Mosfet and BJT switches

Thread Starter

electr

Joined May 23, 2009
49
The following relay driver can use either a BJT or a MOS as a switch (of course that when using the MOS, the base resistor isnt needed).

When using a BJT, you would want the BJT to be saturated, so VCE is minimal, and therefore most of the VCC is applied to the coil.

My question deals with using a MOS as the switch.
You would also want VDS to be minimal in this case, so most the coil would take most VCC.

MOS in Linear region => RDS minimal => IDS*RDS minimal => VDS minimal

So that means that you would want the MOS to be in Linear region , right?

The thing is that i always heard that you'd want the device to be fully turned on (meaning saturated), so therefore i wanted to hear your opinion about it.
Thank you.

 
Last edited:

Thread Starter

electr

Joined May 23, 2009
49
Thanks for the advice.
I fixed the opening message.

I'd like to understand about the Mosfet rather than to get other solutions.
 
Last edited:

Thav

Joined Oct 13, 2009
82
Yes, you would probably want to get the MOSFET in the linear region. The best way to do this is to provide enough gate to source voltage to keep it out of saturation.

I forget exactly why the terms sort of switch between the two devices. It happens with amplifiers too. When making transistor amplifiers, you want the BJT "active" and the FET "saturated".
 

SgtWookie

Joined Jul 17, 2007
22,227
A standard N-channel enhanced MOSFET needs Vgs=0v (Vgs is voltage on the gate using the source as the 0v reference) to be fully turned off, and Vgs=10v to be fully turned on.

Logic level enhanced MOSFETs are fully turned on when Vgs=5v.

If Vgs is somewhere in between those levels, it will be in the "linear region", acting as a resistor, and will dissipate power as heat. MOSFET datasheets normally show a "threshold voltage"; this is a Vgs where the MOSFET will conduct a certain amount of current when Vds (voltage on the drain using the source as a 0v reference) is within a range of values.

Unless you are using an enhanced MOSFET in a linear amplifier, you want to get the gate to the ON or OFF level quickly.

If you are using a bjt (bipolar junction transistor) as a saturated switch, you want to give the base 1/10 of the desired collector current. This helps to ensure a low Vce (voltage on the collector using the emitter as a 0v reference) and minimal power dissipation in the transistor.
 

Ron H

Joined Apr 14, 2005
7,014
There seems to be some confusion about the meaning of "linear region" and "saturation region" as applied to a MOSFET. See the attached graph, which I copied from this file.
When operating the MOSFET as a switch, the operating point needs to be in the extreme lower left corner of the graph.
When operating the MOSFET as an amplifer, the operating point generally needs to be well into the saturation region.
 

Attachments

Thread Starter

electr

Joined May 23, 2009
49
@ SgtWookie :
it will be in the "linear region", acting as a resistor, and will dissipate power as heat.
The power dissipated in the MOS is VDS*IDS, right?
For a given current IDS, the VDS in the linear region is much smaller than the VDS in the saturation region.
There, less power will be dissipated in the MOS in the linear region.

Did you intend to say that when saturated, the power is not dissipated into heat?

@ RON
What makes you think there's a confusion?
 

Ron H

Joined Apr 14, 2005
7,014
@ RON
What makes you think there's a confusion?
You said:
So that means that you would want the MOS to be in Linear region , right?

The thing is that i always heard that you'd want the device to be fully turned on (meaning saturated), so therefore i wanted to hear your opinion about it.
The second sentence seems to be at odds with the first. I thought you were thinking that saturated has the same meaning with MOSFETs as with BJTs.
 

Thav

Joined Oct 13, 2009
82
Well, consider the image Ron posted. Say your peak current in your circuit could be 5A. If you only drive your FET with 2 or 3V (again, referring to that image) your peak current will never get to 5A, you'll hit ~3-4A and then Vds will start to go up without the current increasing much. So you might follow that Vgs=3 curve out to Vds=10 and still only have 3A = 30W.

If you drive Vgs to 7V you'll hit 5A at around Vds=1V for 5W. You do not want your FET to saturate in switching applications!
 

ELECTRONERD

Joined May 26, 2009
1,147
You do not want your FET to saturate in switching applications!
Thav,

Isn't the whole point of saturation used to make transistors act like an automatic switch? Unless you want the transistor (FET in this case) to be in it's linear region, and then if they are composed with other components, the entire circuit would make up the switching application? So the FET wouln't be the switch but the entire circuit would be the switch made up of transistors in their linear region? I think the denotion "switching" has gotten me confused; you would want the FET to saturate when it was in a processor which is basically composed of discrete transistor circuitry all in saturation to act as switches. So would that be considered a switching application?

I think I might be making this harder than it really is. :D

Austin
 

Wendy

Joined Mar 24, 2008
23,053
Thav,

Isn't the whole point of saturation used to make transistors act like an automatic switch? Unless you want the transistor (FET in this case) to be in it's linear region, and then if they are composed with other components, the entire circuit would make up the switching application? So the FET wouln't be the switch but the entire circuit would be the switch made up of transistors in their linear region? I think the denotion "switching" has gotten me confused; you would want the FET to saturate when it was in a processor which is basically composed of discrete transistor circuitry all in saturation to act as switches. So would that be considered a switching application?

I think I might be making this harder than it really is. :D

Austin
Loose the term "automatic" and you almost have it. The two conditions a transistor disappates the least heat is completely on or completely off. Saturated is just another word for completely on.
 

ELECTRONERD

Joined May 26, 2009
1,147
Loose the term "automatic" and you almost have it. The two conditions a transistor disappates the least heat is completely on or completely off. Saturated is just another word for completely on.
So transistors that are in their linear region are only "on" depending on the input of the signal. For example, If I have a sine wave going into a transistor circuit, and if the transistor was in it's linear region, then the transistor would only be "on" when the signal was "on" and when there was nothing (between the peaks of a sine wave) it would be off? That would mean you'd have to think of the input of the sine wave frequency and see what portion in the sine wave is off so you can calculate what part of the transistor should be conducting and what shouldn't (thus, in it's linear region for that frequency)?

Appreciate the help Bill,

Austin
 

ELECTRONERD

Joined May 26, 2009
1,147
I'll also add that when you determine the linear region for that frequency it may not always be correct. Suppose the input signal wasn't a perfect sine wave? How would you know what would be the linear region for the transistor, based on the input signal? I guess the really complex transistor stuff would be calculating when the transistor should be "on" and "off" according to the input signal. This is all a indeterminate conjecture, so I'd appreciate it if someone verified it.

Thanks,

Austin
 

Ron H

Joined Apr 14, 2005
7,014
Thav,

Isn't the whole point of saturation used to make transistors act like an automatic switch? Unless you want the transistor (FET in this case) to be in it's linear region, and then if they are composed with other components, the entire circuit would make up the switching application? So the FET wouln't be the switch but the entire circuit would be the switch made up of transistors in their linear region? I think the denotion "switching" has gotten me confused; you would want the FET to saturate when it was in a processor which is basically composed of discrete transistor circuitry all in saturation to act as switches. So would that be considered a switching application?

I think I might be making this harder than it really is. :D

Austin
Austin, go back and read my previous post, and look at the graph carefully!:)
 

ELECTRONERD

Joined May 26, 2009
1,147
There seems to be some confusion about the meaning of "linear region" and "saturation region" as applied to a MOSFET. See the attached graph, which I copied from this file.
When operating the MOSFET as a switch, the operating point needs to be in the extreme lower left corner of the graph.
When operating the MOSFET as an amplifer, the operating point generally needs to be well into the saturation region.
I don't want to ramble on and on, so I'll ask this directly.
  1. As I see it, the linear region is the smooth lines and the saturation region is the points between the lines and way to the left of the graph (seperated by the first line plot that's graphed). Correct?
  2. If I want to design a Class-A amp and put the transistor in its linear region, all I have to do is choose, let's say, 7.5mA (they don't specifically say whether it's in amps or milli-amps) and then choose 10V. I did that because it was on a straight line which is the linear region (or so I think at the moment). What about the gate and source voltages and currents, what should they be?
More confusion; you say:
When operating the MOSFET as an amplifer, the operating point generally needs to be well into the saturation region.
I thought it was supposed to be in its linear region if you want it to be a adequate amplifier!?

Thanks Ron,

Austin
 

thatoneguy

Joined Feb 19, 2009
6,359
Compare the MOSFET image from Ron with this one from the AAC e-book showing a typical BJT trace:







When operating in the "horizontal line" area of the graph the MOSFET is in "Constant Current" mode for a given VGS, where the transistor is in "Constant Current" mode for a given IBE.

Base Current to switch "ON" a BJT is roughly IC*0.1

Gate Voltage to switch "ON" a normal MOSFET is VGS ≈ VDS

Saturation is not the same as "ON".

Saturation in a BJT is when both internal diodes are fully forward biased, which coincides with "on".

Saturation with a MOSFET is defined as when the "pinch" in the channel between Source and Drain moves towards the Source. This is when the MOSFET is operating in Ohmic Mode, rather than RDSON=min.

 

ELECTRONERD

Joined May 26, 2009
1,147
Compare the MOSFET image from Ron with this one from the AAC e-book showing a typical BJT trace:







When operating in the "horizontal line" area of the graph the MOSFET is in "Constant Current" mode for a given VGS, where the transistor is in "Constant Current" mode for a given IBE.

Base Current to switch "ON" a BJT is roughly IC*0.1

Gate Voltage to switch "ON" a normal MOSFET is VGS ≈ VDS

Saturation is not the same as "ON".

Saturation in a BJT is when both internal diodes are fully forward biased, which coincides with "on".

Saturation with a MOSFET is defined as when the "pinch" in the channel between Source and Drain moves towards the Source. This is when the MOSFET is operating in Ohmic Mode, rather than RDSON=min.
thatoneguy,

I looked in the 2N3904 and BC548 transistors specs and I don't see that kind of graph shown. What should I look for instead, if I can't find it's quiescent point for class A operation?

Austin
 
Top