Bleed resistor/LED discharging indicator for bulk capacitor

Thread Starter

Xavier Pacheco Paulino

Joined Oct 21, 2015
728
Hi,

I don't know if there's much behind this. According to the circuit below, I intend to add a led to indicate whether the bulk capacitor is charged or not. The rectified and filtered voltage is about 170VDC. The voltage drop across the led is around 2V. For a current of 20 mA through the led if the capacitor is fully charged, R = 170-2/20mA = 8.4k ohms. , which throws a power rating around 3W. But I think there's something wrong. Any comments?

 
Last edited:

MaxHeadRoom

Joined Jul 18, 2013
23,140
Is this for your T.M. project? If so I would not worry about it as the motor and controller are usually not accessible to any user.
Max.
 

MaxHeadRoom

Joined Jul 18, 2013
23,140
I have seen this LED in treadmills with AC three phase motors. For example, Life Fitness has a model which includes it, and the interesting thing is that the led is on for more than a minute even when the power is turned off.
Many versions of VFD's such as this generally have it, the Capacitor bank total capacity is generally much higher.
Max.
 

WBahn

Joined Mar 31, 2012
26,398
I just was confused because I saw in another design a 130K resistor.
Using 130 kΩ with 168 V yields a current of 1.3 mA. That may well be visible enough even with the current LED you are planning to use. It also reduces the power draw to under a quarter watt.

If this is for indicating, not bleeding, then choose an LED that can work with a much smaller current. Even a typical "20 mA" LED is probably sufficiently visible at 2 mA to 5 mA and there are many LEDs available today that operate at 2 mA and less as their target current. DigiKey has several 0.5 mA options and you can probably run them at something like 0.1 mA or so. If you run it at 0.2 mA, then you've dropped your 3 W down to around 30 mW.

But if you are using this for bleeding, then even at your 20 mA you have a problem because your RC time constant is already pushing 20 seconds. If you bring the current down two orders of magnitude, then your time constant goes up those same two orders bringing you into the half-hour time frame.

So one thing you need to decide is what your goal is -- indication or bleeding.
 

Thread Starter

Xavier Pacheco Paulino

Joined Oct 21, 2015
728
Using 130 kΩ with 168 V yields a current of 1.3 mA. That may well be visible enough even with the current LED you are planning to use. It also reduces the power draw to under a quarter watt.

If this is for indicating, not bleeding, then choose an LED that can work with a much smaller current. Even a typical "20 mA" LED is probably sufficiently visible at 2 mA to 5 mA and there are many LEDs available today that operate at 2 mA and less as their target current. DigiKey has several 0.5 mA options and you can probably run them at something like 0.1 mA or so. If you run it at 0.2 mA, then you've dropped your 3 W down to around 30 mW.

But if you are using this for bleeding, then even at your 20 mA you have a problem because your RC time constant is already pushing 20 seconds. If you bring the current down two orders of magnitude, then your time constant goes up those same two orders bringing you into the half-hour time frame.

So one thing you need to decide is what your goal is -- indication or bleeding.
To be honest, yes, I was confused. My goal was to indicate if there's charge in the capacitor. I thought that while the capacitor is discharging, the led is fading, so it was kind of bleeding. But I see that I got my wires crossed with regards to the concepts of bleeding and indicating.
 
Top