Hi, I have got into audio electronics (tubes and transistors)...I am very safety conscious because I have little understanding of whats what. I always use a 150 to 250K bleeder resistor in my power supplies. However, unless I put a meter on the power line , I am not aware of the voltage still in the cap. I worry about being caught out.

So I wonder if I could also add / or even substitute a 15k resistor then LED wired in series from the power rail to the ground rail.

If I take a 10000uf smoothing cap as an example

then parallel the 15k resistor and a red led to ground

The forward voltage drop of the led is 1.9volts

So at switch off. As the capacitor discharges, will the LED remain bright until there is 1.9volts of charge left in it?

So if my reasoning is right. In a 10000uf cap which has 1.9 volts still in it (ie the led has dimed) would that capacitor still hurt you if you caused a short?

Is a 10000uf capacitor at 1.9volts more dangerous than a 4700uf capacitor at 1.9 volts ??

Sorry if my question sound stupid

 

I do this on high voltage designs.
The LED will be on until its current drops below about 1mA.
You can work out the voltage from that.

 

Is a 10000uf capacitor at 1.9volts more dangerous than a 4700uf capacitor at 1.9 volts ??

How dangerous is a AAA battery?

 

Well this is why I posted . I don't actually know "mathematically" why I can touch the two poles of the battery. I was hoping someone could give me the maths to work it out. All my messing is from trial and error instead of theory and I want to change this.

If ohms law is applicable here to find the voltage , what value do I use for resistance

the forward resistance of the led or the led resistance + the resistor in series with it?

I am guessing on the second possiblity

 

Hi, I have got into audio electronics (tubes and transistors)...I am very safety conscious because I have little understanding of whats what. I always use a 150 to 250K bleeder resistor in my power supplies. However, unless I put a meter on the power line , I am not aware of the voltage still in the cap. I worry about being caught out.

So I wonder if I could also add / or even substitute a 15k resistor then LED wired in series from the power rail to the ground rail.

If I take a 10000uf smoothing cap as an example

then parallel the 15k resistor and a red led to ground

The forward voltage drop of the led is 1.9volts

So at switch off. As the capacitor discharges, will the LED remain bright until there is 1.9volts of charge left in it?

So if my reasoning is right. In a 10000uf cap which has 1.9 volts still in it (ie the led has dimed) would that capacitor still hurt you if you caused a short?

Is a 10000uf capacitor at 1.9volts more dangerous than a 4700uf capacitor at 1.9 volts ??

Sorry if my question sound stupid

With large electrolytics at high voltage, you may want to keep the LED and bleed resistors apart - calculate a bleed resistor to discharge the caps in a reasonable time without too much dissipation while running. Then calculate the series resistor to light the LED while there's still some juice left.

If you can track down a source of low self heating current PTC resistors that can handle the voltage, they approximate to a constant current, so the LED will stay lit down to a few volts without excessive dropper dissipation at full voltage.

The PTC resistor might be a bit specialty, particularly at presumed 20mA you want for a LED - you could maybe use a higher current part and put a shunt resistor across the LED to bypass some of the current.

 

Certainly 1.9V is perfectly safe. There's little math involved. It's related to the tolerance of the human body to current and the typical body resistance.

The minimum series resistor value will be determined by the maximum current rating of the LED which will occur at the maximum capacitor voltage. Thus the minimum resistor value would be (Vcap (max) - Vled) / Iled(max).

 

Certainly 1.9V is perfectly safe. There's little math involved. It's related to the tolerance of the human body to current and the typical body resistance.

The minimum series resistor value will be determined by the maximum current rating of the LED which will occur at the maximum capacitor voltage. Thus the minimum resistor value would be (Vcap (max) - Vled) / Iled(max).

Apparently 48V is the magic threshold - above that, all sorts of regulations and safety certifications kick in.

AFAIK - phone companies use 48V for that reason.

 

So are we saying then , that a led will not light once the capacitor(s) in a psu drop below its forward voltage .? Or are we saying it can be any voltage but once the current falls below 1ma then it will not light?

 

It's mainly the forward voltage (which is relatively constant). It may still light weakly even below 1mA. There's not as distinct a threshold for current as there is for voltage.

 

Even with the 1.9 volts you may still get a pretty good "pop" from 10,000 Ufd. if you short it. It won't shock you but it might make you knock the cat off the work bench.

 

led's dont handle much current, it would be better to design a lower resistance bleeder to discharge the caps and use an led with a resistor to indicate when the caps are charged.

 

Assuming we're talking high voltage capacitors a good one is a neon lamp and a series resistor in parallel with the bleed chain. The neon will go out when the voltage drops below about 60V. A capacitor across the neon will make it blink (faster at higher voltage).

An LED, 2 resistors and a zener can make a good constant brightness until threshold. 1k in series with the LED, 10V zener across LED and resistor together, then a higher value resistor to the power supply (to suit PSU voltage).