Why Does a Capacitor in Series with a Transistor Base Allow Current to Flow? - Page 1
Forum section: electronics / beginners

author: techie1234
author_rank: Regular Contributor
timestamp: April 08, 2017, 07:47:57 pm
content: As part of my electronics learning curve I’m now beginning to experiment with transistors. First up: I constructed a simple on/off switch setup to drive an LED from a 9V battery. Limited the current on the base of a 2N3904 using a 10M resistor (was experimenting with how low of a current would still turn on) and the LED with a 390 ohm resistor. All was good (although I was surprised at how low of a current at the base could turn on the LED).Then I put a 4700uF cap I had lying around in series with the base current limiting resistor, thinking for some reason it would keep the base running after disconnecting the battery. The lamp turned and stayed on – which after realizing the pointlessness of the capacitor has me confused. Shouldn’t the capacitor be blocking the DC? And if so, why did the LED come on at all (or, at least for more than a fraction of a second as the voltage ramped up)?

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 07:52:58 pm
content: Here's my thinking - when you first flipped the switch, you had an impulse - you essentially went from 0V to 9V at the input to the cap in 0s. Mathematically, an impulse is the sum of infinite frequencies - so some of them passed. At that point, the capacitor had some degree of charge from that impulse, which will take quite a while to bleed off though a 1M resistor. I'm thinking if you just let it run, at some point the LED would turn off again as the capacitor discharged from the initial impulse.

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 08:00:47 pm
content: You should be able to simulate this in LT Spice by having a switch turn on at some point after t=0 to verify. You should see the inpulse hit and then the Vc increase to a non zero value, as well as Ib increase to a non-zero value and bleed off. Also, play with changing the bat->cap->resistor->base to bat->resistor->cap->base.

author: techie1234
author_rank: Regular Contributor
timestamp: April 08, 2017, 08:07:32 pm
content: Quote from: evb149 on April 08, 2017, 07:59:12 pmIt is an electrolytic capacitor (albeit hooked up correctly) - but I hadn't considered the potential leakiness of the capacitor. My learning up to this hasn't involved such small currents so I've been able to safely ignore such problems. But perhaps no more.Quote from: evb149 on April 08, 2017, 07:59:12 pmThat's what I meant to accomplish but didn't have my thinking cap (ha!) on - when I stepped back for a second to consider what I had done, this question arose.

author: techie1234
author_rank: Regular Contributor
timestamp: April 08, 2017, 08:16:23 pm
content: Part IISo, then, for the next part of my experiment I decided to drive a 14V E10 light bulb, so I switched from the 2N3904 to a more robust MJE182 - still with the cap in series with the base. The bulb came on. I took the battery and cap off the base – and the lamp stayed on. Doh - looks like I’d blow the MJE182. So I tried it again with a new MJE182– same result. I then switched back to using the 2N3904 in the same base/cap/resistor circuit that fried the MJE182 – and it worked fine through repeated battery connects and disconnects. Couldn’t get it to blow. Huh.So, my Part II question is:Why am I blowing the (more robust) MJE182?, or perhapsWhy am I not blowing the (less robust) 2N3904?

author: SeanB
author_rank: Super Contributor
timestamp: April 08, 2017, 08:28:46 pm
content: Sure you connected it correctly, it should work as specified. check the 3 pins and see you have 2 diode junctions present, or a short across all 3.

author: james_s
author_rank: Super Contributor
timestamp: April 08, 2017, 08:34:32 pm
content: 4700uF is going to look like a short circuit at very low currents, until the capacitor charges up to whatever potential is placed across it. If you put the cap in series with the base of the transistor and then pass a current through it the transistor will switch on until the capacitor charges. Even then there may be enough leakage current to keep the transistor on in some cases.

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 08:35:17 pm
content: Here's the simulation - I have to replace your 4700uF with a .1uF to have the time domain be more understandable, but the same applies. The switch turns on at 100ms into the simulation, you can see the voltage increase then bleed off, as well as the current through the LED. This simulation assumes 0 bleed current. So even outside of that (which very well may be another factor) your going to have this because of the impulse.

author: techie1234
author_rank: Regular Contributor
timestamp: April 09, 2017, 02:01:43 am
content: Quote from: SeanB on April 08, 2017, 08:28:46 pmUh, does changing the package type change the order of the leads?

author: FlyingHacker
author_rank: Frequent Contributor
timestamp: April 09, 2017, 04:07:37 am
content: Check the data sheet for the particular component (and sometimes even the particular manufacturer of that component, sadly). Or use a meter. There are numerous tutorials on determining the pins of a transistor with a multimeter available online, even videos.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: April 09, 2017, 04:27:26 am
content: Quote from: techie1234 on April 09, 2017, 02:01:43 amHmmm.... according to the data sheets I just dug up ...The 2n3904 is a TO-92 package and looking at the flat side, pins down, the order from L(1) to R(3) is EBC.The MJE182 is a SOT-32 package and looking at the numbered side, pins down, the order from L(1) to R(3) is ECB.Isn't they? Mixing up Base and Collector connections might not be too good for your transistors....

author: SeanB
author_rank: Super Contributor
timestamp: April 09, 2017, 04:04:38 pm
content: Worse is some transistors come in packages with different leadouts. Nominally a TO92 package, but some are BCE and others are ECB while even some of them are CBE.Depends if they are JEDEC, Pro Electron or JIS numbering schemes, and have the exact same die inside there.

author: Ratch
author_rank: Regular Contributor
timestamp: April 09, 2017, 04:57:27 pm
content: Quote from: techie1234 on April 08, 2017, 07:47:57 pmWho says the capacitor blocks only DC? A cap blocks all voltages applied to it. By blocking, I don't mean immediate shut off like a switch does. The cap builds up a reverse voltage as it energizes, which cancels the applied voltage. The rate the reverse voltage builds up is determined by the RC time constant of the circuit. Usually the period of anapplied AC signal is designed to be much less that the time constant of the cap and resistance, so the signal reverses before the back-voltage of the cap has much effect on it. With DC, however, the cap back-voltage will eventually equal the applied voltage and completely cancel it. The 4700uf cap probably has a high RC time-constant, so it will take a while for the cap reverse voltage to equal the applied voltage and stop the charge flow.Ratch

author: Seekonk
author_rank: Super Contributor
timestamp: April 09, 2017, 05:48:13 pm
content: You need to learn a sense of perspective. A LED will light with as little as a couple microamps. That 10M resistor isn't really putting the transistor into full conduction. Putting your meter probes on dry skin will have a lower resistance than that. So I figure you are driving that LED with less than 200 micro amps with that transistor gain. No serious current there. Capacitors aren't perfect and will leak. It can take a week for a new capacitor to fully form and in that time the leakage is higher. In most circuits that leakage doesn't matter.

author: Zero999
author_rank: Super Contributor
timestamp: April 09, 2017, 06:35:37 pm
content: Quote from: kickit2 on April 08, 2017, 08:35:17 pmGoing b y the results, it appears, you forgot to model the capacitor's leakage current, which can be roughly approximated by a resistance in parallel with the capacitor. You can either draw a separate resistor in parallel with the capacitor or right click on the component and enter the value of the equivalent parallel resistance in second from bottom text field.