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zener diode question
Posted by
m3vuv
on 09 Feb, 2022 15:38
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Hi all,i need a 6.2v zenner,but i only have one that zeners at 7.8v,if i add a couple of silicon diodes and a shottky in serries with it ie say 1.5v drop assuming silicon 600mv drop and a shottky at 300mv,would that work as a 6.3v zener ?
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#1 Reply
Posted by
m3vuv
on 09 Feb, 2022 15:39
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6.3v would be near enough .
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#2 Reply
Posted by
TimFox
on 09 Feb, 2022 16:53
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No. Your combination would increase the total voltage across the diodes from 7.8 V to approximately 7.8 V + (2 x 0.7V) + 0.3 V = 9.5 V.
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That will increase Vf to where you want it but it won't be as stable as a zener made for that voltage.
In fact, it will be a very poor zener.
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#4 Reply
Posted by
TimFox
on 09 Feb, 2022 19:11
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He has 7.8 V, but wants to decrease it to 6.2 or 6.3 V.
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#5 Reply
Posted by
ledtester
on 09 Feb, 2022 19:22
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The OP should explain how the zener is being used.
It is possible the OP wants to create a 6.3V power supply by first creating a 7.8V power supply with the zener on hand and dropping the voltage with some diodes. This requires a consistent current draw.
If the intent is to create a reference voltage into a high-impedance input, well, that's another story.
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#6 Reply
Posted by
TimFox
on 09 Feb, 2022 19:29
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He did say how he meant to connect them (in series to make a “6.3 V Zener”) but he apparently thought that the additional diodes would have a reversed polarity voltage, possibly due to the reversed symbols for the forward-biased added diodes and the reverse-biased Zener.
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#7 Reply
Posted by
Zero999
on 09 Feb, 2022 19:46
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What are you trying to do? If you want a voltage reference, then there are various tricks, to do that. If the output impedance is not important, simply add a potential divider. If you need a lower impedance, add a couple of silicon diodes in series with the load.
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#8 Reply
Posted by
m3vuv
on 09 Feb, 2022 20:47
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forgot to say,i also have a 5.1v zener,its for d10 in attatched schematic,the kit came with it missing.
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#9 Reply
Posted by
TimFox
on 09 Feb, 2022 21:07
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Two forward-biased 1N4148 (or similar) PN diodes in series with a 5.1 V Zener would give about 6.3 to 6.5 V. 5.1 V Zeners have a reasonably low temperature co-efficient, and the two forward-biased diodes would degrade that somewhat. (Temperature-compensated reference diodes comprise a Zener and one or more forward-biased diodes in series, where the opposite-direction temperature co-efficients cancel at a specified value of current.)
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#10 Reply
Posted by
m3vuv
on 09 Feb, 2022 22:17
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so would a normal pn diode with the 5.1v zenner work in place of d10 then in this psu that schematic i posted? TIA
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#11 Reply
Posted by
magic
on 09 Feb, 2022 23:26
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It looks to me like this diode sets negative supply to some LM358 and also reference voltage for CC mode.
The latter is dodgy due to thermal drift of such zeners. 5.1V would actually be better for that, or two TL431 in series or TL431 with feedback dvider to get 6.2V.
I can't see anything immediately wrong with reducing supply voltage of those opamps to -5.1V but I haven't analyzed the circuit thoroughly.
Anyone?
BTW, what's J1 and J2?
edit
Nevermind. If you add two diodes to a 5.1V zener, you get 4mV/°C drift (assuming perfect zener) which is 0.06%/°C or 0.6% for 10°C. Who cares, just do it.
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#12 Reply
Posted by
m3vuv
on 09 Feb, 2022 23:40
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As far as i know j1+j2 are the connectors for the voltage and current adjustment pots,i will give the zenner a go,at least it may be ok untill i can order the correct one .
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#13 Reply
Posted by
magic
on 09 Feb, 2022 23:49
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I suppose 5V would work, those LM358 only need to sink current from power transistor's base through a few diodes and it looks like their output voltage is near that floating ground (which tracks the OUT+ voltage). The CC one needs to overcome Vf of an LED, so it's a bit harder.
Worst thing, there will be no regulation. And the range of CC adjustment may be a bit narrower if regulation does work. That could be fixed by tweaking those resistor dividers.
edit
And whoever drew that schematic must have been drunk. Why are LM358 power supplies upside down?
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#14 Reply
Posted by
magic
on 10 Feb, 2022 15:04
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As far as i know j1+j2 are the connectors for the voltage and current adjustment pots
It's a dodgy circuit. Two problems with it:
1. As pots wear out, they may lose contact between the wiper and the track. When that happens, this PSU will output maximum voltage.
2. If you manage to set the CV pot to 0Ω very fast while there is still voltage on the output (or if something fails and the output loses regulation), the output will discharge through D10, a parasitic diode inside LM358 and the CV potentiometer - not a great idea.
I would mod it: connect the pot's track to J2 as intended, then separate VR2 and connect it to the wiper with a third wire.
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#15 Reply
Posted by
TimFox
on 10 Feb, 2022 15:24
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A quick look at Mouser indicates that, for example, 6.2 V 500 mW Zeners in onesies cost $0.16 (USD). That one, BZX79C6V2 is available from several manufacturers, in stock at Mouser, with different suffices. Perhaps that is a better solution to your repair.
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#16 Reply
Posted by
magic
on 10 Feb, 2022 15:58
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What makes you think that a 6.2V zener would be a better repair than a 5.1V zener?
Reminder: the designer of this PSU kit probably wasn't entirely sober
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#17 Reply
Posted by
TimFox
on 10 Feb, 2022 17:20
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The words "for example" in my post indicate that Zener diodes in that voltage range and power are available at very low prices. The 6.2 V device was an example. If you read his original post, he was trying to produce a 6.2 V Zener from multiple components.
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#18 Reply
Posted by
w2aew
on 10 Feb, 2022 21:02
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#19 Reply
Posted by
TimFox
on 10 Feb, 2022 22:13
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Vbe multipliers need three components and are temperature sensitive. With that number of components, a TL431 would be a better choice.
Replacing a Zener diode (two wires) with a cheap new Zener diode will fit the PCB better.