Rds(on) of mosfet. - Electronic Circuits Projects Diagrams Free

alphadog · 14th February 2009, 05:29 PM

I read in a datasheet of mosfet that its Rds(on) 1.35ohm @ VGS=10V, IDS = 2.7A.

When in ON state, is IDS = VDS / RDS(on) ?

Thank you.

kchriste · 14th February 2009, 07:11 PM

Basically, yes. VDS should be fairly low at that point and IDS should be within spec and limited by the design of the circuit.

alphadog · 14th February 2009, 08:48 PM

Thanks.

Correct me please if i'm wrong, in saturation state, Rds is very large, then how come IDS is maximal (for constant VGS) in sturation state?

ronsimpson · 14th February 2009, 09:04 PM

In saturation state RDS is very small.

alphadog · 14th February 2009, 10:01 PM

But in stauration state, Rds = VA / IDS, isnt it?
And VA (Early Effect Voltage) is very large, therefore Rds is very large also.

Where am i wrong?

ronsimpson · 14th February 2009, 10:45 PM

The numbers from a data sheet do not mean the FET will all wise work like that.

In your example from the first post: 10 volts is applies from G-S. A current source (2.7A) is applies to the D. The FET looks like a 1.35 ohm resistor. The voltage D-S is 3.6 volts, under these conditions.

Appling 10 volts to the gate will not cause 2.7 amps. It will cause the FET to turn on and look much like a 1.35 volt resistor. The current is a function of what is on the D.

Early Effect Voltage???? Are you talking about the first nano-seconds of turn on? Saturation is talking about being on (DC) not AC.

Roff · 15th February 2009, 01:09 AM

Quote:

Originally Posted by ronsimpson

In saturation state RDS is very small.

This is a common misconception for people who cut their teeth on BJTs. Saturation in a MOSFET is basically opposite of that in a BJT. In a MOSFET, saturation is the high-impedance region (active mode) of the characteristic curves.

Chaerl · 16th February 2009, 04:20 PM

I know the early effect voltage (VA) is normally applied with BJT's. I don't know if it is applicable with FET's. It is actually the variation of the width of the base as the reverse voltage between collector base varies. Typically, the gate of the FET's as virtually isolated, high impedance input.

Roff · 16th February 2009, 04:30 PM

Quote:

Originally Posted by Chaerl

I know the early effect voltage (VA) is normally applied with BJT's. I don't know if it is applicable with FET's. It is actually the variation of the width of the base as the reverse voltage between collector base varies. Typically, the gate of the FET's as virtually isolated, high impedance input.

A search finds several references to the Early effect in MOSFETs. It's a different effect, but with similar results.

Chaerl · 16th February 2009, 04:57 PM

Quote:

Originally Posted by Roff

A search finds several references to the Early effect in MOSFETs. It's a different effect, but with similar results.

I see, I try to dig on that for more details.

alphadog · 17th February 2009, 03:48 PM

similar to the way the VA describes the linear dependency of IC on VCE in saturation region in BJT, it also describes the linear dependency of IDS on VDS in saturation region in MOSFET.

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14th February 2009, 05:29 PM	#1
alphadog	Rds(on) of mosfet. I read in a datasheet of mosfet that its Rds(on) 1.35ohm @ VGS=10V, IDS = 2.7A. When in ON state, is IDS = VDS / RDS(on) ? Thank you.

14th February 2009, 07:11 PM	#2
kchriste	Basically, yes. VDS should be fairly low at that point and IDS should be within spec and limited by the design of the circuit. __________________ Inside every little problem, is a big problem trying to get out.

14th February 2009, 08:48 PM	#3
alphadog	Thanks. Correct me please if i'm wrong, in saturation state, Rds is very large, then how come IDS is maximal (for constant VGS) in sturation state?

14th February 2009, 09:04 PM	#4
ronsimpson	In saturation state RDS is very small.

14th February 2009, 10:01 PM	#5
alphadog	But in stauration state, Rds = VA / IDS, isnt it? And VA (Early Effect Voltage) is very large, therefore Rds is very large also. Where am i wrong?

14th February 2009, 10:45 PM	#6
ronsimpson	The numbers from a data sheet do not mean the FET will all wise work like that. In your example from the first post: 10 volts is applies from G-S. A current source (2.7A) is applies to the D. The FET looks like a 1.35 ohm resistor. The voltage D-S is 3.6 volts, under these conditions. Appling 10 volts to the gate will not cause 2.7 amps. It will cause the FET to turn on and look much like a 1.35 volt resistor. The current is a function of what is on the D. Early Effect Voltage???? Are you talking about the first nano-seconds of turn on? Saturation is talking about being on (DC) not AC.

16th February 2009, 04:20 PM	#8
Chaerl	I know the early effect voltage (VA) is normally applied with BJT's. I don't know if it is applicable with FET's. It is actually the variation of the width of the base as the reverse voltage between collector base varies. Typically, the gate of the FET's as virtually isolated, high impedance input.

17th February 2009, 03:48 PM	#11
alphadog	similar to the way the VA describes the linear dependency of IC on VCE in saturation region in BJT, it also describes the linear dependency of IDS on VDS in saturation region in MOSFET.