Designing a "discrete" pot

[ADWSystems]

This is not a good application for bipolar transistors due to the "high" voltage drop of the junctions.

If I was to drive the gate with 12V I'd be good?

I the battery tester circuit a while back, the design used an N-channel and gate was driven between 2.5 and 5.0 volts for the grounded load. I'm not sure the difference between them.

 

[ADWSystems]

Unfortunately pots do not come as inexpensively as resistors for the wattages I need.

 

[Jaguarjoe]

Unfortunately pots do not come as inexpensively as resistors for the wattages I need.

This is like playing 20 questions.

How much current are we talking about? If its less than a dipswitch but more than a trimpot, it can't be a lot.

How about an active load, like a MOSFET and a current sense resistor? Use an opamp to compare the CS value with your desired value and let the MOSFET control at that point.

 

[ADWSystems]

Only because you guys are venturing into options that I haven't considered, did know existed, or doesn't meet the design requirements. The load will be handling at least 5W but not more than 10W.

I already tried the active load but then realized that I am not trying to keep a constant current but a constant load. The active load that I tried (like your description and based on the battery tester in another thread) turned out to fight with a voltage source that was upstream. I thought it would be a neat idea, but didn't work.

 

[crutschow]

This is not a good application for bipolar transistors due to the "high" voltage drop of the junctions.

If I was to drive the gate with 12V I'd be good?

I the battery tester circuit a while back, the design used an N-channel and gate was driven between 2.5 and 5.0 volts for the grounded load. I'm not sure the difference between them.

The saturation voltage drop of a power bipolar transistor can be quite low (much less than a diode drop but not quite as low as a large MOSFET).

!2V gate-to-source (not just gate) voltage would be enough to fully turn on most power MOSFETs.

If the N-channels only had 2.5 to 5.0V on their gate and they were driving a grounded load attached to their sources, then the output voltage would be very low or zero. If they were N-channels then the load must have been connected to the transistor drains and the battery common was not connected to the circuit common.

 

[ADWSystems]

Where are you "measuring" the "output" voltage? At the source pin?

The actual circuit performs as the simulation predicted. The source is connected to the load resistor but never exceeds 1 volt for a single cell. and the gate runs from 2.5 to 5 volts (Vcc).

The battery common is definately connected to the common of the remainder of the circuit and the load resistor is definately not connected to the drain.

 

[Jaguarjoe]

Beware that an ancient 741's output does not swing to Vcc and Vee. You'll get within 1-1.5 volts or so. It's listed in the datasheet.
What are the two 5k resistors (R01, R02) for?

 

[ADWSystems]

What are the two 5k resistors (R01, R02) for?

I knew that. I made a typo (actually looked at the wrong trend) the Opamp out swings 2.5 to 4.0 (not 5.0).

Offset compensation per the datasheet. They should not be left floating.

 

[ronv]

Spec.

Seems like we are chasing things around because there is no spec.. You talk of 5 to 10 watts minimum / maximum, then show a 4 ohm load, then talk of 100 ohm resistors. Why don't you set the requirements? Are they unknown?

 

[ADWSystems]

Seems like we are chasing things around because there is no spec.. You talk of 5 to 10 watts minimum / maximum, then show a 4 ohm load, then talk of 100 ohm resistors. Why don't you set the requirements? Are they unknown?

The requirements are: (unchanged from the OP but more defined thanks to the brainstorm power here)

For one of the newest prototypes, I need a "discrete" pot. Basically a rotary switch changing the selected resistors (8 values total). The idea is to increase current draw in known steps (8 total) using 100 ohm resistors. The loads needs to be resistive (ie., PN junction drops are out, so no diodes or bipolar junctions) and capable of handling 5-10 watts, so pots are ruled out again. So relay and/or FET switch matrix of-sorts are in. The selection needs to be by a single input (ie. not set by dip switch) so any random technician can operate it accurately.

The simpliest thing would be to connect an increasing number of resistors in parallel at the contact on the switch (ie., 1 on the first, two on the second, three on the third, etc.). That would add to 32 resistors, some could be reduced if the resistors in parallel equate to a typically available value. I also have sketched a pot driving a 3914 and ULN2003 with relays to substitute for the dip switch, which is an option. I'm working some of the viabilities of the diode and FET configuration as mention in a previous string.

The other stuff to which you refer (the circuit with the 4 ohm resistor) was a string of thoughts on the principals of using N-FETs and P-FETs with reference to a previous design.

 

[ronv]

So let me convert. If the lowest value is around 14 ohms and this yields 10 watts the current must be around .85 amps and the voltage source around 12 volts?

 

[ronv]

Switch

Seems like one of these should work.

 

[carbonzit]

What RonV is getting at, in his usual patient way, is that it would really really help us if you could give us a few details (like voltage and current) about your circuit.

 

[ADWSystems]

The lowest resistance would 12.5 ohms (8 parallel 100ohm resistors) and the current through each resistor will be around 1/4 amp. Overall the system will have a common ground, but the circuit we are discussing will have it's own power source.

 

[ADWSystems]

What are your thoughts on this?

 

[dougy83]

What are your thoughts on this?

Looks like it might work... but as you obviously have a power source available, I'd do it this way (see attached) as I feel it's a neater and simpler solution.

It's an opamp-stabilised resistor multiplier. The string of 10k resistors around the rotary switch provides the tap points for the different multiplication factors from 1 to 8. The 12.5 ohm should be rated for the maximum dissipated power and the mosfet must be able to dissipate at least a quarter of that amount.

 

[ronv]

I already tried the active load but then realized that I am not trying to keep a constant current but a constant load. The active load that I tried (like your description and based on the battery tester in another thread) turned out to fight with a voltage source that was upstream. I thought it would be a neat idea, but didn't work.

If where you have "out" in your schematic is ground that should work if you pick a fet that is well turned on after the diode drops. You don't say the voltage at the rotary switch.

Or, the straight switch with 11 ohms and 1.5 ohms as a stock value.

 

[crutschow]

I like dougy83's circuit. Simple and elegant with only one power transistor and one power resistor.

Note that the op amp positive supply voltage will need to be equal/greater than the maximum test voltage plus the gate-source voltage needed to fully turn on the transistor, plus the difference between the op amp maximum output voltage and its positive supply voltage (if it's not a rail-rail type).

 

[ADWSystems]

As M1 and R10 form the load for the system being tested, V1 would represent the customer system. Thus the resistor string for setting the load resistance is based on the test system output. I'll need to digest that.

Won't the opamp out vary as the voltage at the junction of the MOSFET Drain and 12.5 ohm resistor varies?

What is R9 for?

 

[crutschow]

Won't the opamp out vary as the voltage at the junction of the MOSFET Drain and 12.5 ohm resistor varies?

What is R9 for?

Of course the op amp varies the voltage in response to the switch setting. That's how it provides the variable load resistance values.

For example, if the switch is set to the half-way point, then the op amp feedback loop will adjust the transistor gate voltage so that the voltage across the 12.5Ω resistor is 1/2 the load voltage. The other half of the voltage is dropped across the transistor drain-source connection. This makes the load resistance appear to be 25Ω.

I think R9 was just added for test purposes.