madbeanpedals::forum

Projects => How Do I? Beginner's Paradise. => Topic started by: 287m on March 13, 2015, 11:15:36 AM

Title: just simple ask
Post by: 287m on March 13, 2015, 11:15:36 AM
(http://3.bp.blogspot.com/-jXFMvmht6Rw/VQMZDukYmZI/AAAAAAAAACc/ZyhZs99jmCQ/s1600/mb.jpg)


what 'the best' for circuit protection and why?


what the best 'position' for diode, must have right after +9V or free?
i mean free, can place anywhere, as long is +9V signal/route?


what difference for pinout position? (1st and 2nd stage)
any difference or nothing?



a billion thanks for your answer

best regards
hermawano

PS : sorry for my bad english. hope you can understand straightly.
and sorry if open this thread in wrong sub forum[/list]
Title: Re: just simple ask
Post by: midwayfair on March 13, 2015, 11:31:22 AM
In A, if positive voltage appears on the anode of D1, it will be allowed to conduct to the circuit. If negative DC appears on the anode, the diode will not conduct to the circuit it unless it exceeds the reverse breakdown voltage of the diode (provided in the datasheet, and LOTS of volts). If you are using a DC adapter of the wrong polarity, it will not work. If you are using the right polarity, you will lose an amount of voltage equal to the forward voltage of the diode (provided in the datasheet -- it's about 0.2V in practice for us).

In B, any negative voltage that appears on the cathode of D1 will conduct to ground. The diode will conduct as best it can, which in this case will use all available current from the power supply. If you use the right polarity of power supply, all voltage will be conducted to the circuit. If the power supply is a battery connected backwards, the battery may not provide too much current and you would just drain the battery, until the battery dies or until you connect it the right way. If the power supply is a DC adapter of the wrong polarity, the diode will attempt to conduct as much current as possible, which, given that the adapter is sourcing hundreds or even thousands of mA from a wall socket, will be so much that the diode will burn up and become a dead short, permanently connecting the ground and power connections, shorting the power supply pins together, and eventually destroying the power supply.

C is the same as A, except that a 1N4001's forward voltage is ~0.5V, more than twice that of a 1N5817. You will lose more voltage.

Quote
what 'the best' for circuit protection and why?

Which of those descriptions above sounds the safest for the circuit and the power supply you are using?

In D, there is no difference between IC1_A and IC1_B or IC2_A and IC2_B. Look up the datasheet on the op amp you are using to make sure the pin numbers are correct for your op amp. A dual op amp is simply two of the same circuit inside the same chip.
Title: Re: just simple ask
Post by: 287m on March 13, 2015, 12:20:48 PM
wow. thanks Jon

you explained very clearly.

almost all my circuit used figure C
the power supply use like this (http://cdn.kaskus.com/images/2014/03/14/2317770_20140314120554.jpg)

last, thanks Jon for some small yours layout. hats off to you
Title: Re: just simple ask
Post by: Coda-effects on March 20, 2015, 05:24:50 AM
Really useful stuff to know! I read it with attention!