05-03-2023, 12:27 AM
Hi Guys
Champ81, do you know Ohm's Law? That is what guides you here and there is no need for any active circuitry when it comes to discharging capacitors, or in assuring that they share voltage when wired in series.
Ohm's Law tells us the relationship between voltage (V in volts), current (I) in amperes, and resistance ® in ohms. When a voltage is impressed across a resistance, a current will flow through the resistance. The magnitude of current is inversely proportional to the resistance This gives us an equation:
I = V / R
You can think of V as pressure, I as flow and R as what limits the flow. What fllows here are electrons.
You can look at it another way, and think that a current through a resistance generates a voltage across the resistance, as:
V = I x R
If you know the voltage across the resistance and also know the current, then you can calculate the resistance:
R = V / I
So, if you have 400V across 100k, the current is:
I = V / R = 400V / 100,0000R = 0.004A
If we reduce R to 10k, there is a lot more current:
I = V / R + 400V / 10,000R = 0.04A
The Power equations tell us how much heat (W) in watts will be generated by the voltage (V) in volts, and current (I) in amperes, as:
P = V x I
which can be re-arranged as :
P = I*2 x R
and
P = V*2 / R
The *2 means "squared".
So, with the 400V across 100k, the power is:
P = V x I = 400V x 0.004A = 1.6W
and with 10k:
P = V x I = 400V x 0.04A = 16W.
In each case, you need a resistor with a power rating in excess of the heat that is being dissipated, and preferably it is a flame-proof type, either wire-wound or metal-oxide - NEVER carbon.
If you look at schematics, you see high-value bleeders across filter caps, typically 220k or higher. If we use the equations above, a 220k resistor will dissipate 0.72W with 400V across it. We could use a 1W but it will get hot, so maybe use a 2W so it will be a bit physically larger and run cooler. There is no need to use a low-value resistor in this high-voltage position as it simply wastes power and heats up the inside of the chassis for no reason.
If the supply voltage were only 40V, a 10k resistor would dissipate 0.16W and you could use a half-watt resistor without worry.
Our book Ready, Set, Go! (RSG) explains all the high-school science that is the foundation of understanding electronic quantities and basic circuits.
Champ81, do you know Ohm's Law? That is what guides you here and there is no need for any active circuitry when it comes to discharging capacitors, or in assuring that they share voltage when wired in series.
Ohm's Law tells us the relationship between voltage (V in volts), current (I) in amperes, and resistance ® in ohms. When a voltage is impressed across a resistance, a current will flow through the resistance. The magnitude of current is inversely proportional to the resistance This gives us an equation:
I = V / R
You can think of V as pressure, I as flow and R as what limits the flow. What fllows here are electrons.
You can look at it another way, and think that a current through a resistance generates a voltage across the resistance, as:
V = I x R
If you know the voltage across the resistance and also know the current, then you can calculate the resistance:
R = V / I
So, if you have 400V across 100k, the current is:
I = V / R = 400V / 100,0000R = 0.004A
If we reduce R to 10k, there is a lot more current:
I = V / R + 400V / 10,000R = 0.04A
The Power equations tell us how much heat (W) in watts will be generated by the voltage (V) in volts, and current (I) in amperes, as:
P = V x I
which can be re-arranged as :
P = I*2 x R
and
P = V*2 / R
The *2 means "squared".
So, with the 400V across 100k, the power is:
P = V x I = 400V x 0.004A = 1.6W
and with 10k:
P = V x I = 400V x 0.04A = 16W.
In each case, you need a resistor with a power rating in excess of the heat that is being dissipated, and preferably it is a flame-proof type, either wire-wound or metal-oxide - NEVER carbon.
If you look at schematics, you see high-value bleeders across filter caps, typically 220k or higher. If we use the equations above, a 220k resistor will dissipate 0.72W with 400V across it. We could use a 1W but it will get hot, so maybe use a 2W so it will be a bit physically larger and run cooler. There is no need to use a low-value resistor in this high-voltage position as it simply wastes power and heats up the inside of the chassis for no reason.
If the supply voltage were only 40V, a 10k resistor would dissipate 0.16W and you could use a half-watt resistor without worry.
Our book Ready, Set, Go! (RSG) explains all the high-school science that is the foundation of understanding electronic quantities and basic circuits.


