BJT saturation

saturated if increasing Ib no longer increases Ic. For example, if I have a circuit where the the collector of NPN is at +5V through a 1K resistor, maximum Ic is 5ma. Assuming hfe is 100, above 50uA of Ib doesn't increase Ic due to 1K resistor. Is this the saturation point for the transistor? If this is the case

something here?

The definition is a bit fuzzy. A transistor can be said to be saturated if the Vc:Ic curve is in the ohmic region, or, maybe, if the collector voltage is below the base voltage.

I don't much like the definition that you quote, "increasing Ib no longer increases Ic." That's too restrictive.

be saturated if increasing Ib no longer increases Ic. For example, if I have a circuit where the the collector of NPN is at +5V through a 1K resistor, maximum Ic is 5ma. Assuming hfe is 100, above 50uA of Ib doesn't increase Ic due to 1K resistor. Is this the saturation point for the transistor? If this is the case

something here?

You're missing an objective. What are you trying to accomplish? Discussing a definition does little to make an undefined circuit work.

be saturated if increasing Ib no longer increases Ic. For example, if I have a circuit where the the collector of NPN is at +5V through a 1K resistor, maximum Ic is 5ma. Assuming hfe is 100, above 50uA of Ib doesn't increase Ic due to 1K resistor. Is this the saturation point for the transistor? If this is the case

something here?

Saturation depends on V_CE, which is mostly governed by the external circuit. There's a built-in junction potential due to carrier diffusion across the junction, and when V_CE get small enough to cancel out that potential, the collector is unable to attract any more carriers, so I_C stops increasing.

What exact terminal voltages that happens at depends on the external circuit and also on the transistor characteristics, e.g. the bulk resistances of the emitter and collector.

But saturation is more of a gradual phenomenon--for practical purposes, a transistor goes into saturation when its beta drops by a factor of maybe 10.

Cheers

Phil Hobbs

saturated if increasing Ib no longer increases Ic. For example, if I have a circuit where the the collector of NPN is at +5V through a 1K resistor, maximum Ic is 5ma. Assuming hfe is 100, above 50uA of Ib doesn't increase Ic due to 1K resistor. Is this the saturation point for the transistor? If this is the case

Oh, in that specific case, increasing Ib likely does decrease Vc so also increases Ic a little. It would probably take a lot more base current to hit the minimum Vc point.

Get a couple of transistors and try it.

On a sunny day (Mon, 25 Mar 2013 08:47:51 -0700 (PDT)) it happened snipped-for-privacy@gmail.com wrote in :

The maximum current does. But the transistor is supposed to be saturated if Vce drops to a few hundred mV (approx 250 mv) for a Si transistor. It will not go much lower, no matter how much more base current you put into it. It is important to notice that this is less than the approx .7 volt Vbe, so the Vce in saturation is lower than Vbe.

In your example if the collector resistor is low enough the transistor may self-destruct before Vce can get that low (Ic x Vce + Ib x Vbe = dissipation). Also beta may be different at different values of Ic.

be saturated if increasing Ib no longer increases Ic. For example, if I have a circuit where the the collector of NPN is at +5V through a 1K resistor, maximum Ic is 5ma. Assuming hfe is 100, above 50uA of Ib doesn't increase Ic due to 1K resistor. Is this the saturation point for the transistor? If this is the case

something here?

A simplistic view, adequate for circuit analysis, is that the B-C junction forward biases and "steals" base drive, sort of like a "too late" Baker clamp ;-) ...Jim Thompson

In some cases, you can get Vc down into the single digits of millivolts. In the over-driven emitter follower config, you can get saturation voltage either side of zero.

id t=

I h=

isto=

reas=

r? I=

ssin=

dred mV

If you are seriously interested in getting a low saturation voltage, you use an inverted high-current-gain transistor, driving your base current through the base collector junction and treating the emitter as the collector.

Old-fashioned textbooks still cover this.Trevor. H. Wilmshurst's "Analog Circuit Techniques" seems to be one of them.

t=

h=

resisto=

increas=

I=

missin=

mV

Once upon a time, such devices were made in a lateral layout configuration, were symmetrical, and had low drop, used for choppers... IIRC they were called INCH?

Some discussion here...

...Jim Thompson

When looking for a definition I can easily explain and others can precisely understand without equivocation, that's the view I take.

Jon

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

Power transistors don't saturate?

Silly Jimmy ;-)

Tim

Well, he *said* he was simplistic.

mV

But s*it for gain..

Jamie

t=

h=

resisto=

increas=

I=

missin=

mV

That's not what I meant.

Imagine an NPN emitter follower: collector at some positive rail voltage, and some load like a resistor from emitter to ground. Now start pulling the base up. It acts like a regular emitter follower until Vb gets about 0.5 above the collector. Keep pulling the base positive. For many transistors, Ve can "saturate" to Vc plus or minus some tens of millivolts, tweakable to zero by trimming the base drive. The "forced beta" will be low, maybe less than 1.

I once built a bunch of 16-bit DACS using pnp-npn pairs like that, each pair feeding a wirewound resistor R-2R ladder network. We trimmed the base currents on the four MSBs to fine-tune the accuracy.

said t=

if I h=

resisto=

increas=

stor? I=

missin=

hundred mV

We'll forgive your ignorance.

said t=

I h=

resisto=

increas=

transistor? I=

missin=

hundred mV

I showed you a way to make a transistor saturate to exactly zero volts, something that you didn't know before, and you respond by being the tedious ass that you love to be.

That's a nice technique.

said t=

I h=

resisto=

increas=

transistor? I=

missin=

hundred mV

ass

Can you explain how that works? Where you tie the collector can't affect the saturation characteristics of the transistor???

Stated another way, can't you do the same thing with two diodes. Saturation not involved?????

said t=

I h=

resisto=

increas=

transistor? I=

missin=

hundred mV

ass

In the over-driven emitter-follower case, any extra base current flows into the collector, not the emitter. More base current increases the voltage drop in the c-b junction, in the ohmic bits of the silicon, and in the collector wire bonds, all of which pull up the emitter voltage. That lets you get to zero saturation voltage, which you can't do with a regular grounded-emitter config, even if you flip the transistor over.

Sure, but it wouldn't be as stiff or as accurate. I'm guessing the TC would be a lot worse. It's sort of academic these days, now that mosfet switches have been invented.