What's a good diode (preferably some jelly bean part) to use when one needs an extremely low reverse leakage current?
The reverse voltage is normally between 1.33 and 1.50 VDC. The diode is connected to a high resistance voltage divider that only has 3 uA going through it. Its only function is to protect a comparator (TS3V393C) from having a negative voltage applied to it's inputs under non-operating conditions.
The diode (D1)is shown in the lower left hand corner of the first schematic at:
A small-junction NPN transistor, base-emitter junction can be used to get leakage down where you appear to need it. Tie the collector and base together.
In that circuit, I see no need for the diode. The current is strictly limited, and the comparator surely has its own input protection. Why do you believe more is necessary? Is there an exremely high voltage on the other end of that 100M resistor?
The other end is connected to a capacitor at -300 VDC, the other end is connected to Vcc which is normally at +4-6 VDC, except if the battery is disconnected while the flash's capacitor is still charged, in which case, the -300 VDC is going to force the 3 uA into something.
My guess is that the Vcc rail would go negative with respect to ground; as to whether the IC would withstand having 3 uA forced through it the wrong way I don't have clue.
Could some familiar with the inner workings of the TS393C care to speculate about this?
Hmm, would putting a reverse diode between Vcc & ground suffice?
(Keep in mind that I learned most of my what I know about electronics from Forest Mims's books!)
i tihnk what you mean is a low cut off voltage. most diodes are good for reverse leakage for basic things unless your talking about RF? in any case i think what your looking for with spec's like that is a germanium type diode.
According to Bob Pease in "Troubleshooting Analog Circuits" (page 66) it's the base-collector diode you want to use. He suggests a 2N3707 or
2N930, although he also mentions 'some 2N3904', saying that there may be a problem with some because of gold doping. His claim is 1pA leakage at 7V.
He also mentions JFET devices, in particular the 2N4117A and PN4117A, which apparently have very small junctions, leading to even lower leakage of typical 0.1pA, and guaranteed 1pA max.
Later in that chapter, he also mentions LEDs as having incredibly low reverse leakages, but only if you keep them dark, and only reverse-bias them up to 1V.
base-collector diode you want to use. He suggests a
may be a problem with some because of gold doping. His
With all due respect for Mr. Pease, I'm going to stick my neck out and disagree. Most of the leakage of a diode comes from surface leakage and thermally generated carriers in the depletion region. For a normally built BJT, the B-E junction has less exposed surface than the B-C junction and also has a smaller volume depletion region, being both narrower and occupying somewhat less area. So, unless one requires a reverse breakdown higher than the several volts available from a B-E junction, it is the better choice. (I am willing to be proven wrong on this, but it will take some evidence.) I suspect Mr. Pease's advice was directed toward the case where more than 3 to 6 Volts of reverse standoff is needed.
apparently have very small junctions, leading to even
I suppose those might be considered the "jelly bean" part the OP requested. I was trying to not overdo the low leakage requirement because the circuit in question can tolerate a few nA harmlessly.
leakages, but only if you keep them dark, and only
I almost mentioned that, but the LED forward drop would make it useless in the OP's circuit. There, the diode is supposed to conduct so that a "protected" circuit's input protection network does not have to source a few uA when gently pulled below ground.
No need to speculate. The manufacturer's schematic shows protection diodes to each rail from each input. There is no harm in asking those diodes to conduct a few uA. They are rated to take 50 mA, so your few uA are not abusive.
Those would be the Vishay (Siliconix) PAD (TO-18b) and JPAD (TO-92) parts. Farnell stocks the PAD5 (5pA) and JPAD50 (50pA)- the former costs a couple of dollars in small quantities, and the latter around $1. There is a PAD1 part.
The diode is not needed. As you say , the flash capacitor is referenced to the BATT(+) rail. When the battery is removed, the capacitor will discharge through the BATT(+) rail, R1 , R2, and the 100M. The comparator terminals will be no more than 500K/100Mx300V= -1.5V maximum relative to the GND terminal. Since the supply voltage across the capacitor will be only a few tens of mV, this will not damage the IC, any substrate conduction will be limited to 3uA. If you reduce R2 to 50K , R5=1M ,R3=27K, R4=1M, then you retain the same adjustment range while reducing the maximum reverse discharge voltage to -150mV.
base-collector diode you want to use. He suggests a
may be a problem with some because of gold doping. His
Well you know that alpha_F x Ieo = alpha_R x Ico, so that right there tells you that Ico is an integer multiple or two of Ieo. Pease is
*never* wrong, so it must be that the Early effect narrowing of the EB depletion region causes that junction reverse saturation current to surpass the BC diode leakage in some exponential way with voltages smaller than breakdown.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.