Hello. I have obtained a copy of this above circuit which is built using a blinking LED. The capacitor will slowly charge and eventually become an open circuit equivalent as no more direct current can flow through it, which causes the transistor to turn "off" meaning no current flows from the collector junction to the emitter junction because no more base current reaches the transistor. While this is happening, the blinking LED receives a voltage from the DC source to motivate the IC contained therein to blink repeatedly. The brightness of the blinking diminishes gradually as the transistor cuts "off" and the circuit is broken.

My question is how can I derive the DC and AC analysis of this circuit mathematically? For DC analysis, the capacitor can be replaced by an equivalent open circuit after a long time, which means that transistor is "off" and there is no conduction through any circuit elements. The voltage at the top of the LED is 14V. For AC analysis, there is no AC signal source, so do I have to replace the battery with a cosinusoidal source? How can I derive the differential equation relating the base current as a function of time to known circuit parameters? I found that capacitor current equals current through the bottom resistor plus current through the base junction, but all of these are functions of time and cannot be related to each other in any way.

Thank you for reading and good day.

 

With such large resistors you can't assume that the base current is a negligible fraction of the current in the RC circuit, so you will need a model for the base-emitter junction diode characteristic.

You aren't looking for an AC analysis, but for a transient analysis using the large signal models.

What is the purpose of the reverse-biased diode? It seems like it would only have whatever reverse leakage current is in it.

You can probably make a reasonable estimate that the 1 MΩ resistor has negligible current and that all of the capacitor current will flow into the base of the transistor until well after the LED turns off. That will leave you with a first order circuit.

 

Is the base emitter junction modeled as a Silicon diode? The current through the diode increases exponentially when forward biased above 0.7V.

The reverse-biased diode functions as a 0.7V drop in the forward biased region, so conducts no current when the voltage source is applied. If the voltage source is removed, and the capacitor is fully charged, the diode enters the breakdown region and conducts current in the opposite direction effectively discharging the capacitor. This ensures proper operation of the circuit when power is applied again.

If I apply conservation of energy using KVL:
V_capacitor + I_capacitor *100kohm +0.7V -14V = 0
dVc/dt +dic/dt *100k = 0
since ic = C dVc/dt, dic/dt = 47u *100k *d^2 Vc/dt^2
dVc/dt +4.7 d^2 Vc/dt^2 = 0
d^2 Vc/dt^2 +1/4.7 dVc/dt = 0

Is this the correct differential equation?
Vc(t) = k1 +k2 *e^(-t/4.7)

 

I think the reverse biased diode is a LED with an integrated blinker.

From J_Rod's post (emphasis added by me):
"While this is happening, the blinking LED receives a voltage from the DC source to motivate the IC contained therein to blink repeatedly."

 

I think the reverse biased diode is a LED with an integrated blinker.

From J_Rod's post (emphasis added by me):
"While this is happening, the blinking LED receives a voltage from the DC source to motivate the IC contained therein to blink repeatedly."

I don't think so. First, it is reverse biased. Second, then what is the point of the transistor in the circuit at all? I think the blinking LED is the top-right component. Note that he says that the blinking diminishes as the transistor is cut off. But hopefully he will clarify things.

 

Is the base emitter junction modeled as a Silicon diode? The current through the diode increases exponentially when forward biased above 0.7V.

The reverse-biased diode functions as a 0.7V drop in the forward biased region, so conducts no current when the voltage source is applied. If the voltage source is removed, and the capacitor is fully charged, the diode enters the breakdown region and conducts current in the opposite direction effectively discharging the capacitor. This ensures proper operation of the circuit when power is applied again.

If I apply conservation of energy using KVL:
V_capacitor + I_capacitor *100kohm +0.7V -14V = 0
dVc/dt +dic/dt *100k = 0
since ic = C dVc/dt, dic/dt = 47u *100k *d^2 Vc/dt^2
dVc/dt +4.7 d^2 Vc/dt^2 = 0
d^2 Vc/dt^2 +1/4.7 dVc/dt = 0

Is this the correct differential equation?
Vc(t) = k1 +k2 *e^(-t/4.7)

How does the diode enter breakdown when the voltage source is removed? If you removed the source and shorted it, then the diode would be forward biased and discharge the cap (most of the way).

The base current increases exponentially with Vbe both above and below 0.7 V, there is nothing magical about 0.7 V. At some point the exponential increase becomes more linear as the parasitic resistances begin to dominate.

 

I think the reverse biased diode is a LED with an integrated blinker.

From J_Rod's post (emphasis added by me):
"While this is happening, the blinking LED receives a voltage from the DC source to motivate the IC contained therein to blink repeatedly."

Thanks for the reply. I tried to indicate the LED with the squiggly rays in series with the 220 ohm resistor. I'm not really sure what the IC is, perhaps a 555 timer, but the datasheet indicated there was an IC built in the diode which was novel to me.

The gallium phosphide blinking LED should be forward biased since applied voltage 14V > 3.5V threshold, but the silicon diode below should not conduct while the capacitor charges since the p material connects to ground voltage reference while the n material connects to a positive voltage level. The transistor acts like a switch to turn the blinking functionality on and off.

 

How does the diode enter breakdown when the voltage source is removed? If you removed the source and shorted it, then the diode would be forward biased and discharge the cap (most of the way).

The base current increases exponentially with Vbe both above and below 0.7 V, there is nothing magical about 0.7 V. At some point the exponential increase becomes more linear as the parasitic resistances begin to dominate.

By remove the source, I mean that the battery is replaced by an open circuit. Since the capacitor fully charges, wouldn't this mean that the diode has -14V drop from +ve to -ve terminals?

 

By remove the source, I mean that the battery is replaced by an open circuit. Since the capacitor fully charges, wouldn't this mean that the diode has -14V drop from +ve to -ve terminals?

No. Why would it?

If the capacitor is fully charged then there is no current flowing out of the battery (since the transistor is off, no current flows through that path, either). So the circuit is in equilibrium with the battery and if you were to put a switch between the battery and the rest of the circuit it would have no current flowing in it and no voltage across it, so you could open and close it all day long and the circuit wouldn't know or care.

 

Oh okay, so to discharge the capacitor, the leads have to be connected or the diode has to provide a conductive path to discharge the capacitor.

 

Oh okay, so to discharge the capacitor, the leads have to be connected or the diode has to provide a conductive path to discharge the capacitor.

Yep. You could put a bleed resistor across it sized to give it a time constant about 10x the decay time when the circuit is in operation, too.