Is this an acceptable circuit (using BJT) for draining a capacitor? - Page 1
Forum section: electronics / beginners

author: sparkydog
author_rank: Regular Contributor
timestamp: April 20, 2023, 06:03:04 pm
content: Inanother thread, CountChocula helpfully provided the following circuit for discharging a capacitor that's part of an RC delay (thanks again!):The intended function is that SENSE delivers some voltage which (slowly) charges CD. While SENSE is present, Q1 is on and blocks current. When SENSE goes open-circuit, Q1 turns off and allows CD to drain rapidly through R2. The goal is for CD to only hit some threshold voltage if SENSE stays on for some period of time, and for any interruption to reset the 'clock'.We then got into a disagreement whether D1 is necessary, with bothCircuitLab's time domain simulation (tip: block cookies if it wants you to log in) andCircuitJS(interactive) insisting that, not only is it not needed, but omitting it allows the capacitor to drain from "very low" to "zero" much more quickly. (In both cases, it drains from "high" to "very low" quickly through Q1. However, Q1 becomes high resistance as the voltage nears zero. Without the diode, the capacitor can still discharge through RD+R1.)Now, in CountChocula's defense, itseemslike D1 is indeed necessary to prevent Q1 turning on and blocking current. What the simulations seem to show, however, is that Q1's resistance is proportional to the difference between VEand VB, so long as VBis non-trivially less than VE... which it will be because RD/R1 effectively form a voltage divider.So... "is this correct?" is an obvious question. (Note, I'm asking about the circuitwithoutthe diode.) More importantly, though,is this going to fry Q1?

author: mikerj
author_rank: Super Contributor
timestamp: April 20, 2023, 06:43:28 pm
content: What voltage is going to be applied at SENSE? If it's more than about 5v you need to consider Vbe reverse breakdown as there would be nothing limiting current other than source impedance of the sense voltage.

author: sparkydog
author_rank: Regular Contributor
timestamp: April 20, 2023, 08:35:26 pm
content: Quote from: mikerj on April 20, 2023, 06:43:28 pmYou know, if you looked at the simulation, you would have answers? 🙂VSENSE= 12 V. This, however, is where I'm confused; if a BC327 has a V(BR)EBO= -5 V, what does that mean? It releases magic smoke if VB> VE+ 5 V? It releases magic smoke if VE> VB+ 5 V? It releases magic smoke if |VB- VE| > 5 V?What does "nothing limiting current" mean? Nothing limiting currentfrom where to where?While charging, VB= 12 V and VEslowly climbs to 12 V, with current through Q1 measured in picoamps (i.e. negligible). While discharging, VBis about 0.7 V lower than VEand ~72 µA goes from E to B, and ~7.2 mA goes from E to C. All of these aremaximumsthat decrease as the capacitor discharges.

author: mikerj
author_rank: Super Contributor
timestamp: April 21, 2023, 08:19:32 am
content: Quote from: sparkydog on April 20, 2023, 08:35:26 pmIf you marked the schematic I wouldn't have to click on unknown links...Quote from: sparkydog on April 20, 2023, 08:35:26 pmIt means this is the guaranteed maximum reverse bias voltage you can apply to the BE junction before it breaks down (like a reversed biased Zener diode). Reverse biased means the base is at a lower potential then the emitter.Quote from: sparkydog on April 20, 2023, 08:35:26 pmExactly what it says. You have a voltage source connected to SENSE, which will reverse bias the base emitter junction when the capacitor is at a lower voltage than SENSE, and you have no resistive circuit element in that path to limit the current flow if the base emitter junction does break down. The current will cause the capacitor to charge rapidly, so the voltage should drop below the breakdown threshold fairly quickly but it doesn't take much energy to damage the BE junction like this.Even if the current is limited, BE breakdown should be avoided unless it's deliberate (e.g. noise generation) as it can cause a permanent reduction in transistor performance.Quote from: sparkydog on April 20, 2023, 08:35:26 pmIf the BE junction breaks down the emitter voltage will not climb slowly, it will be a rapid increase. You could add a diode in series with the base circuit to prevent the possibility of breakdown, but this will add another ~0.6v drop so the transistor will start turning off when the capacitor is around ~1.2v instead of ~0.6v.

author: sparkydog
author_rank: Regular Contributor
timestamp: April 21, 2023, 06:05:25 pm
content: Reminder: this is what I'm simulating:...and I'm confused. In one place, you said:Quote from: mikerj on April 21, 2023, 08:19:32 amBut in another place you said:Quote from: mikerj on April 21, 2023, 08:19:32 amThe latter only makes sense if the base is at ahigherpotential than the emitter. So, as I asked previously,which is it? Is VB≫ VEbad? Is VE≫ VBbad? Arebothbad?Here is the behavior of the circuit according to CircuitLab:The instantaneous 12A spike when SENSE goes high is a little concerning, but other than that, IB≈ -20 pA while charging, and spikes briefly to ≈ -100 µA when SENSE goes open-circuit. (Note that that value is based on the capacitor charge < 6 V as shown, and may be higher if the cap is allowed to charge to a higher voltage.)CircuitJS is not exhibiting the initial spike; otherwise, as far as I can tell it is entirely consistent with CircuitLab.Are the simulations correct, or arebothneglecting BE junction breakdown?

author: wasedadoc
author_rank: Super Contributor
timestamp: April 21, 2023, 06:14:16 pm
content: Quote from: sparkydog on April 21, 2023, 06:05:25 pmThe clue is in the word "reverse". For a PNP transistor the base-emitter junction is reverse biased when the base is more positive than the emitter.

author: TimFox
author_rank: Super Contributor
timestamp: April 21, 2023, 08:26:18 pm
content: Simple description of BE behavior:By itself, the base-emitter junction acts as a Zener diode:-- in the forward direction, acting as a forward-biased diode, it conducts at roughly 0.7 V, behaving as a silicon diode-- in the reverse direction, acting as a Zener breakdown, it will conduct (strongly) at an amazingly small voltage: typically 5 to 7 V for a normal silicon BJT.If I remember correctly, before silicon planar construction became the norm, BE breakdown was comparable to BC breakdown.For Si BJTs now, even for high-voltage BJTs (VCEO> 100 V), the BE breakdown voltage remains low.Specifically, the ancient 2N404 germanium PNP transistor has maximum ratingsVCE= 25 V, andVEB= 12 V.An early silicon transistor, 2N697 NPN, has maximum ratingsVCE= 40 V, andVEB= 5 V.