Could you send me a schematic?

That depends on your still-awaited answer to the question in post #4 (and repeated in post #11).

 

The voltage output will be somewhat sensitive to load and temperature.
LTspice simulation of example circuit below for load varying from about 1A to 0.1A:

The output voltage variation with load is significantly smaller than I thought it would be.

Nice sim.

ak

 

Can I use a voltage divider instead of a potentiometer. If so, what voltage do you think is best

The gate-source operating voltage can typically vary a volt or more from unit-to-unit of the same part type, so you can either use a potentiometer to compensate for that, or select the voltage divider resistor values by trial and error (select at test) to get the output voltage you want over the desired load current range with the actual MOSFET you will use.

What do you think the tolerance is on a cell phone voltage wise?

I would expect you want the output voltage to be within ±5% of 5V (±0.25V).

 

That depends on your still-awaited answer to the question in post #4 (and repeated in post #11).

I don't have any good BJTs. The supply voltage is a good 12.5V switcher from a medical device. I'm sure there is very little, if any variation.
The 2 fets are>
D30NF06L (surface mount)
And
H7P0601DS
?

That depends on your still-awaited answer to the question in post #4 (and repeated in post #11).

 

The gate-source operating voltage can typically vary a volt or more from unit-to-unit of the same part type, so you can either use a potentiometer to compensate for that, or select the voltage divider resistor values by trial and error (select at test) to get the output voltage you want over the desired load current range with the actual MOSFET you will use.

I would expect you want the output voltage to be within ±5% of 5V (±0.25V).

The gate-source operating voltage can typically vary a volt or more from unit-to-unit of the same part type, so you can either use a potentiometer to compensate for that, or select the voltage divider resistor values by trial and error (select at test) to get the output voltage you want over the desired load current range with the actual MOSFET you will use.

I would expect you want the output voltage to be within ±5% of 5V (±0.25V).

Thank you very much for the info and Sims! The fets I have are>
D30NF06L
And
H7P0601DS

 

I don't have any good BJTs.

Well what bjts do you have?

 

Small L6 npns. Surface mount. They have a 5.V Max at the base. Can't compensate for the 0.6V drop.

 

This is a common misunderstanding. The 5V is a limit on the negative voltage on the base with respect to the emitter.

The voltage from the base to the emitter is normally around 0.6 to 1V when the transistor is on, and less when off. Anything higher than that will cause enough current to flow to damage the transistor. The base is normally protected by a resistor, which may then be connected to a higher voltage.

Bob

 

This is a common misunderstanding. The 5V is a limit on the negative voltage on the base with respect to the emitter.

The voltage from the base to the emitter is normally around 0.6 to 1V when the transistor is on, and less when off. Anything higher than that will cause enough current to flow to damage the transistor. The base is normally protected by a resistor, which may then be connected to a higher voltage.

Bob

So your saying I can put 5.6 V to the base of that L6? To get 5V out?

 

No. You drive current, limited by a resistor, through the base-emitter junction in the forward direction and the base-emitter voltage developed is about 0.6V. The junction must never be reverse-biased by more than 5V or breakdown will occur.

 

No, read it again. Even 1V would destroy a BJT.

You can put a resistor on the base and 5.6V on the other side of the resistor.

Do you understand why a resistor is needed with an LED? The transistor base emitter junction is just like an LED with a forward voltage of about 0.7V.

Bob

 

Wait maybe we are misunderstanding you. In an emitter follower circuit you can put 5.6V on the base and the emitter will rise to about 5V, depending on the load.

Bob

 

The output voltage variation with load is significantly smaller than I thought it would be.

Large power MOSFETs have a high transconductance which makes for a small Vgs change with a change in output current for currents much smaller than their maximum.

 

No. You drive current, limited by a resistor, through the base-emitter junction in the forward direction and the base-emitter voltage developed is about 0.6V. The junction must never be reverse-biased by more than 5V or breakdown will occur.

That's what I thought. I've read the data sheet on those a while back and still remember some of the numbers. But here's a question! Will they work on Mr Crutchows differential regulator? It has a voltage divider setting a reference to 5 volts at the base.

 

Large power MOSFETs have a high transconductance which makes for a small Vgs change with a change in output current for currents much smaller than their maximum.

Will those little L6's work on your differential regulator?
By the way. My switcher supply voltage is actually 12.5V, so I'll have to change the value on the positive side of the divider. Correct? If the tiny bjt's will work for me that is.

 

Will those little L6's work on your differential regulator?
By the way. My switcher supply voltage is actually 12.5V, so I'll have to change the value on the positive side of the divider. Correct? If the tiny bjt's will work for me that is.

Also; is there a similar design for the use of a N channel POWER mosfet?

 

If you are allowed or have access to BJTs, normal silicon diodes and passive components you could make a simple switch-mode buck converter.

OK, so what exactly is a switch'mode buck converter? Will the L6's work for that?

 

OK, so what exactly is a switch'mode buck converter? Will the L6's work for that?

Here's a sim of a crude buck converter.
A standard multivibrator comprising two L6s drives a P-MOSFET to chop the 12V supply . The chopped supply has an average voltage set by the duty cycle of the multivibrator and is smoothed by L1 and C3 to give 5V out to the load. The duty cycle is adjusted by varying R2 or R3.


I would not recommend this for driving critical loads, since there is no negative feedback to regulate the output voltage.

 

Will those little L6's work on your differential regulator?

Probably.
What are their characteristics and part number?

 

Probably.
What are their characteristics and part number?

They are npn
2SC1623-L6
There are different types. These are out of a Sony Walkman. I think this is the correct #. L6 is all that's printed on it. You have my FET info?