I haven't been able to locate a schematic for this board anywhere and it just arrived in a plain bag with no documentation. Thank you anyhow for checking it out for me. I tried a resistor to calculate the current but it just knocked the voltage down too far for it to run. No worries, I'll just locate a working ammeter and find it that way.

MisterBill2 mentioned using a timer circuit of some sort. I believe I came across some type of delay circuit that used the 555 timer and a capacitor. Would that work like the MOSFET circuit you proposed as far as getting around the 50mA switch limit?

 

Really, if you use a mosfet switch to turn things on, and are switching a positive bias to the gate to switch it on, then just add a capacitor gate to source for the timer, and change the switch to a momentary button. That wil charge the capacitor and things will stay on until the charge leaks off. About as simple as it can get.The cap may need to be 100mfd though, + side to the gate, negative to the source.

 

I tried a resistor to calculate the current but it just knocked the voltage down too far for it to run.

If you're trying to measure how much current the board draws, I'd use a power supply with a current meter. A power supply with one should measure current in a way that doesn't affect the regulated voltage. Or you could use a clamp on current meter. That's what I use to measure charging current from a battery maintainer.

 

Really, if you use a mosfet switch to turn things on, and are switching a positive bias to the gate to switch it on, then just add a capacitor gate to source for the timer, and change the switch to a momentary button. That wil charge the capacitor and things will stay on until the charge leaks off. About as simple as it can get.The cap may need to be 100mfd though, + side to the gate, negative to the source.

Using dl324's schematic would the capacitor need to be between Q1 and ground or between R1 and ground? Would you need to use a different capacitor value to get the delay time you need, 2min 45 sec, and if so how would you calculate for the correct value?

 

Using dl324's schematic would the capacitor need to be between Q1 and ground or between R1 and ground? Would you need to use a different capacitor value to get the delay time you need, 2min 45 sec, and if so how would you calculate for the correct value?

Fir the circuit in post #2, the capacitor would replace R1. Because the current drawn by leakage in the mosfet and the capacitor is unknown, calculations would be difficult. So there would be some experimenting required.

 

Thank you MisterBill2, that helps.

Onto another issue that just came up during testing. I'll also include my first ever attempted drawing of what I'm talking about so no laughing please. OK you can laugh.

I'm trying to drop the voltage down on another power rail to 4.5v in order to run the sound player. I've run a 20ohm 5watt resistor on my 9v positive rail and it hasn't changed a thing. I checked the resistor and it reads as a 20ohm but is showing no current. There is no load, just measured the voltage between the + and - after the resistor.

 

The voltage drop across a resistor is proportional to both the resistance and the current through the resistor. So when the current is close to zero, as it is with only the meter drawing current, the voltage drop is also close to zero. V= I x R:
voltage = current x resistance. Volts= amps x ohms.

 

The voltage drop across a resistor is proportional to both the resistance and the current through the resistor. So when the current is close to zero, as it is with only the meter drawing current, the voltage drop is also close to zero. V= I x R:
voltage = current x resistance. Volts= amps x ohms.

I thought it might be something like that. I just wanted to check it before putting a load on so as not to damage the device. So if I have nothing drawing the current through the resistor it's going to still read the full 9v after the resistor but if I attach the device that only needs 4.5v then it'll show as 4.5v after the resistor. Does that sound about right?

 

I thought it might be something like that. I just wanted to check it before putting a load on so as not to damage the device. So if I have nothing drawing the current through the resistor it's going to still read the full 9v after the resistor but if I attach the device that only needs 4.5v then it'll show as 4.5v after the resistor. Does that sound about right?

The votage at the device will depend on how much current it draws. You could also use a three terminal voltage regulator.

 

The votage at the device will depend on how much current it draws. You could also use a three terminal voltage regulator.

Looks like I'm back to the drawing board. The 20ohm resistor on 9v should have given me 4.5v but I get only 9v. I tried plugging in the player and it won't work at all so it's not even attempting to draw any current but works fine on it's 4.7 volt power supply.

Well, no results are still results so I'll just work the problem and hopefully something clicks. Once again I thank everyone for your help, it may not seem like it but I have learned a lot from this group which is really appreciated.

 

Looks like I'm back to the drawing board. The 20ohm resistor on 9v should have given me 4.5v but I get only 9v. I tried plugging in the player and it won't work at all so it's not even attempting to draw any current but works fine on it's 4.7 volt power supply.

Well, no results are still results so I'll just work the problem and hopefully something clicks. Once again I thank everyone for your help, it may not seem like it but I have learned a lot from this group which is really appreciated.

I suggest measuring the current, because that will allow knowing what resistor to use. OR, if you can use an adjustable voltage supply you can use it with the 20 Ohm resistor in series, and adjust the supply until you have 4.5 volts at the device, and then measure the supply voltage. Another way to gain the same information without measuring current. And if the supply was at nine vots, then you are all set.

 

I suggest measuring the current, because that will allow knowing what resistor to use. OR, if you can use an adjustable voltage supply you can use it with the 20 Ohm resistor in series, and adjust the supply until you have 4.5 volts at the device, and then measure the supply voltage. Another way to gain the same information without measuring current. And if the supply was at nine vots, then you are all set.

I found a buck converter from an old project, took out the resistor, lowered the voltage to 4.7 and it works. Because of the different volume levels being played the voltage was all over the place from 1-4.8 volts. For whatever reason it just would not work with just using the resistor to drop the voltage.

My complete guess as to what happened is that the player only needed 1 to 2 volts when it started which wasn't enough to drop the voltage through the resistor. Since the the voltage didn't drop perhaps there's something on the sound board that protected the chip from over voltage and wouldn't operate at 9v. I'm completely guessing but that's all I've got.

 

Resistor voltage dropping is best for constant loads. That is why I suggested using a three terminal 5 volt regulator IC. An LM7805, possibly, if the load is less than an amp.

 

Fir the circuit in post #2, the capacitor would replace R1

I built the circuit provided by dl324 in post #2 and it works great, turns on and off with a momentary button. I replaced R1 with a 100uf capacitor, for timing, and it won't shut off. The capacitor doesn't seem to be loosing it's charge.

 

The capacitor doesn't seem to be loosing it's charge.

It'll likely take days because leakage current will determine how quickly it'll discharge.

 

I thought it might be something like that. I just wanted to check it before putting a load on so as not to damage the device. So if I have nothing drawing the current through the resistor it's going to still read the full 9v after the resistor but if I attach the device that only needs 4.5v then it'll show as 4.5v after the resistor. Does that sound about right?

Check out this online circuit simulator. Under the menu 'Circuits' then 'Basics' there is an example called 'Ohm's Law'. Change some values and hover over stuff to see its voltage, current and power. This simulator has enough examples to keep you busy for a while.

Circuit Simulator Applet (falstad.com)

 

OK, the capacitor is very low leakage. I had not anticipated that. So if you have a one megohm resistor put that across the capacitor. Usually the cheap capacitors leak down in a few minutes or less.

 

OK, the capacitor is very low leakage. I had not anticipated that. So if you have a one megohm resistor put that across the capacitor. Usually the cheap capacitors leak down in a few minutes or less.

Thanks for the followup. I put a 1M ohm resistor across(parallel) the capacitor and it did discharge the capacitor in approximately 3min 45sec. It slowly dimmed the LED until it was fully discharged. Is there a faster way to go from the on/off transition so I go from full voltage to nothing?

 

Is there a faster way to go from the on/off transition so I go from full voltage to nothing?

What are your constraints on solution space? Are you open to using a 555 timer to control the on time?

 

What are your constraints on solution space? Are you open to using a 555 timer to control the on time?

I should definitely have enough room on the board for a timer. Just so happens I have some 555 timers. BTW that MOSFET chip was super tiny. Soldering wasn't a problem but seeing the chip was another story...lol.