Here's the (non ideal) diode equation I referred to.

Following is a graph I made of the ∆V across a real pair of diodes in a rectifier, using the diode equation to fit the data. You can see that in the "normal" range of use (up to about 1A for this recitifer), the ∆V is nearly constant. But at high or low current, it deviates more.

 

Most transistors turn on well when the base current is 1/10th the collector current.

Got the graph, but I'm a little confused about the following. With a 1K resistor on the LED, the transistor turned full on. As I'm increasing the value of that resistor, is there a point at which the BE voltage drops below 0.6? If yes, what is that resistance? In this case for example?

 

I'm a little confused about the following. With a 1K resistor on the LED, the transistor turned full on. As I'm increasing the value of that resistor, is there a point at which the BE voltage drops below 0.6? If yes, what is that resistance? In this case for example?

When the value of the 1k resistor is increased then the LED and transistor collector current is reduced. Then the base current is reduced. Then the VBE of the transistor is reduced.

The lines on the graph appear to bend upward at higher currents but become straight at lower currents. So I spliced together two logarithmic graphs and sketched a guess. The collector current when VBE is 0.5V is about 0.5uA then the CD4026 output voltage is 9.0V and the led voltage drop is another guess at 1.2V. Then the resistor in series with the LED is about (9V - 1.2V - 0.5V)/0.5uA= 14.6M ohms.

 

Your transistor is used as a saturated switch, not as a linear amplifier so its base current should be 1/10th the collector current.

If you're talking about the emitter follower configuration, the transistor always operates in its linear region and is not saturated. The base current that it draws is defined by the collector-emitter current and the transistor's current gain, and saturation is not involved.

 

If you're talking about the emitter follower configuration, the transistor always operates in its linear region and is not saturated. The base current that it draws is defined by the collector-emitter current and the transistor's current gain, and saturation is not involved.

You are correct and I am wrong, BUT I think the current gain is reduced when the VCE is a little less than 1V. The current gain also reduces at low currents.

Years ago I learned that a saturated transistor has its collector-base junction forward-biased.

 

You are correct and I am wrong

Well, that's a very dignified response
I often misspeak too. We have to get used to burying our egos. Thank you.

BUT I think the current gain is reduced when the VCE is a little less than 1V. The current gain also reduces at low currents.

Right.

Years ago I learned that a saturated transistor has its collector-base junction forward-biased.

Right. I think of a saturated transistor as one whose VCE is lower than its VBE, which amounts to the same thing, or whose IB is much higher than its IC divided by its HFE. Or a transistor whose IC will not increase significantly if its IB is increased significantly. "Saturation" is the perfect word for it.

An emitter follower can saturate if its base is brought higher than its collector, but generally the collector is connected to the highest voltage in the circuit so this can't happen, and I'd argue that it would stop being an emitter follower in that situation.

 

The lines on the graph appear to bend upward at higher currents but become straight at lower currents. So I spliced together two logarithmic graphs and sketched a guess. The collector current when VBE is 0.5V is about 0.5uA then the CD4026 output voltage is 9.0V and the led voltage drop is another guess at 1.2V. Then the resistor in series with the LED is about (9V - 1.2V - 0.5V)/0.5uA= 14.6M ohms.

Cool! I got it. By the way, what is the program you used to draw the LEDs and the 4026?

 

Cool! I got it. By the way, what is the program you used to draw the LEDs and the 4026?

I draw schematics with ordinary Microsoft Paint program.
I copy parts from other schematics and paste them into a new schematic.

 

Got it. Is Qucs good?

 

I have never seen QUCS before. Unfortunately there are many schematic drawing and simulation programs that people use and they are all different.