# Buffered Voltage Source

#### Teg Veece

Joined Mar 14, 2005
21
This forum has been a great help in the past. So I thought I post this problem I'm having. I'm trying to make a device that steps 15 V down to 5 volts using a voltage divider and then buffers this voltage so I can attach whatever load I want onto the end and it won't change the 5 V. I have to use only resistors and MOSFETS.

Any ideas? I think it can be done using only 1 MOSFET. An attached diagram would be really helpful.

Thanks.

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by Teg Veece@Mar 3 2006, 09:19 AM
This forum has been a great help in the past. So I thought I post this problem I'm having. I'm trying to make a device that steps 15 V down to 5 volts using a voltage divider and then buffers this voltage so I can attach whatever load I want onto the end and it won't change the 5 V. I have to use only resistors and MOSFETS.

Any ideas? I think it can be done using only 1 MOSFET. An attached diagram would be really helpful.

Thanks.
[post=14569]Quoted post[/post]​
Do you have a specific p-channel or n-channel mosfet in mind for your design?

hgmjr

#### Papabravo

Joined Feb 24, 2006
19,309
Originally posted by hgmjr@Mar 3 2006, 06:15 PM
Do you have a specific p-channel or n-channel mosfet in mind for your design?

hgmjr
[post=14591]Quoted post[/post]​
As I think about this problem I am bothered by "then I can attach whatever load I want". I can imagine a situation where a load attached to the 5V output would draw so much current through the upper leg of the divider that it would let the magic smoke out. I think a garden variety 7805 voltage regulator would be cheaper than a MOSFET and do a better job of making 5V from 15V up to about 1 Amp. So back of the envelope, (15V-5V)*1 Amp = 10 Watts The TO-220 package with a heat sink can effectively dissipate that much power.

If this is only a homework challenge then ignore this response.

#### windoze killa

Joined Feb 23, 2006
605
Originally posted by Papabravo@Mar 4 2006, 10:42 AM
As I think about this problem I am bothered by "then I can attach whatever load I want". I can imagine a situation where a load attached to the 5V output would draw so much current through the upper leg of the divider that it would let the magic smoke out. I think a garden variety 7805 voltage regulator would be cheaper than a MOSFET and do a better job of making 5V from 15V up to about 1 Amp. So back of the envelope, (15V-5V)*1 Amp = 10 Watts The TO-220 package with a heat sink can effectively dissipate that much power.

If this is only a homework challenge then ignore this response.
[post=14607]Quoted post[/post]​
He is not allowed to use regulators, only MOSFETs and resistors.

But. this ofcourse can be done very easily by biasing the MOSFET as a regulator. I would give you an example but I am not at work so I don't have protel at the moment. But if to do a dredge of the net i am sure there will be a thousand designs for it. Of course you must remember that there is a finite limit to how much load you can put on any power supply.

#### Spoggles

Joined Dec 2, 2005
67
Hello:

As long as you select the approproate devicie, you should be able to supply a wide varation of loads. The problem with using the "source follower" circuit attached (also see emittter follower) is the regulation is only as good as your 15volt source, and JFETS need at least 2 or 3 volts to bias the gate-source enough to get source to drain current.

You need to observe P=IE and keep the device's power specs in mind. One of the problems with emitter (source) followers is that they do not saturate, so you need to allow for this. Most power handling specs assume that no voltage is dropped in the device. Using P=IE should work well for you tho. Also be aware that you can drop some voltage in the reistor feeding the Drain.

Hope this helps.

I am having trouble gettin to this attachment. oh well later

Spoggles