It seems by the characteristic curve of a BJT, that it behaves as a resistance in saturation region.

BJT characteristic curve

Is this conclusion true? and if so how can I calculate this resistance (seen from collector) from datasheet values?

  • \$\begingroup\$ A BJT is NOT a FET or MOSFET. a FET has a drain-source resistance. if you know the load you gonna connect between the Vcc and the collector (assuming NPN). and you can calculate the voltage drop on your load on a given current. you can calculate a virtual resistance by (Vcc - Vload) / I if you have perfect resistor as load (Vcc - R*I)/I ==> Vcc/I - R \$\endgroup\$ – on8tom Mar 4 '16 at 8:11

Indeed that is true, a BJT behaves somewhat like a resistor when it is in the saturation region.

But it is not a very "well behaved" resistor, it is a very non-linear one. You can explain this by considering what would happen if you would apply an increasing voltage across this resistor. Then you would be increasing \$Vds\$ right ? Above a certain voltage \$Vdssat\$ the transistor will leave the saturation region and it will start behaving like a current source. Of course there's no sharp change from saturation and current mode behaviour, it's a gradual change.

The "value" of the "saturation resistance" of a transistor is therefore very dependent on \$Vbe\$ and \$Vds\$ and also the transistor's own properties like \$Hfe\$ (beta).

So in practice you cannot rely much on this "saturation resistance" having a certain value. If you just want to use the transistor to switch something on and off, the actual value does not matter so much as long as it is low enough. And that is also how saturation mode is used in general, a mode you use when you use a BJT to switch something on/off or short a signal to ground for example.

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  • \$\begingroup\$ In other words if you are switching current to a lamp or motor or relay you are fine (up to the rated current) but if you are switching an audio signal, you are adding some distortion, and if you are switching an RF signal you may be compromising its sensitivity. \$\endgroup\$ – Brian Drummond Mar 4 '16 at 12:50

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