Slave stepper motor

[mirocha]

You could add an LED to let you know when the dump has been activated, but you will probably only get very brief flashes. Fractions of a second is all one would expect.

The circuit you are showing looks wrong to me. You have the current from the supply going through a 1kohm resistor - that would never work at any power level! If you changed the 1k resistor to 1 ohm it might work - maybe.

It is a bit hard to say how to add an LED without more details. Assuming that we are talking about a +70 VDC supply, that means the collector of the transistor may rise to about +70 V. In that case you could put a LED in series with a resistor across the 33 ohm resistor. The anode of the LED goes to the collector, and the resistor should be about 70 V / 0.01 A = 7,000 ohm - say 6.8 kohm ( to ground).

Cheers
Roger

The circuit is off the gecko website and it does work, the current goes through the diode to the stepper motor gecko drive.

[RCaffin]

Gaarrk !
I overlooked that. Sorry. Frustrating morning.
The bit about the LED remains.

Cheers
Roger

[mirocha]

Built the Returned energy dump and added an LED to let me know it is working. I added a diode, capacitor, 6.8 K resistor and a LED to the original circuit. When the there is returned energy the capacitor charges, it lights the LED and the LED gradually dims as the capacitor discharges, if you have returned energy again while the LED is dimming the LED comes back to full brilliance and starts dimming again. I built the circuit to mount above my Gecko drives, I will be adding one to each drive, also have the LED attached to a terminal strip, I may wire it to LED's on control pane cover. Protects the Gecko drives and eliminates voltage spikes to the power supply.

[m_c]

You should of added the diode to the opposite side of the 33ohm resistor. The operation of the original circuit, is the capacitor absorbs the return energy quickly, and it's then dissipated through the 33ohm resistor.
With your circuit, the capacitor still absorbs the energy, however it's now only draining that charge via the LED and 6k8 resistor, as the 33ohm resistor is now behind a diode. If you get lots of sustained energy dumps, there is a real possibility the drive is going to go overvoltage due the energy not getting bled of quick enough.

[RCaffin]

The operation of the original circuit, is the capacitor absorbs the return energy quickly, and it's then dissipated through the 33ohm resistor.
With your circuit, the capacitor still absorbs the energy, however it's now only draining that charge via the LED and 6k8 resistor, as the 33ohm resistor is now behind a diode.
I don't think so.
In fact, I am quite sure that is completely and utterly wrong.
The 1000 uF cap is just there to stretch the visibility of the LED glow. It has nothing to do with the operation of the dump - nothing at all.

Cheers
Roger

[mirocha]

Now you have me concerned. The following is the description of the original circuit from Gecko's website:

This circuit protects stepmotor and servomotor drives from damage during periods of energy return when operating near maximum rated voltage.

Energy return occurs when a large inertial load is rapidly decelerated from a high speed. The energy stored in the moment of inertia (kinetic energy) must be removed and dissipated. The drive will return this energy back to the power supply as a current, which will cause a voltage rise of the power supply voltage. If the power supply voltage is already near the drive’s maximum rated voltage, this additional voltage will destroy the drive.

This circuit senses the direction of current flow. When it is from the supply to the drive, it does nothing. If it reverses, then the current is shunted to ground.

CIRCUIT DESCRIPTION:

Normally power supply current flows thru rectifier D1, biasing Q2 off via R2. If the direction of current reverses, D1 shuts off and a reverse voltage develops across R1. This voltage now turns Q1 on. The returned current now flows from the drive, thru Q1 and to ground via R3. The main purpose of R3 is to keep Q3 within its secondary-breakdown limits (SOA).

The way I read this is the Q1 transistor turns on when the output voltage is higher than the input voltage, taking the voltage higher than the input voltage to ground through the transistor to ground through the 33 ohn resistor. They have a 1000 cap which I do think would absorb the instantaneous hit but as the motor continues to wind down generating reverse voltage the capacitor is fully charged and the higher voltage continues and the voltage is shunted to ground through the transistor and 33 ohm resistor for as long as the motor is winding down.

When Q1 transistor starts to conduct in the circuit I added I believe it instantaneously charges the capacitor I added and lights the LED, the diode I added was so as the reverse voltage dissipates through the 33 ohm resistor and the motor is no longer giving reverse voltage because the motor has stopped my diode keeps my added capacitor from discharging through the 33 ohm resistor, remains powering my LED till it is discharged through the 6.8 K resistor and does not affect the original circuits operation.

Previously you said to use a 6.8k resistor with an LED across the 33 ohm resistor and said it would be a very fast blink of the LED, what I did was add the Capacitor and a diode to block the added Capacitor from feeding back into the original circuit.
I could be totally wrong could you look at the circuit again I'm not schooled in electronics and was guessing based on my research of diodes.

The original 1000 cap is still there not behind the diode I added, I added a second 1000 cap to keep the LED lighted longer.

[RCaffin]

The Gecko description is dead on.

The original 1,000 uF cap on the right does not do very much. The amount of charge it can hold is very small compared to the amount of energy flowing through to the motor. It may be useful for noise and stability reasons, and to limit the ability of one motor drive from affecting another one. Certainly, I would keep it there.

Previously you said to use a 6.8k resistor with an LED across the 33 ohm resistor and said it would be a very fast blink of the LED, what I did was add the Capacitor and a diode to block the added Capacitor from feeding back into the original circuit.
I could be totally wrong could you look at the circuit again I'm not schooled in electronics and was guessing based on my research of diodes.
The original 1000 cap is still there not behind the diode I added, I added a second 1000 cap to keep the LED lighted longer.
Your circuit is just fine. The 1,000 uF cap you added does exactly what you suggest: it keeps the LED alight for longer. It will heve very little effect on the dump circuit.

Cheers
Roger

[mirocha]

The Gecko description is dead on.

The original 1,000 uF cap on the right does not do very much. The amount of charge it can hold is very small compared to the amount of energy flowing through to the motor. It may be useful for noise and stability reasons, and to limit the ability of one motor drive from affecting another one. Certainly, I would keep it there.

Previously you said to use a 6.8k resistor with an LED across the 33 ohm resistor and said it would be a very fast blink of the LED, what I did was add the Capacitor and a diode to block the added Capacitor from feeding back into the original circuit.
I could be totally wrong could you look at the circuit again I'm not schooled in electronics and was guessing based on my research of diodes.
The original 1000 cap is still there not behind the diode I added, I added a second 1000 cap to keep the LED lighted longer.
Your circuit is just fine. The 1,000 uF cap you added does exactly what you suggest: it keeps the LED alight for longer. It will heve very little effect on the dump circuit.

Cheers
Roger

Great the circuit is ok, I like that the LED let's me know the transistor is conducting and the dump circuit is working. This dump circuit is cheap insurance to prevent damage to the drives.

[m_c]

Ignore what I said.
I'll blame being tired and too lazy to check your diagram against the original. I wrongly assumed looking at your diagram, that you'd drawn the extra diode over an existing line, but having checked, I see you added the extra capacitor as well.

I'm glad there's usually somebody around to tell me when I post nonsense!