# discharging capacitor to ground

Discussion in 'Electronic Basics' started by Peter, May 22, 2004.

1. ### PeterGuest

Hi folks,

I want to discharge a 1uF monolytic capacitor charged at 5V through a 2N3904
(collector to cap, emitter to ground). The transistor driven into saturation
to do that.
It is not an oscillator, the cap is permanently charged through a 1K
resistor to a 5V source.

But should I add a limiting resistor (like 33 ohms) to limit peak current
under 200mA? Or the limiting resistor is not needed because of capacitor
internal resistance or transistor resistance?

ThankS!
-- Peter --

2. ### John LarkinGuest

No need. The energy stored here is far too small to damage the
transistor. Let it discharge as fast as the transistor can yank it
down.

John

3. ### JamieGuest

yes, limit the current with a resistor.
a cap can really generate some heavy
current in very short periods.
and some transistors can not handle that.
they will actually blow/crack the case sometimes
with hardly no heating due to the sudden current
impact that can cause pressure expansion.

4. ### PeterGuest

Good! I don't like adding unecessary components, even a single resistor

Thanks!

5. ### John PopelishGuest

As an exercise you might calculate the die temperature rise above
ambient if the transistor dumps the cap and innediately turns off and
repeats this process as fast as the 1k pull up resistor can recharge
the capacitor. By the way, the instantaneous peak current is limited
by the beta of the transistor times the base current. There are a few
simplifications in doing this math.

6. ### John LarkinGuest

Ballpark 5 milliwatts; not enough to worry about.

John

7. ### Kevin AylwardGuest

Well, if it were me, I would use a resistor to limit the current to the
max specified of the transistor, which I think is around 1A peak. One
always wants to be able to actually *do* the sums to ensure that the
data sheet specs are met. Without doing this, all bets are off. V/0 is
infinite. For example, a 2N3904 is not speced to be able to take, say,
10A, so make *sure* that it is never in a situation that will cause this
to occur. Its simply good design practise.

Unless one has an absolutely overriding reason *not* to ensure that a
transistor is within what is specked for the transistor, not doing the
basic V/I calculations is pretty daft. It certainly gets one into the
wrong mindset for doing reliable design in general, even if one can get
away with cheating in a one off situation. If I was doing a design
review/sign off on this, I would be demanding a worst case analysis that
guaranteed that the transistor was indeed within its most *basic* of
specs (V, I, Po), even if I were to let some of the others slip through
for practical reasons (e.g. a 1u gate leakage spec, when it is really
1pa).

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

8. ### John LarkinGuest

I figure that, if you can do a 10-second quick mental estimate of the
situation, and it turns out to be orders of magnitude away from any
sort of danger, then do it and move on to something more important.
Save the serious math for when it really matters.

12 microjoules dumped from 5 volts isn't going to hurt any transistor.
And there's only 6.25 milliwatts available, max. You can do both those
calcs in your head. That's what I would say at a design review.

John

9. ### Geir KlemetsenGuest

It may help to put a coil with recistance next to zero in serie with the
capacitor. I don't know how to calculate the lowest posible value (Henry)
the coil must have in order to keep the peak current below the transistors
maximum peak current limit.

10. ### Kevin AylwardGuest

Well, the point of the resistor is to prevent doing any serious maths.
Its the simplest way to ensure that it is within spec. If one needs to
avoid the component, that's when a bit more work is required to see if
one can cheat reliably.
But if you can't guarantee that the transistor is within *all* rated
specs, i.e. say < 1A in this case, you still have no recourse to the
manufacture if they all start failing.

I agree, that in this case its doubtful if there will be problems, but
in general, transients into caps are an issue in transistor failures.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

11. ### John LarkinGuest

I see people using huge caps - as, say, 1000 uF in 12-volt
applications - as the timing caps on 555 thingies. That's scairy. 1000
mikes charged to 8 volts is 32 mJ, enough to fire a spark plug.

John

12. ### Bill BowdenGuest

How do you drive the transistor into saturation, do you use
a resistor on the base?

If so, you might just recalculate the value of the base
resistor so the HFE of the transistor limits the collector
current. That way you can do it all with one resistor.

-Bill

13. ### PeterGuest

But the HFE can vary widely from one transistor to the next, even in the
same batch. Can be anywhere between 75 and 300 or so, and the collector
current further changes the HFe.
Well! I could take the maximum HFE as a start, then be sure the energy
stored can be dissipated by the junction in this short amount of time; as
John said, it is quite small with 1uF. As long as there is no other hidden,
hazardous phenomena for the transistor in the scenario of course.

-- Peter --

14. ### John LarkinGuest

The beta of a 2N3904 peaks at about 8 mA, and is dropping fast fast -
down by about 10:1 - at 200 ma. So the peak current will sort of
self-limit, given reasonable base drive. Not that it matters, with
this low energy available.

John