Considering the circuit below.:
What is the purpose of capacitor C5? I know it provides positive feedback to sustain the LC oscillations, but how does this work ?
Electronic – Capacitor between emitter and collector
capacitorcircuit analysisfmtransistors
Related Solutions
Subset summary:
I = excess current to be provided.
T = time to provide this extra current.
V = acceptable drop in voltage during this period.
C = capacitance in Farad to meet this requirement.
Then:
- C = I x T / V
In theory, and close enough to be useful in real applications:
One Farad will drop in voltage by one volt in one second with a 1 Ampere load.
Scale as required.
The results are not encouraging :-(.
(1) Providing a capacitor to do everything
For over current of I ampere, droop of V volt over time T seconds (or part thereof) Capacitor C required is, as above)
C = I x T / V <- Cap for given VIT
ie more current requires more capacitance.
More holdup time requires bigger capacitance.
More acceptable Voltage droop = less capacitance.
or droop given CIT is, simply rearranging
- Vdroop - (T x I) / C
or time a Cap C will hold up given C I V, simply rearranging =
- Time = T = V x C / I
So eg for 1 amp overload for 1 second and 2 volt droop
C = I x T / V = 1 x 1 x/2 = 0.5 Farad = Um.
Supercaps may save you as long as required peak current can be supported.
SUPERCAP SOLUTION
A Supercap (SC) solution looks almost viable.
These 3F, 2.5V supercaps are availale ex stock from Digikey for $1.86/10 and under 85 cents in manufacturing volume.Prices
For the 3F, 2.7V unit the acceptable 1 second discharge rate to 1/2 Vrated is 3.3A. Internal resistance is under 80 milliohms allowing about 0.25V drop due to ESR at 3A.
Two in series gives 1.5F and 5.4V Vmax. 3 in series gives 1 Farad, 8.1V Vmax, same 3A discharge and 0.75V drop due to ESR at 3A.
This would work well for surges in the tenths of a secnd range. For the specified wort case 3A, 5 seconds requirement perhaps 15 Farad is needed.
The same family 10F, 2.7V $3/10, 26 milliohm looks good. 10A allowed discharge. Two in series drooping from 5.4 to 5 volts at 3A gives
T = V x C / I = 0.4 x 5 / 3 = 0.666 seconds.
Getting there.
(2) IF the droop causes system reset etc and one wishes to avoid this (as one usually does :-) ) an often useful solution is to provide a sub supply for the electronics with cap that hold them up over the dropout period.
eg electronics need say 50 mA. Holdup time desired = say 3 seconds (!). Acceptable droop = 2V say.
From above
- C = I x T / V = 0.05 x 3 / 2
= 0.075 Farad
= 75,000 uF
= 75 mF (milliFarad)
This is large by most standards but doable. A 100,000 uF supercap is reasonably small. Here the 3 second holdup is "the killer". For a more typical say 0.2S dropout the required cap is
75,000 uF x 0.2/3 = 5000 uF = very doable.
(3) A small holdup battery for the electronics can be useful for obvious reasons.
(4) Boost converter: In a commercial design where 4 x C non rechargeable batteries were used, to provide 5V, 3V3 and motor drive battery (exercise equipment controller) end of life Vbattery got well below needed 5V during end of battery life and much much below when motors operated. (The primary design was not mine). I added a boost converter based on a 74C14 hex Schmitt CMOS inverter package to provide 5V to the electronics at all times plus 3V3 regulated to the microcontroller. Quiescent current of boost converter and 2 x LDO regs and electroncs under 100 uA.
E&OE - may have got something on wrong side somewhere there, easily done. If so, somebody will tell me about it :-).
ADDED:
Query: It has been (quite understandably) suggested that
I am not sure you are answering the users main question.
To stop from overloading a power supply it does not seem feasible.
It is not a case of power supply cutout, it is a case of wanting to allow higher current for short periods(on the order of 5 or more seconds).
This seems like a case of needing another power supply
Response
I believe that I am addressing the question completely, as asked, BUT I am also addressing what I believe is liable to be the larger question as well.
Consequently, there seem to be tangents and irrelevant material here.
I have addressed points unasked as well as points asked based both on my own experiences in closely analogous applications and also on general expectations.
The issues are
"What if demand exceeds supply" and
"What if supply falls below demand".
These are one and the same in practice but may have different causes.
Note that my answer (1) specifically says
- "For over current of I ampere"
and his question was
- " ... but there are cases (when all servos are suddenly loaded) when I think the power draw will exceed 2A for a short period of time.
ie dealing with overcurrent is exactly what he is asking.
BUT overcurrent is caused by overload and, when the "cost" of trying to deal with overcurrent is seen (0.5 Farad caps or whatever) then the perspective may well turn to "what can we do to ride out this overload differently". The next most obvious "solution" is to accept the hit on motor performance, let the supply rail fall BUT maintain a local supply to keep the eectronics sane. Another solution which I didn't bother addresssing is to deloa the system by eg slowing servo rates when all are on at once. Whether this is acceptable depends on the application.
The reason that we can TRY to address the short term overcurrent situation is that the supply has spare capacity most of the time and this is used to charge the caps prior to the surge event. The caps do not magically manufacture extra current, just save up spare current for arainy day.
To supply current the capacitor MUST lose voltage so I specify the acceptable limit for that too. I think you'll find that if you couch his requirement in numbers and then plug them into my formulae that his question as-asked will be answered.
Re on geometrikal post.
- But it is not a case of 6V*3A*5s. You need enough capacitance to stop the output from sagging low enough to cause the output of the power supply to need to host more current. It is really just not going to happen in a good way.
What happens depends very much on the original supply characteristics.
Imagine an LM350 was being used. Datasheet here. This is essentially an LM317 on steroids. Good for about 3A in most conditions and 4.5a IN MANY, deep-ending on application. 3A guaranteed. Fig 2 shows that it is good for 4.5A for a Vin-Vout differential of 5 to 15V depending on other issues. It can be run up near its current limit with good regulation. If being run at 3A and if the drop across it is not too high and it is well heatsunk it will not be hot and intermittent peaks of 4.5A will be provided. Do this too often and the temperature will rise and figs 1,4,5 and a few things unshown will affect how it behaves. First off Vout will start to droop on peaks and a capacitor on the output will help it serve the load. Increasing drOop and longer peaks and the capacitor will be called on to do more. If the IC decided to completely cut out for a moment (which it is unlikely to ever do) as long as T x I / C does not exceed the voltage droop which is acceptable the capacitor will do the whole job. Restore Iout to 3A and the capacitor will recharge until next time.
There really isn't a single '2N2222' transistor, since each manufacturer just seems to use XYZ_2N2222 as a part number for a generic 'NPN silicon amplifier transistor'.
Hence the 2N2222 pin-out is sadly not standardized. Most 2N2222s have the orientation which your diagram shows, where if the flat side is facing you, the pins are E-B-C respectively. However, some manufacturers have the pins going C-B-E for the TO-92 package. For example, the ON Semi P2N2222A variant: datasheet
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Best Answer
For a circuit to oscillate it is necessary to have a LOOP GAIN of unity (Barkhausen criterion). Hence, the amount of output voltage that is fed back to the input of the amplifying unit depends on the gain (magnitude and phase!). In the present case, only a part of the output voltage is created across the impedance seen at the emitter node because of the voltage division rule. This emitter node impedance is \$re=r6||(1/gm)\$ with \$gm=transconductance\$.
For the circuit shown, the exact calculation of loop gain is rather involved since the feedback circuit (C5-re) acts as an additional load for the transistor and, hence, must be considered for gain calculation (in parallel to the LC tank). Something similar applies to the phase. Because the feedback path (high-pass C5-re) creates a certain phase shift (the capacitor C5 CANNOT be regarded as a short for "high frequencies") the resulting oscillation frequency will NOT be identical to the resonant frequency of the LC tank. Instead, there will be a small frequency shift.