Different ways to discharge circuits?

[bonedoc]

Different ways to safely discharge in circuit components?

I am wanting to learn different in-circuit procedures for discharging. Im thinking 400V or less. I have heard of people using heating elements. I have also heard of having some 220V bulbs in parallel. This would also allow you to visualize to some extent.

However, these ideas are bulky. I have learned a lot about electrical equations in the past couple weeks on here.I am pretty excited, since there are not many college classes here.

I wondered what a massive resistor would do? I have some radial mount resistors that are rated to 700V and 10W. They are 200K Ohm. If I placed this between a load and a grounded button, would the button also have to be rated at the 400 or whatnot DC volts? Or, would the resistor buffer this an allow for a smaller button rating?

All the high voltage stuff is high dollar!

Any ideas appreciated!

 

[Harald Kapp]

Think about the voltage across the open switch. If the switch is open, there is no current. If there is no current, what is the voltage drop across the resistor? Whatever voltage doesn't drop across the resistor must be acrosss the open switch contacts.

The required power rating of a resistor can be calculated from P=U²/R= I²*R.

Help us understand what you want to do so we can help you better:
What do you want to discharge?
Which operating conditions do you have (voltages, currents, capacitances etc.)
What is the goal of this discharge-thing? There might be other (better) ways to achieve the same goal.

Harald

 

[bonedoc]

Hey, I wanted a safety switch on my board to discharge a cap bank. Max power would be 1500w. That is 300v and a peak current of about 5A. I understand a switch with a low voltage rating may see sparks jump across it, but I though a massive resistor would allow for a smaller switch.

Think about the voltage across the open switch. If the switch is open, there is no current. If there is no current, what is the voltage drop across the resistor? Whatever voltage doesn't drop across the resistor must be acrosss the open switch contacts.

The required power rating of a resistor can be calculated from P=U²/R= I²*R.

Help us understand what you want to do so we can help you better:
What do you want to discharge?
Which operating conditions do you have (voltages, currents, capacitances etc.)
What is the goal of this discharge-thing? There might be other (better) ways to achieve the same goal.

Harald

 

[bonedoc]

Just to clarify, the resistor is between the load and the switch ;-)

300V------>220kOhm----->switch------>Ground

I know that resistors do not regulate voltage. So, I suppose that all that happens is that I will severely limit the current, but the potential will be the same? There must be a clever way to do this?

Maybe make a voltage divider and divide the voltage to 2 equal parts, and then go to a 2 pole switch?

 

[Resqueline]

I don't see the problem here, but it would be nice if you could supply some more information.
For example how big that cap' bank is, how fast you need it to discharge, and what kind of limitations you have on the switch.
To dissipate 10W in a 220kΩ resistor you'd need over 1400V. At 300V it will pass 1.36mA and dissipate 0.41W which will likely take forever to discharge the bank.
To discharge with 5A peak like you mentioned you'll need a 60Ω resistor. Switches & relays that can take this are readily available. Even a transistor can do it.

 

[bonedoc]

My voltage is 300V and the capacitance is is 480uf.

E = .5 x .000480 x 300 x300 = 21.6J

Not sure on the formula to calculate discharge time with a given resistance.

Maybe I should get a high voltage PNP transistor and tie it to the power supply voltage, so that it is closed when the unit is powered. When it looses power, it will discharge to ground? The NPN version I am using is the IRF840. Does anyone know of a PNP version?

 

[Resqueline]

A 98% discharge is 5 * C * R so 220kΩ results in 528 seconds (= 9 minutes) and 60Ω results in 0.144 seconds discharge time (to 6V).

You could use a relay (a very common solution, considered failsafe) or you could build the attached circuit which I believe should do the job.

 

[bonedoc]

I didnt see that was an NPN on Q2. Would 2n2222 be fine? I have tons of them too.

http://en.wikipedia.org/wiki/2N2222

Also, it looks like I dont need 10W or even 220k Ohms to use that circuit. I wish I could use .25 watt resistors. Here are my calculations for R2:

-use .25w resistors

-disapate .20w to be safe

W = V X I

.20 = 300 x I

I = 0.067A max

V = IR

R = V/I

R = 300/.067 = 4478 Ohms

Discharge time = (C X r)/2

.00048 X 4478 2 = 3.6 ms = just fine

So, the question is this....I am a little confused about the schematic you showed. What is the purpose of the zenner and the 220k? To pull the IRF840 down until Q2 is activated?

 

[GonzoEngineer]

I crowbar as much asmuch as 1700V with IGBT's.....no burned contacts to worry about.

 

[bonedoc]

If I dont figure this out soon, I may smash my desk with a crowbar. LOL.

 

[(*steve*)]

2N2222 should be fine for Q2. The zener not only protects the gate of Q1, but also Q2 from voltages higher than 16 volts.

 

[Resqueline]

Bonedoc, didn't you learn anything from my previous posts? I have now idea where you got your math from but it's not of this world..
The 220kΩ will dissipate what I said earlier, so if you want to use a 1/4W R1 you calculate I = 0.25W / 300V = 0.83mA and so R1 = 300V / 0.83mA = 360kΩ.
The zener keeps the gate from going above 18V (over 20V will damage the gate) and also prevents the need to use a 400V transistor for Q2.
This high gate voltage makes Q1 conduct hard, exept when the 12V presence makes Q2 conduct, which effectively shorts out the Q1 gate - keeping it off.
The total discharge time will be as per my previous calculation; 5 * R2 * C = 144ms.
R3 & R4 are completely non-critical values, and you can even increase R3 to 100kΩ or reduce R4 to 1kΩ.
This ratio change will only adjust how much the 12V has to drop before the discharge starts. (To 6V in the latter case, instead of to 1.1V with a 10/10 ratio.)
Yes, 2N2222 will be just fine.

 

[bonedoc]

Would .25W be fine for R1? Im not real sure how to calculate the value, but here is my guess:

300V/220,000 = 1.37ma

W = V X I

W = 300 x .00137 = .41W

So, I assume not. To be safe and dissipate .20W:

.20W = 300 x I

I = 0.67mA

V = IR

R = 300/.67 = 448K

I dont think the IRF840 will even turn on at with that current, will it? If I must use high wattage resistors, I can get radial mounts.

 

[bonedoc]

Lets just say for the sake of saving space, I do radial caps, and I use the 220kOhm resistor. That would come to about .41W.....no problem....

But, isnt the 60Ohm resistor in parallel, so getting 300V? That means 5A is going through it!? And....5A x 300V = 1500W! Isnt this high wattage? Either I am looking at this wrong, or 1500W is easy to find.

 

[(*steve*)]

Ignoring the fact that the 220k resistor is not there to discharge the capacitor, lets examine your math.

W = V X I

.20 = 300 x I

I = 0.067A max

Actually I = 0.2/300 = 0.00067

V = IR

R = V/I

R = 300/.067 = 4478 Ohms

300/0.00067 = 447k

Discharge time = (C X r)/2

.00048 X 4478 2 = 3.6 ms = just fine

Well, it's closer to 4 x 480e-6 x 447e3 = 858 seconds, or about 15 minutes (to get to a little under 6V -- actually it will be over 16 due to the zener)

So, the question is this....I am a little confused about the schematic you showed. What is the purpose of the zenner and the 220k? To pull the IRF840 down until Q2 is activated?

The purpose of the zener and the 220k (1/2W at least) is to keep the mosfet turned off until you want to discharge the capacitor. the 60 ohm resistor is used to discharge the cap.

 

[(*steve*)]

The 60 ohm resistor will be carrying that high current for only a few milliseconds and the power averaged out over a second or so is far lower.

To be safe you'd probably want to check the specs to see the maximum current the resistor can handle, but it's likely to be safe.

the 220k resistor has about 284 volts across it, so the dissipation is V^2/R = (284 x 284)/220000 = 0.36W. This brings up an interesting topic of resistor voltage ratings. I would recommend that you use two 120K resistors in series because 284 volts is probably getting uncomfortably close to the voltage rating of a typical small resistor.

 

[bonedoc]

LOL...I guess I should not type in the dark!

If the 60W is just to discharge, It will see a huge wattage.1500W, right?

Is is just such a fast discharge that it wont burn the resistor up? Do I need to get a high wattage resistor on the 60Ohm resistor, if not, what are the guidelines? Thanks again for the help and the fast responses!


EDIT - I just saw your second response. Thanks!

 

[bonedoc]

I am looking through some resistors and they are saying:

5 x rated power (3.75 W and smaller), 10 x rated power (4 W and larger) for 5 s

so I may just get a larger wattage to be safe.

 

[bonedoc]

Ok, I am stupid ;-) One reason was going for smaller power was to save space, but I recall some (expensive) circuitry where they mount axial resistors on a 45 degree angle so that a smaller pcb stance can be used. I guess this is fine, and long as they are not too close to arc.

I just have one LAST question ;-)

As far as zeners to protect transistors:

-Are they good to use with mosfets for protection in higher voltage cases, even when optocouplers are used?

-How does one best decide the value? I am mainly using the IRF840 and a NTE2987 logic level.

-In general, is the anode always on ground, and cathode to the gate?

 

[(*steve*)]

Regarding the Zener, it's main job is to prevent the gate of the mosfet from becoming more than 16 volts more positive than the source. The gate is very easily damaged by excess voltage.

As a consequence, the voltage across the bipolar transistor also never exceeds 16 volts (which is important because a 2N2222 has a maximum voltage well under 300V).

A zener is almost always connected reverse biased, with the cathode more positive than the anode. Between gate and source you *could* connect the zener the other way around if you wanted to allow the gate to go more negative than the source (but not so much as to damage it). This would be the case with a P channel mosfet (this one is N channel) or with a somewhat rarer N channel depletion mode mosfet.

So, no there are no simple rules. You need to know what you want to achieve.