high-power voltage to current converter

[Robert]

I've been told the torque produced by a D.C. motor is quite
proportional to the current drawn. I have a low-voltage, low-power
signal voltage (say 0-10 V at 1 mA) that I want to use to produce a
proportional torque in a large D.C. motor (say about 1 H.P. and 100 V,
hence around 15 A). So what I want is a circuit that adjusts the
voltage fed to the motor in such a way that the current drawn is
proportional to the input signal. Does any one have an idea how to do
that? Robert

 

[Robert]

In case it makes things easier, my signal source is actually a
potentiometer, so if it is more convenient to have resistance rather
than voltage as the input, it is available. However, the
potentiometer cannot handle high currents. I was thinking of simply
attaching the potentiometer to the base of a large transistor, and
have the collector/emitter in series with the motor. But transistors
aren't linear near cutoff, so this won't work.

Robert

 

[[email protected]]

In case it makes things easier, my signal source is actually a
potentiometer, so if it is more convenient to have resistance rather
than voltage as the input, it is available.  However, the
potentiometer cannot handle high currents.  I was thinking of simply
attaching the potentiometer to the base of a large transistor, and
have the collector/emitter in series with the motor.  But transistors
aren't linear near cutoff, so this won't work.

You should repost this question on sci.electronics.basics

The basic answer to your question is an operational amplifier - Apex
make parts that can deliver a few amperes, or you can add one or more
high current boosters to a regular operational amplifier.

In either case you use negative feedback around an amplifier with a
relatively high open loop voltage gain to make the booster transistors
look linear even near cutoff.

 

[Mook Johnson]

In case it makes things easier, my signal source is actually a
potentiometer, so if it is more convenient to have resistance rather
than voltage as the input, it is available. However, the
potentiometer cannot handle high currents. I was thinking of simply
attaching the potentiometer to the base of a large transistor, and
have the collector/emitter in series with the motor. But transistors
aren't linear near cutoff, so this won't work.

Robert

Depending on the range where you want to operate this will produce a HUGE
amount of heat. 1HP is ~750Watts. That 7.5A @ 100V If you run at low RPM
or in a stall condition, that would put 750W on the transistor and surely
blow it.

Your best approach is a PWM method.

You're looking for PWM torque control of a DC motor.

 

[legg]

Depending on the range where you want to operate this will produce a HUGE
amount of heat. 1HP is ~750Watts. That 7.5A @ 100V If you run at low RPM
or in a stall condition, that would put 750W on the transistor and surely
blow it.

Your best approach is a PWM method.

You're looking for PWM torque control of a DC motor.

Yeah....driving any uncoupled inductive load produces the same loss in
a linear driver as occurs when its driving a short circuit.

RL

 

[Robert]

Depending on the range where you want to operate this will produce a HUGE
amount of heat. 1HP is ~750Watts. That 7.5A @ 100V If you run at low RPM
or in a stall condition, that would put 750W on the transistor and surely
blow it.

Your best approach is a PWM method.
You're looking for PWM torque control of a DC motor.

That's a good point that I overlooked. Even if I get sufficiently
large transistors that can handle the load, the expense, cooling
demand, and power waste seem ridiculous. The PWM idea seems much more
efficient, if more complicated to implement.

 

[Robert]

You're looking for PWM torque control of a DC motor.

Hey, here's an idea that might bring back memories if they're any old-
timers out there. A motor-generator set. The motor runs constantly
and turns the generator. The control voltage adjusts the field of the
generator, which in turn controls the generator's electrical output.
That in turn runs another motor, whose torque is then proportional (?)
to the control voltage!

 

[bg]

Robert wrote in message

I've been told the torque produced by a D.C. motor is quite
proportional to the current drawn. I have a low-voltage, low-power
signal voltage (say 0-10 V at 1 mA) that I want to use to produce a
proportional torque in a large D.C. motor (say about 1 H.P. and 100 V,
hence around 15 A). So what I want is a circuit that adjusts the
voltage fed to the motor in such a way that the current drawn is
proportional to the input signal. Does any one have an idea how to do
that? Robert

This is a very common requirement that I've seen in quite a few "off the
shelf" servo amlifiers. Westamp comes to mind as one that did have a 1 or 2
HP 100vdc PWM amplifier. I don't know if westamp is still around, but you
could do a search for servo amplifiers to see what's available. Bill
Sloman's answere is dead on, but I'll add to it. All you need is a low
valued resistor in series with the motor. The current thru the resistor is
equal to the current thru the motor , which develops a voltage proportional
to motor current. You now have a voltage representing motor current, and it
is a simple matter of using op amps to compare the feedback voltage to the
input voltage, process it as you desire. BTW a .667 ohm resistor would give
you a 0 to 10 volt feedback for a 0 to 15 amp motor current.

 

[gearhead]

Robert wrote in message



This is a very common requirement that I've seen in quite a few "off the
shelf" servo amlifiers. Westamp comes to mind as one that did have a 1 or  2
HP 100vdc PWM amplifier. I don't know if westamp is still around, but you
could do a search for servo amplifiers to see what's available. Bill
Sloman's answere is dead on, but I'll add to it. All you need is a low
valued resistor in series with the motor. The current thru the resistor is
equal to the current thru the motor , which develops a voltage proportional
to motor current. You now have a voltage representing motor current, and it
is a simple matter of using op amps to compare the feedback voltage to the
input voltage, process it as you desire. BTW a .667 ohm resistor would give
you a 0 to 10 volt feedback for a 0 to 15 amp motor current.

A few obvious considerations:
You probably can't put the sense resistor and the PWM drive both on
the low side. One of them has to go on the high side -- so choose
your poison. You need a freewheel diode connected so the freewheel
current goes through the sense resistor. And I think you need to low-
pass filter the voltage from the sense resistor before it goes to the
op amp.

 

[Mook Johnson]

This is a very common requirement that I've seen in quite a few "off the
shelf" servo amlifiers. Westamp comes to mind as one that did have a 1 or
2
HP 100vdc PWM amplifier. I don't know if westamp is still around, but you
could do a search for servo amplifiers to see what's available. Bill
Sloman's answere is dead on, but I'll add to it. All you need is a low
valued resistor in series with the motor. The current thru the resistor is
equal to the current thru the motor , which develops a voltage
proportional
to motor current. You now have a voltage representing motor current, and
it
is a simple matter of using op amps to compare the feedback voltage to the
input voltage, process it as you desire. BTW a .667 ohm resistor would
give
you a 0 to 10 volt feedback for a 0 to 15 amp motor current.

That would also generate 150W Watts of heat!!! (15^2 * .667 ohms) Your
sense resistor would likely be something like a 0.005 ohm run to a
differential amp with a gain of 133. That would produce a 10 V signal at
full current and would dissipate ~1 Watt. Make sure the resistor is low
inductance (prefereably surface mount)


You might want to look at something like a UCC1800 from TI configured as a
forward converter. Just replace your transformer with a motor with a fast
diode in parallel. by fast I mean a diode with a reverse recovery time less
than 50nS. the old 1n4000 series won't get it. For motor use, you should
look at 20KHz operation so the PWM wine i nout of the audible range (if that
is a concern). Don't go much higher than this or you're only heating up the
mosfet more with no gain in performance.