make an variable resistor using transistor or ?

[spoton]

Hello, I have something im working on and have talked to many but cant seam to get placed in the right direction. What I have is a large potentiometer rated for 250 watts, I want to replace it with a circuit that uses a small motorized potentiometer to drive the base on a transistor (or something) that will emulate the larger variable resistor. The big pot is 0-60 ohms and the voltage is 93 VDC passing through it when the pot is on 0 ohms and drops to around 50 vdc at 60 ohms. The load that the dc current goes to after passing through the pot is 42 ohms. (what i want to do) I have a remote that triggers a small motor, fwd and rev, and has an ajustable timer to set travel distance than stop, this will drive the motorized pot, but the small motorized pots available are small and wont handle the watts. Any help would be great. I have tried using a power transistor but im not sure how to correctly hook up. It think I need a emitter follower set up because the dc is positive in relation to the load and I cant acess the ground on the other end so I need to limit the current going in and by doing this the 42 ohms plus the current will give me the voltage i need to see on the load. I need to vary a 93 VDC from .80 amps to 2.21 amps using a small motorized pot. At the 2.21 amps the transistor should be saturated and like an open switch. When I tried this I was fine as long as I had the transistor saturated, but when I started to turn down the bias my transistor becam really hot. I know this is long but I want to try and comunicate what I need to do. Any help would be great. I can't believe how much i forgot in 16 years

[Major_Problem]

The problem with the transistor you are describing is common for the problem at hand. The need for this large 250W resistor is related to the current flowing through it. Replacing the resistor with e.g. a transistor will produce the same amount of heat as the resistor.

If you want to use the same technique as the resistor, you need to cool it. A small circuit diagram would be very helpfull for me to solve the problem. As I understand it, you want to slow down a DC-motor by lowering the voltage across it. What I don't understand is the voltages and resistances. It seems that your powersupply can not handle the load and droppes in voltage.

If this is your goal, then there is an other solution. You could use a Solid State Relay for speed regulation.

http://pcbheaven.com/scripts/imageprese ... 790375.jpg

This is just an example how to look at it.

The basic principle is P=UxI (Power (heat produced) = U (Voltage across the load that produces the heat) x I (Current flowing throug this load).

If the current is Zero, the heat is zero. If the voltage across the load (The resistor of transistor) is zero, also the heat produced is zero. Only if you are in those between, the power, thus the heat increses.

In your example it would be 93V x 0.80 A = 74.4 Watt. Or, 93 x 2.21 = 205,53 W (but this would be the power consumed by the DC-Motor i presume).

Hope to have pointed you in the right direction.

Please excuse my English, I'm from the Netherlands and it's a bit rusty

Greets,
Wim

[Major_Problem]

Hi,
I'm from te Netherlands so please excuse my english.
What you describe (as I understand it) you've got a varable resistor an DC electromotor in series. The voltage acrossthe both of them is 93 Vdc. Resistor at 0 Ohms the motor consumes 2.21 Amp. Resistor at 60 Ohms the current is 0.8 Amps. This would be:

R = U/I: 93V/2.21A = 42 Ohms
R = U/I: 93V/0.8A = 116 Ohms (116-60 = 56 Ohms for the motor). The resistance of the motor is variable due to the rotational speed.

Anyway, The power consumed in the resistor equals to P=U x I. U we don't know but that would be U = I x R. Together tis would be:
P= U x I
P= (I x R) x I. Thus: P = I x I x R = I^2 x R.

At 60 Ohms this is 0.8^2 x 60 = 38.4 Watts
At 0 Ohms this is 2,2^2 x 0 = 0 Watts.

In between the two, the power is at a maximum.

In any case, if you replace the resistor by e.g. a transistor you need to cool it very well because the power is still produced in the transistor, same value as the resistor.

An other way of changing motor speed is by the use of a Solid State Relay. If this is the way you want do go, please respond, then i'll send you a schematic how to do this.

I've been designing motor controllers for doors in rolling stock (Subway, trains etc).

Greets,
Wim

[spoton]

I am trying to limit current to an exciter on a DC welder. I ohmed the exciter and it was 42 ohms. When the pot is set on 0 ohms the open voltage is 93VDC is pulling 2.21a. As the pot is increased in ohms it limits the current to the exciter and then the voltage drops across the 42 ohm exciter, at this point the pot heats up. I want to replace the pot and limit the current electronicly, I guess the question is, should a transistor act the same way as the pot? if I am saturated there is little voltage droped across the transistor which equals less watts at the transistor and less heat, as i limmit the current to the exciter the voltage goes up and generates heat on the transistor? I will try to do a drawing this weakend and post. Thank you for your help. the 42 ohms is static on the excitor,I think its basiclly a coil and the resistance is in the windings so less current equals less voltage drop, kinda like a votage divider with 2 resistors, now I want to do this with a small pot and some kind of current limmiting device.,

[pebe]

A better (and probably an easier) way of doing what you want is to vary the 93VDC supply.

How is the 93V obtained in the first place?

[spoton]

[quote="pebe"]A better (and probably an easier) way of doing what you want is to vary the 93VDC supply.

How is the 93V obtained in the first place?[/quote]
It is from a pure DC gen. The gen is controled by a gov, so I cant changs the voltage.

[pebe]

Does the generator use permanent magnets for the field, or shunt coils?