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bunalmis
Joined: 03 Jan 2003
Posts: 254
Helped: 5
Location: Turkey
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 24 Feb 2006 12:38 direction of current about nmos |
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You see simple schematics. Gate voltage of MOS is higher than threshold value.
Current source reverse connected and body diode conduct.
I have a question. Does Mos conduct ? ( Is<> 0 ?)
If yes,
What we can says ratio for Id/Is ?
Can MOS take all current to self body? ( ID=0 ?)
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VVV
Joined: 26 Nov 2004
Posts: 1584
Helped: 290
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25 Feb 2006 4:05 Re: NMOS and reverse direction current |
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Yes, the MOSFET will conduct in the reverse direction.
Tthe body diode has exactly the same current rating as the MOSFET, so it can take the entire current.
However, if you turn on the MOSFET (normally, by applying a gate voltage higher than the threshold with respect to source for an N-ch device), then the current will flow through the channel rather than the body diode.
If the current is high enough so that the voltage drop across the channel resistance is sufficient for the body diode to turn on, then the diode will conduct some of the current. The actual percentage can be determined approximately from the diode VI curve.
Generally, however, the Rdson is selected such that the diode does not turn on when the MOSFET is fully enhanced.
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bunalmis
Joined: 03 Jan 2003
Posts: 254
Helped: 5
Location: Turkey
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25 Feb 2006 15:17 Re: NMOS and reverse direction current |
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Thanks vvv
Can we say?
if Rds on value is very low and (Idmax * Rd) << 0.6 (I accepts 0.6V is forward voltage drop of diod)
Rds*Is = ? Rds*(-Is) at the full range
Are 1 and 2 Qaudrant fully symetric?
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VVV
Joined: 26 Nov 2004
Posts: 1584
Helped: 290
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26 Feb 2006 5:11 Re: NMOS and reverse direction current |
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| The first and second quadrant are not fully symmetrical, because of the diode. Eventually, it will turn on. Without the MOSFET being one, the diode can still conduct. Therefore the two quadrants are not symmetrical.
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tekno1
Joined: 26 May 2005
Posts: 85
Helped: 5
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02 Mar 2006 1:46 Re: NMOS and reverse direction current |
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Bunalmish 
This circuit is a non practical trick circuit (i.e. academic)
There is a current loop where a current source injects a current to n-ch transistor's Source and Substrate, Drain is at lower potential connected to current source to close the current loop.
Fact 1: Gate is at a higher potential than Source. Since Fact 1 there is an active channel between Source and Drain i.e. channel is on so it acts like a resistor between Source and drain (Vgs > VT).
Note: Gate potantial is higher than Source and Drain potentials. Source potential is higher than Drain potential since current flowing throuh Source to Drain. In this configuration Drain acts as Source and Source acts as Drain. So actual Vgs is between Gate and Drain. Vgs is actualy Vgs + Von (details details...).
Fact 2: Vgs voltage source just supply a gate potential but no current flows throug it (ignoring Gate leakage here. Even if we consider it, it is very small unless current source is in the same magnitude we can ignore it).
If we visualize 3-D layout with relevant parasitics what we see is that n+ Source shorted to P+ Substrate externally (so no substrate to Source diode action) and there are two current paths to Drain (where current source connected so that current can flow through)
First path is a pn diode between Substrate(Source) to Drain and second path is through an active n+ channel between Source(Substrate) and Drain.
External diode also paralel connected to internal Source Drain pn diode. Now since channel is a Resistor so unless IxR > Vt of pn diode (or external diode) all the current flows through the channel (voltagedrop through channel will be low). (Id/Is=0, Id takes all the current)
If IxR > Vt of internal pn diode or external diode than most current flow through diodes and assuming that they have same Vt, current amount in each diode will be determined by internal diode resistors acting as ballasts. (Id/Is depends. channel, internal diode and external diode all conduct. Since part of the Id goes through internal diode. If we assume internal and external diodes have same Vt and their resistances are the same -- including connection resistances--- than Ichannel+Iinternal_diode=Id and I internal_diode=Is)
If IxR=Vt than diodes and channel will all conduct. ( Id and Is need to be calculated and for this two diode's and channel's equivalent circuits must be considered. Two diodes and channel re all parallel connected)
I think questioner means to ask the first case where IxR < Vt of diodes (other cases complicated and need more variables being determined).
I hope this explanations will take you out of bunalmis situation. I felt sorry.
If you have any question please ask.
regards,
tekno1
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