andyr Posted July 9, 2017 Posted July 9, 2017 All you electronics-knowledgable people - like @Zaphod Beeblebrox or @VanArn - can you advise? I wish to use a diode to drop about a volt, in a DC power supply; the device being fed power sucks about 700ma. Someone suggested to me that I use a 1N4002 … but I don't have any in stock - and it looks like you have to buy them by the thousand! I actually have a couple of 1N5404 diodes in stock - can I use one of these instead? All it seems to be (from reading the spec sheets) is a beefier version of a 1N4002? (Takes 3a, not just 1a.) Thanks, Andy
CraigC Posted July 9, 2017 Posted July 9, 2017 (edited) hi Andy silicon diodes have a forward voltage drop across the junction of 0.7 V. Ensuring your other electrical parameters are satisfied one or 2 in series could help. otherwise you can use a zener diode of the appropriate value as they act to 'clamp' the voltage to that value.. I would prefer the silicon diode method or possibly a series resistor if you know the current. Edited July 9, 2017 by LockedoutofCraigC
andyr Posted July 9, 2017 Author Posted July 9, 2017 12 minutes ago, CraigC said: hi Andy silicon diodes have a forward voltage drop across the junction of 0.7 V. Ensuring your other electrical parameters are satisfied one or 2 in series could help. otherwise you can use a zener diode of the appropriate value as they act to 'clamp' the voltage to that value.. I would prefer the silicon diode method or possibly a series resistor if you know the current. Thanks, Craig ... but my Qu was ... can I use a 1N5404 instead of the 1N4002 that was suggested to me? Regards, Andy
Addicted to music Posted July 9, 2017 Posted July 9, 2017 Like @CraigC said a single diode regardless of rating will drop 0.6-0.7V. Have 2 in series and it will drop 1.2-1.4v.
Zaphod Beeblebrox Posted July 9, 2017 Posted July 9, 2017 1 hour ago, andyr said: Thanks, Craig ... but my Qu was ... can I use a 1N5404 instead of the 1N4002 that was suggested to me? Regards, Andy Yes. In fact, better, as 700ma will cause the diode to run quite warm.
andyr Posted July 9, 2017 Author Posted July 9, 2017 3 hours ago, Zaphod Beeblebrox said: Yes. In fact, better, as 700ma will cause the diode to run quite warm. Thanks, Trevor. Andy
andyr Posted July 10, 2017 Author Posted July 10, 2017 On 09/07/2017 at 11:19 AM, Addicted to music said: Like @CraigC said a single diode regardless of rating will drop 0.6-0.7V. Have 2 in series and it will drop 1.2-1.4v. Actually, Peter, I've just soldered it in place and measured the voltage drop ... it's only 0.4v - that's with 500ma flowing through it (using a 22ohm ceramic resistor as a load, to get that current). Andy
soundbyte Posted July 10, 2017 Posted July 10, 2017 The Fairchild data sheet mentions 1.2V forward voltage @ 3A Onsemi lists 1V max @ 3A so different manufacturers say different things.
andyr Posted July 11, 2017 Author Posted July 11, 2017 On 10/07/2017 at 2:56 PM, soundbyte said: The Fairchild data sheet mentions 1.2V forward voltage @ 3A Onsemi lists 1V max @ 3A so different manufacturers say different things. Yes, I had found that comparative spec sheet - and others. Some refer to that Voltage number as the maximum voltage drop ... all I can say is the actual voltage drop I get is only 0.4v. Andy
Addicted to music Posted July 11, 2017 Posted July 11, 2017 (edited) Andy, there are numerous reasons why you measure 0.4v. Depends on: the type of diode you are using the accuracy of that multimeter and its internal impedance. temperature. all these will effect the measurement you are conducting. If you measure 0.4v then that's good because it require less voltage to turn on, also lowers heat due to a lower voltage drop. Add 2 in series and you will measure . 0.8v. Wheather its 0.8 or 1.2v it really doesn't matter unless you want super accuracy. alternately if you want the accuracy then use a Zener diode with a clamp down so it's exactly 1.0V Additional Edited: that spec that saids 1.2v @ 3.0amps, literal means if you are delivering 3.0amps then there will be a voltage drop of 1.2V across it, very different to what's required for conduction. Edited July 11, 2017 by Addicted to music
soundbyte Posted July 11, 2017 Posted July 11, 2017 (edited) 3 hours ago, andyr said: Yes, I had found that comparative spec sheet - and others. Some refer to that Voltage number as the maximum voltage drop ... all I can say is the actual voltage drop I get is only 0.4v. Andy Just ran a few checks here with numerous diodes, results are as follows. 13.75V regulated supply with two nominally 9.1ohm 25watt resistors in series (~18.2ohms) gives me ~710mA. Diodes connected were. 1N4004 - .856V 1N5408 - .79V UF5408 - .99V FR607 - .775V BYX20/200 - .872V 1N5819 .384V Schottky for something different and as a check of the setup. From this list I would look into a UF5408 for your requirements, the spec sheet says 1.7V @3A for maximum instantaneous forward voltage see note (1) at the bottom left. I assume we are both measuring steady state at ~700mA, not as tested and perhaps that may be the difference. Edited July 11, 2017 by soundbyte
andyr Posted July 12, 2017 Author Posted July 12, 2017 (edited) 6 hours ago, Addicted to music said: Andy, there are numerous reasons why you measure 0.4v. Depends on: the type of diode you are using the accuracy of that multimeter and its internal impedance. temperature. all these will effect the measurement you are conducting. If you measure 0.4v then that's good because it require less voltage to turn on, also lowers heat due to a lower voltage drop. Add 2 in series and you will measure . 0.8v. Wheather its 0.8 or 1.2v it really doesn't matter unless you want super accuracy. alternately if you want the accuracy then use a Zener diode with a clamp down so it's exactly 1.0V Additional Edited: that spec that saids 1.2v @ 3.0amps, literal means if you are delivering 3.0amps then there will be a voltage drop of 1.2V across it, very different to what's required for conduction. 3 hours ago, soundbyte said: Just ran a few checks here with numerous diodes, results are as follows. 13.75V regulated supply with two nominally 9.1ohm 25watt resistors in series (~18.2ohms) gives me ~710mA. Diodes connected were. 1N4004 - .856V 1N5408 - .79V UF5408 - .99V FR607 - .775V BYX20/200 - .872V 1N5819 .384V Schottky for something different and as a check of the setup. From this list I would look into a UF5408 for your requirements, the spec sheet says 1.7V @3A for maximum instantaneous forward voltage see note (1) at the bottom left. I assume we are both measuring steady state at ~700mA, not as tested and perhaps that may be the difference. Thanks for going to all that trouble, guys - but I can live with the result I got. I thought I would be working with a 13.2v supply - and hence thought I would need to drop a volt or so - but the Sbooster LPS has a middle option (between 12v & 13.2v) ... 12.5v. So, to end up with ~12v, if I use the 12.5v supply into the diode ... I end up with 12.1v out. (And, yes, this is steady-state current.) Andy PS: Thanks for those voltage drop figures, sb - I will keep that info for future reference. Edited July 12, 2017 by andyr
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