# Help with explanation on this transistor behaviour

Hi, I found the above circuit on a website where it was said to produce fade in effect for the LED's. I tested the circuit out in my breadboard with R2 as 1K and C1 as 1000uF and R3 as 100 ohms.

The circuit indeed works. In that website it was explained as when the power is ON capacitor will start charging through R1 and once it reaches the base bias voltage of 0.7V. It activates the base of transistor Q1.

1) But I couldn't figure why LED's are fading in rather than lighting up in an instant. Because as far as I know when the base bias voltage is reached transistor switches ON instantly allowing the flow of current from collector to emitter, How LED's are fading in ?

2) Which state transistor is acting in this circuit? For saturation base voltage should be greater than collector and emitter but neither of this is true in the above circuit but still current flow exists between collector and emitter?

• It's an emitter follower design. When the base voltage reaches two LED drops plus one base-emitter drop, the BJT turns on. Once that happens, and as the voltage on $C_1$ continues to rise, the remainder voltage appears across $R_3$ and yields an increasing current for the LEDs. Once the base reaches the collector voltage (or a little bit below it) the increasing current stops and the LEDs are on at full brightness. – jonk Dec 22 '18 at 3:50
• Look at the Ve node and work out what happens there as the capacitor charges to bring Vbe near 0.7V. Remember as Vbe nears the threshold voltage, current will start to flow. That should start to give you some ideas on further analysis. – Daniel Dec 22 '18 at 3:51
• FYI tinyurl.com/ybhgtnhg press reset when running – Sunnyskyguy EE75 Dec 23 '18 at 19:44

## RC timing

Ignoring the BJT and LED portions of the circuit, the following shows you how to approximate the RC charging process. The left side is from your schematic. The right side is the Thevenin equivalent (same thing, just slightly simplified):

simulate this circuit – Schematic created using CircuitLab

Looking at the right side, find that $$\V_\text{TH}=9\:\text{V}\cdot\frac{R_2}{R_1+R_2}\$$ and that $$\R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$$. It's common practice to invent (create) a new variable: $$\\tau=R_\text{TH}\cdot C_1\$$ (multiplying resistance in Ohms times capacitance in Farads yields units in seconds.) This is done because much is known from this simplification. In $$\1\tau\$$ the voltage will rise to about 63% of the supply voltage (the Thevenin value in this case); in $$\2\tau\$$ it's 86%; and in $$\3\tau\$$ it's about 95%. Most of the work has been done in about $$\3\tau\$$ and this fact can be used as a guide to determine when there's very little change remaining.

(This ignores the fact that the base of the BJT will also require some current and therefore affects the timing just mentioned. But if $$\R_\text{TH}\$$ is low enough in value, the base current of the BJT won't matter enough to worry about.)

If the voltage across $$\C_1\$$ starts at $$\0\:\text{V}\$$ when $$\t=0\$$, then $$\V_{\text{base}\left(t\right)}=V_\text{TH}\cdot\left(1-e^{\left[\frac{-t}{\tau}\right]}\right)\$$.

(This allows you to work out what values of $$\V_\text{TH}\$$, $$\R_\text{TH}\$$ and $$\C_1\$$ you want, once you have decided how much current your LEDs will require and what kind of base current will be needed for that.)

Let's add the so-called emitter-follower BJT arrangement, now:

simulate this circuit

Here, the BJT's emitter will simply "follow" the base voltage, less one diode drop. So as the voltage at the base follows the $$\V_{\text{base}\left(t\right)}\$$ equation and rises upward, so also will the emitter rise upward too. This arrangement, though, uses the collector to source a lot of added current, if needed, because the collector can source up to $$\\beta\$$ times the base current. So a tiny base current permits a very large, added collector current. This means that the emitter can source a great deal of current if the load attached to the emitter wants it, without significantly affecting the emitter voltage which is following the base voltage. At this point, you can consider adding some kind of load to the emitter, knowing that the BJT will boost the current compliance.

The only caveat here is that the emitter follows the base and cannot exceed a voltage that is higher than one diode drop below the base voltage. And since the base voltage cannot exceed $$\V_\text{TH}\$$, the emitter voltage cannot exceed $$\V_\text{TH}-V_\text{BE}\$$. Luckily, you can arrange for different values of $$\V_\text{TH}\$$ by modifying the relationship between $$\R_1\$$ and $$\R_2\$$. But this is still a limitation to be aware of.

Here is the effective circuit you'd started with:

simulate this circuit

So what happens now?

Well, the LED diode drops must be met before anything happens. For old-style, typical red LEDs operating at $$\20\:\text{mA}\$$, this is about $$\2\:\text{V}\$$ per LED. For high intensity modern LEDs, this can easily reach $$\3.8\:\text{V}\$$ per LED, or more (when the currents are rather high, too.) So the base voltage must exceed the LED requirements (you show two of them, so twice the voltage requirement of just one.) It must also exceed the $$\V_\text{BE}\$$ required by the BJT, itself, as well. Any residual voltage that is left, after accounting for the two LEDs and $$\V_\text{BE}\$$, will appear across $$\R_3\$$ and this will set the current.

If you think about this for a moment, you'll realize that it will take some time for the base voltage to reach a point where the sum of the two LED voltages and $$\V_\text{BE}\$$ is finally met. Up until this point, nothing much happens and the LEDs will not be lit up. Once it is reached, $$\R_3\$$ will begin to limit the current as the base voltage continues upward (assuming it does so.) So the LED currents will "follow" this rise, once the voltage requirements have been met.

Ultimately, the final current in the LEDs will be $$\I_\text{LED}=\frac{V_\text{TH}-2\cdot V_\text{LED}-V_\text{BE}}{R_3}\$$.

A significant problem in this design is that the final current depends a great deal on (mostly) $$\V_\text{LED}\$$ and the BJT's $$\V_\text{BE}\$$ (less so.) And since the value of $$\V_\text{LED}\$$ can vary quite a bit from one LED to another (at the same current), this is a source of management error in the circuit. It's not terrible, since most folks won't really notice too much. But if $$\V_\text{TH}-2\cdot V_\text{LED}-V_\text{BE}\$$ doesn't leave much margin for $$\R_3\$$ to do its work, then the results can vary quite a lot from one circuit's part selections to another circuit with different parts.

This means this isn't really a good circuit. It may works after a fashion, assuming $$\V_\text{TH}\ge 2\cdot V_\text{LED}+V_\text{BE}\$$. But you may not get consistent results, time after time. Also, we've completely neglected what happens when the power supply is turned off (how does $$\C_1\$$ discharge itself so that it starts again where it left off, the last time?)

## Simple improvement

One way to improve the circuit is to remove the dependency of the LED current upon the LED voltage. To do that, simply move the two LEDs into the collector of the BJT:

simulate this circuit

This only works well if your BJT collector can remain at a voltage that is above $$\V_\text{TH}\$$. But since you can design things so that $$\V_\text{TH}\lt 9\:\text{V}-2\cdot V_\text{LED}\$$, it doesn't pose an insurmountable problem.

Here, the only delay until the LEDs start experiencing current is the short delay required until the voltage on $$\C_1\$$ (the base of $$\Q_1\$$) exceeds $$\V_\text{BE}\$$. (This is a shorter delay than before.)

Also, this removes a dependence upon the variability of your LED voltages. So that's a good thing for the circuit. So this is actually an improvement over the prior design.

## Final commentary

There are many additional ideas you could apply, if you wanted to make the circuit still better.

One problem is how to deal with what happens when the power supply is turned off. So far, we've assumed at the capacitor is completely discharged when the power is turned on and that the power supply is "instantly" turned on (it's fast, relative to the circuit shown.)

Perhaps also you'd like the LED to gradually decline in intensity when the power is turned off. This may require some temporary storage of charge to supply current to the LEDs once the power is removed, to allow a gradual decline. Or, perhaps, you'd like an instant turn-off of the LEDs.

There are a lot of such details that may complicate a practical circuit. But each of them can be resolved, if you clearly specify the behaviors you'd like in all cases. It's just a matter of carefully detailing what you expect in all cases and then working out how to achieve those details, one by one.

• Maybe the OP is depending on the cathode the lower LED connected to ground, then the 'improved circuit' can be converted to one with a PNP transistor. – joe electro Dec 25 '18 at 1:49
• @joeelectro Too bad the OP didn't specify that requirement. In that case, a PNP would be a good approach to try. There are a few other requirements the OP also didn't state, which would materially impact the design. I mentioned a few, but there are more of course. It's one of reasons I recommend writing much more than the bare minimum, when writing a question here. Even if it seems excessive, more information is better as it focuses the answers. Otherwise, others are left writing out a book in order to prepare an answer covering myriad unstated criteria. – jonk Dec 25 '18 at 1:57

This cct. acts both as a delay and a weak fade-in.

Since R1,R2 are equal =1k the base sees 4.5V after a time constant of 0.5k*1mF which reaches 64% of the 4.5V target = 3V in 0.5s . The emitter load of at least 1K * hFE of say 100 makes the load 100kohm have no load effect on the base voltage.

If the emitter had 2 blue or white LEDs of 3V each @20mA or 6V they would never turn on as the emitter cannot exceed 4.5-0.6=3.9V with Red or Yellow being around 2V x2 =4V it might conduct 0.5 mA in 5 seconds and not be very bright with the 1k current limiting emitter resistor.

A big improvement would be to remove R2 then the Vb rise to 9V and emitter can reach ~ 8.3V from Vbe drop and conduct (8.3V-4V)/Re=1k and thus reach 4.3mA LED current. But change R3 from your choice of 100 to 330. Otherwise 50mA!!

Then try reducing C to 10uF and increasing R1 to 50k to use a more feasible cap size.

• a better improvement is to instead move the LEDs to the collector (but keep the resistor on the emitter) – Jasen Dec 22 '18 at 11:42
• @Jasen - Actually, that's not true. Using the transistor as an emitter follower reduces the gain, which would otherwise provide the sort of fast action the OP expects. Plus, the fact that the cap voltage has to get up past the threshold of the emitter-base junction and the LEDs means that turn-on occurs when the cap voltage is not rising as fast as it does initially. And the presence of R2 enhances the effect by reducing the peak voltage applied to the transistor. – WhatRoughBeast Dec 23 '18 at 16:25
• @WhatRoughBeast Actually is is true , as moving the LEDs to the collector reduces the dim threshold voltage to 0.6 instead of 0.6+2Vf and the Re as a current limiter stills raises Zin and reduces the inverting voltage gain to < 1 since Rc (equiv) / Re <1 at some dim threshold. – Sunnyskyguy EE75 Dec 23 '18 at 16:35
• R2 actually degrades the performance by reducing the current and time constant and does not improve the linearity by reducing peak voltage. The Emitter voltage is your current sink that follows the base voltage above 0.6. But I'm sure you realize this. – Sunnyskyguy EE75 Dec 23 '18 at 17:21
• @WhatRoughBeast, R3 still provides degenreation, - just the load has been moved).moving the LEDs mostly reduces the threshold,(because leds ar mostly voltage sources) so the gain is basically the same (it has been reduced by 1 as I_b do longer thow through the LEDs) but the main effect is the threshiold reduces down to where the charge rate of the capacitor is more linear, yeah the charge rate is also faster, but that can be fixed by using a larger R1 and R2, the only disadvantage is the slight dip in brightness after saturation. – Jasen Dec 23 '18 at 19:16