timestamp: February 27, 2013, 03:37:33 pm
content: I have an capacitively coupled AC source. There is a 1k Ohm Resistor used as a dc return(from book), this is in parallel with a forward biased diode in series with a 10k ohm resistor.Q: what is purpose of the dc return resistor?(google searches and my textbooks come up with nada)Q: why is there a voltage drop across the 10k ohm resistor with a dc return?Also for future problems what is the best recommended way to insert circuit so people can actually see what I have Q aboutTY

author: IanB
author_rank: Super Contributor
timestamp: February 27, 2013, 03:44:27 pm
content: A circuit schematic would help.For something simple like this you could use ASCII art.For more complicated circuits, draw them out using any one of the free drawing tools or simulation tools like LTspice and attach an image.

author: olsenn
author_rank: Frequent Contributor
timestamp: February 27, 2013, 04:30:33 pm
content: I think the capacitive coupled AC source just means that the AC source has no DC offset (centered at 0 volts). So the circuit would just look like:AC_Source_>-----------------------------------| |()DIODE |--- || \ Resistor\ Resistor / 1k/ 10k |GND----------------------------------------------

author: robrenz
author_rank: Super Contributor
timestamp: February 27, 2013, 04:41:27 pm
content: Where is the capacitor?

author: fcb
author_rank: Super Contributor
timestamp: February 27, 2013, 04:48:35 pm
content: Do you think Dave needs a section labelled "Homework"

author: kripton2035
author_rank: Super Contributor
timestamp: February 27, 2013, 04:55:55 pm
content: you can use a program like aacircuit to make ascii drawings of schematics and paste them herehttp://www.tech-chat.de/ascii-circuits.html

author: M0BSW
author_rank: Frequent Contributor
timestamp: February 27, 2013, 04:57:35 pm
content: Quote from: fcb on February 27, 2013, 04:48:35 pmI think he should have a section called "ELECTRONICS FOR DUMMIES" That'll be for me

author: gotwood
author_rank: Contributor
timestamp: February 27, 2013, 05:58:32 pm
content: Source------------------------diode---------------out| | || | |ground 1k 10k| |ground groundNot sure what is meant by capacitively coupled source....none of my texts explain what is meant by it or purpose....but the schematic shows a source ac voltage peak of 10v 1khz with a 1 microF cap in series with it.....this is in a dashed box to illustrate its a component??What we use to provide the intended source is a function generator with a Peak to Peak V of 20

author: Christe4nM
author_rank: Unknown
timestamp: February 27, 2013, 06:46:28 pm
content: The AC signal from the source first passes the capacitor, so if any DC (offset) were to be found in the source signal, that capacitor blocks it.The 1k resistor can be seen as an 'input impedance' to the circuit that follows. Remember that to DC a capacitor is effectively an "open". As in opposite to short circuit. Thus the whole DC part of the voltage will fall across the capacitor, yet no DC current can flow through. If you would measure the voltage across the 1k resistor you'd only see the AC signal.Next up is the diode. As you should know by now, the diode blocks current in one direction. So depending on the orientation of the diode, only the positive, or negative part of the AC current can pass the diode. You effectively are removing one half of the AC signal. The AC current that does pass the diode can now flow through the 10k resistor and return to the source. This is called a half wave rectifier (as opposed to full wave rectifier). Mind that the half of the AC voltage that appears across the 10k resistor will be lowered by about 0.6 or 0.7 V since that voltage will fall across the diode to let it conduct.So: signal source -> DC blocking capacitor -> input impedance -> half wave rectifier -> load resistor. Look for these terms on Google or in de index of your books and you should find more info.Edit: even though the capacitor is part of the source that doesn't change anything. Only now you can be sure that the source's output is pure AC.

author: gotwood
author_rank: Contributor
timestamp: February 27, 2013, 06:51:39 pm
content: you answered while I was fixing the diagram.....so in the circuit I do not have a capacitor.... source from generator......into the circuit...the diode is forward biasedI am getting .5 V less to my peak output when i open the 1k DC Return resistorI am wondering why I get the reduced voltage, the 1k ohm resistor is in parallel with the rest of the circuit so I assumed the V to each is the same i.e. voltage same over parallellooking to clarify what is happeningvoltage output is 9.38 V without the dc return....just what I expectedoutput drops to 8.875V when dc return is there......thought the output should be same per above "law"