Current source with BJT

That is correct.

Show me the circuit.

What immediately comes to mind with a single BJT for this is:

+V | ,---, | L | | O | | A | | D | '---' | |/ Vin >----| NPN |>

| \\ / Rset \\ | --- ///

Where the LOAD current is determined by (Vin - 0.7V) / Rset, roughly.

If so, that's saturated operation and I don't imagine that working well for a current source purpose. In the rare case where the load needs almost all of the available voltage headroom in the above circuit, for example, you lose the ability to properly control the current through it as that excess saturation current flows through Rset as well and leaves a smaller and unknown portion for the load.

Which seems true to me, too.

It seems to me that it would be important for predictable design, if nothing else.

Jon

Hmmm, looks like I was wrong, when the circuit is set up and working properly CB is reverse-biased.

I stumbled upon this problem reading a book, it gives the schematic for a current source:

I wondered why he uses a negative voltage on the emitter and if I could get the same effect (constant current) if I do it like this:

It doesn't work. Both junctions are forward-biased which means that the Ic will be nowhere near Ie. Ib is very big, which brings the transtor in the core.

The above is my understanding. My question now is why? Why doesn't it work with the latter circuit but does with the former?

Is it because I connected R1 with R4?

Sorry if I sound too much like a newbie, it's s.e.basics after all :-)

In effect, you redefined ground as the voltage on the emitter resistor, rather than the voltage on the collector and base divider.

But you left the collector connected to this newly defined ground, instead of moving it to the voltage you now have on the base divider. You switched the voltage around on only a part of the circuit. You have to swap all power and ground connections when you switch from a negative to a positive supply (with respect to ground).

That's it.

Yes, it is. :-)

Your first circuit, the working one, is exactly the same as this:

: +V +V +V : | | | : | | | : | | | : | \\ \\ : | / R3 / R1 : ,--+ \\ 100 \\ 20 : | | / / : gnd | | | : | | | : | | | : | | | : | | | : --- | |/c Q1 : - V1 +----| 2N3904 : --- 15 | |>e : - | | : | | | : | | | : | | | : | | | : | \\ \\ : | / R2 / R4 : | \\ 100 \\ 200 : | / / : | | | : | | | : | | | : -V -V -V

Note the differences in schematic drawing, though. I almost always arrange for electron flow to go from the bottom of the sheet to the top. Think of this as a kind of "environment" which then allows signal to flow from left to right across the flow. I also almost always remove excess bus lines of voltage rails. To me, they make you "think" wrongly -- imagining that there is signal on those lines you may need to be aware of, which isn't true. There is nothing going on there that is worth knowing about when analyzing a circuit (usually), so putting wires there just means that your mind is distracted by them.

There are times when the exact path of a supply needs to be called out on the schematic, because some detail about it is important to note -- for example, that one circuit section needs to have some rail and signal tied to another rail and signal of another circuit section via a single conductor pair for noise immunity, etc. But this is rare and, in general, doesn't really help in understanding the circuit. So I generally remove supply buses.

Another small note is that you've tied the ground to the (+) side of V1. That's fine, just so long as you realize that this is atypical. Usually, folks tie the ground in a circuit like this to the other side. That noted, it's no problem and just sets the reference point for the node voltages.

Simple analysis of the working circuit might be the following approach (there are others):

Start with the R2/R3 divider. It sets Q1's base to 7.5V. With your reference point, this would be -7.5V. With low resistors like these (many tens of mA), I start by assuming a Vbe for Q1 of .7V to .8V. Let's go with .8V. So this means that Q1's emitter is about .8V less (well, with your ground, I mean 'more'), or about -8.3V. This sets the current for R4 at ((-15V)-(-8.3V))/200 or about 33.5mA. That current, if R1's implied voltage drop is possible, would flow through R1, too (less the base current.)

So what's the base current? About 1/200th of the collector current, if not saturated. Is it saturated? Well, let's assume that all the emitter current of 33.5mA flows through R1. This produces a voltage difference across it of 33.5mA * 20 or 0.67V. With your reference point, this is -.67V, of course. This means that the collector is reverse biased against the base, so it's definitely NOT in saturation. So we go with the beta of 200 and figure that the base current really is negligibly small (1/200th). So the collector current really is about the emitter current and our calculation for no-saturation holds.

The R2/R3 divider has 15V/(100+100) or 75mA. The tiny base current won't affect it much, so the actual base voltage is also likely to be the -7.5V that we first estimated it to be. Things are working out.

So that's about it.

Now for the second one:

: +V +V : | | ,----, : | | | | : | | | gnd : | \\ \\ : | / R3 / R1 : | \\ 100 \\ 20 : | / / : | | | : | | | : | | | : | | | : | | | : --- | |/c Q1 : - V1 +----| 2N3904 : --- 15 | |>e : - | | : | | | : | | | : | | | : | | | : | \\ \\ : | / R2 / R4 : | \\ 100 \\ 200 : | | / : | | | : | | | : | | | : gnd gnd gnd

Do you see what you did differently, now that the rats nest has been unwound? You connected R1's other end to the other side of the power supply. In doing so, you are pulling the collector into a forward biased condition with respect to the base. Start out with the same idea that the base is 1/2 of the supply, or 7.5V. (Reference has changed, so this is now +7.5V.) But the collector sees only the negative side of V1 through R1 so it cannot go reverse biased with respect to this, only forward biased. Which it does do.

Jon

Wow! Thanks for the very detailed reply!

I understand my error now, it was fairly obvious after all.