I have used the attached circuit with good success, but it did not have the bleeder resistor, R2, and so C1 took a few seconds to discharge. If I make R2 1MΩ and install it as shown, will it reduce the discharge time to less than 1 second, is a 1/4 watt resistor satisfactory, and is there any downside to having it in the circuit?

Thanks.

 

You R2*C1 would give a time constant of 1000s which would mean it would take that resistor about an hour to bleed to 10% of the original voltage. It looks like you are probably bleeding mostly through the LED right now.

Power is not a factor. V^2/R yields less than a quarter of a milliwatt for the 1MΩ resistor.

If you used a 10kΩ resistor you would only increase your current draw by 1.5mA, (which may or may not be significant for your app), and still only be at a couple dozen milliwatts of power in it. But this would only reduce your time constant to 10s so you might notice some improvement, but not a lot.

I would recommend reconnecting your switch so that the diode's anode is hard tied to the input terminal and the common pole of the switch connects to the top of the cap. Then, in one position, it connects the cap to the cathode of the diode to turn things on. But in the other position it connects a 100Ω resistor to ground to discharge the cap in a fraction of a second. True, your peak power in the resistor will be a couple of watts, but it will be so transient that it won't matter and a 1/4 W resistor should work fine. If you want to play it safer, use anything between about 220Ω to 750Ω.

 

Wow! I missed the calculation for that by several orders of magnitude. I thought it was T = R * C * 5 for T to be less than 5 seconds.

Anyway, I like your idea with the switch. I'll redraw it and post it to be sure that I didn't misunderstand. Thanks.

Oh, and you are correct; the LED does discharge the cap, but it takes a few seconds.

You R2*C1 would give a time constant of 1000s which would mean it would take that resistor about an hour to bleed to 10% of the original voltage. It looks like you are probably bleeding mostly through the LED right now.

Power is not a factor. V^2/R yields less than a quarter of a milliwatt for the 1MΩ resistor.

If you used a 10kΩ resistor you would only increase your current draw by 1.5mA, (which may or may not be significant for your app), and still only be at a couple dozen milliwatts of power in it. But this would only reduce your time constant to 10s so you might notice some improvement, but not a lot.

I would recommend reconnecting your switch so that the diode's anode is hard tied to the input terminal and the common pole of the switch connects to the top of the cap. Then, in one position, it connects the cap to the cathode of the diode to turn things on. But in the other position it connects a 100Ω resistor to ground to discharge the cap in a fraction of a second. True, your peak power in the resistor will be a couple of watts, but it will be so transient that it won't matter and a 1/4 W resistor should work fine. If you want to play it safer, use anything between about 220Ω to 750Ω.

 

Wow! I missed the calculation for that by several orders of magnitude. I thought it was T = R * C * 5 for T to be less than 5 seconds.

Me, too.

I had a post all ready to hit "submit" where I was going to suggest reducing it to a 100kΩ resistor so that you would bleed out in a fraction of a second (that just sounds so ... violent... somehow ). But my Rule #2 -- Always ask if the answer makes sense -- saved the day! If you are currently bleading through a 270Ω resistor and it takes several seconds, then how could a 100kΩ resistor bleed it in a fraction of a second. It was on the third look before I slapped by head and realized that there was this little factor of a 1000 that I was marching right past.

 

Here's a revised schematic. I think it is what you recommended.

Thanks again for your help.

 

Yep, that's what I had in mind. I would recommend flipping the switch or swapping the connections in order to avoid the crossing wire in the schematic, but of course that's pure aesthetics.

 

Yep, that's what I had in mind. I would recommend flipping the switch or swapping the connections in order to avoid the crossing wire in the schematic, but of course that's pure aesthetics.

Thanks for the response and the help. The configuration of the switch is such that I need the connections on the pins as shown in the schematic so that on the actual switch, ON will be to the right and OFF will be to the left. Attached is a drawing of my PCB layout; it's a plug-in PS for a solderless breadboard. I am ordering PCB's tonight.

 

How about moving the LED to the input side of the regulator.

 

How about moving the LED to the input side of the regulator.

If I did that, the LED would only indicate when I had power in, not out. Given the LM78xx regulators' ability to shut down if they get too hot, I would prefer the LED on the output. I did think about putting the power switch on the output side of the regulator, but decided against it.

Thanks for your input.

 

I thought about recommending that and I think it's a toss up. On the output side he can count on it being 5V but on the input side the 15V may vary by quite a bit giving different a brightness that might be annoying. But, on the other hand, it's also relaying a limited amount of information.

I don't think it would bleed the capacitor any faster because the resistor is going to be resized to result in about the same current. But it would shift the heat associated with the 10V drop to the LED over to the resistor instead of the regulator and make that 10mA or so available to the load.

I would probably be the type of person to put an LED on both sides.

And whichever side it's on, it won't bleed the capacitor much below a couple of volts (i.e., once the LED turns off).

 

I thought someone might be interested in seeing the finished PCB. I wanted a well filtered and regulated PS that would securely attach to a solderless breadboard without covering too much of the breadboard. The two header shunts are there to provide for disconnecting power from one rail when required.

ETA: It's not my best soldering. Someone had turned my iron down by 50°, and I didn't notice until I was almost finished. (I am the only one who uses my iron. )

 

That's cool! I was lucky enough to find a powered breadboard, has 5V and 15V built into it. Old but still works,or wait a minute, is that me?

 

In a way, my design is a solution in search of a problem. I like PCB layout and I like solderless breadboards, so I just did it. Kind of like a hobby...oh, it is my hobby.

 

Started as a hobby for me too, but has turned into an obsession!

 

What is the PH1 and PH2 on you revised drawing where you added the switch for the bleed resistor? I followed every step of the power through the reg and the decoupling caps but the PH items threw me for a loop. Could you elaborate what they are please?

Thanks!!
Wheelchair Bob

 

Kind of nice design touch how you connect the supply to the bread bord. Is it our own idea?

 

What is the PH1 and PH2 on you revised drawing where you added the switch for the bleed resistor? I followed every step of the power through the reg and the decoupling caps but the PH items threw me for a loop. Could you elaborate what they are please?

Thanks!!
Wheelchair Bob

PH1 and PH2 were added in order to provide a means of disconnecting power and ground from one set of rails on the solderless breadboard. This is for those instances where one set of rails is used for a different voltage than the on-board power supply is providing. For example, a microcontroller is running on 5V from the on-board supply, and controlling a MOSFET that is switching 12V. In the photo, the red shunt is to connect the 5V to the top rail, and the black shunt connects ground. One or both can be removed to isolate the top rail from the on-board supply.

 

Kind of nice design touch how you connect the supply to the bread bord. Is it our own idea?

I want to say it is my idea, but the truth is, I saw it somewhere else.

 

I want to say it is my idea, but the truth is, I saw it somewhere else.

But it was still a good PCB job making the sockets fit into the bread board

 

But it was still a good PCB job making the sockets fit into the bread board

Thanks. Here's another view showing the supports for the back edge of the PCB.