Convert this circuit to be powered by DC 9v?

SgtWookie

Joined Jul 17, 2007
22,230
A 7805 might be used, but it also may require an additional load resistor depending upon how much the logic circuit draws. A 7805 requires a minimum load of 5mA for guaranteed regulation; that would be a 1K Ohm resistor.

Another thing; an NE555's output decreases about 1.7V with a 100mA load, due to the Darlington voltage follower arrangement in it's output circuitry, and may decrease as much as 2.5v with a 200mA load. If you are considering using a 9v "transistor" battery, these frequently have an 8.6v output when new when tested under light load (25mA, or about 350 Ohms). So, 8.6v - 1.7v = 6.9v remaining; 8.6v-2.5v=6.1v remaining. There are some "industrial" type 9v "transistor" batteries available that output around 10v; they have seven internal cells instead of six.

A 7805 regulator has a specified dropout of 2v. If you are supplying it 6.9v, the most it can output is 4.9v; if you are supplying it 6.1v, the most it can output is 4.1v. This may not be enough to power the logic.

Earlier, eblc1388 mentioned (basically) that the blue LEDs would not light if more than two were placed in series due to their higher Vf, but in the subsequent schematic, three are shown in series. That won't work if you're only going to have 6.9v left to work with; blue LEDs may have a Vf of 3.4v to 4v.
 

Thread Starter

critiera119

Joined Nov 21, 2008
66
A 7805 might be used, but it also may require an additional load resistor depending upon how much the logic circuit draws. A 7805 requires a minimum load of 5mA for guaranteed regulation; that would be a 1K Ohm resistor.

Another thing; an NE555's output decreases about 1.7V with a 100mA load, due to the Darlington voltage follower arrangement in it's output circuitry, and may decrease as much as 2.5v with a 200mA load. If you are considering using a 9v "transistor" battery, these frequently have an 8.6v output when new when tested under light load (25mA, or about 350 Ohms). So, 8.6v - 1.7v = 6.9v remaining; 8.6v-2.5v=6.1v remaining. There are some "industrial" type 9v "transistor" batteries available that output around 10v; they have seven internal cells instead of six.

A 7805 regulator has a specified dropout of 2v. If you are supplying it 6.9v, the most it can output is 4.9v; if you are supplying it 6.1v, the most it can output is 4.1v. This may not be enough to power the logic.

Earlier, eblc1388 mentioned (basically) that the blue LEDs would not light if more than two were placed in series due to their higher Vf, but in the subsequent schematic, three are shown in series. That won't work if you're only going to have 6.9v left to work with; blue LEDs may have a Vf of 3.4v to 4v.
The 1kohm resistor will be put on the vIn or Vout of the 7805? Where in eblc1388's drawing would it go?

Also, I am just going to use all one color low voltage LEDs just to see this working at first to make it easier.
 

eblc1388

Joined Nov 28, 2008
1,542
A 7805 regulator has a specified dropout of 2v. If you are supplying it 6.9v, the most it can output is 4.9v; if you are supplying it 6.1v, the most it can output is 4.1v. This may not be enough to power the logic.
The 78L05 is connected to the battery direct instead of the 555 output. It will be struggling to give 5V if the 9V battery is loaded down to 7V.

That's why I have advised the OP to go for 12V gel cell instead.

Earlier, eblc1388 mentioned (basically) that the blue LEDs would not light if more than two were placed in series due to their higher Vf, but in the subsequent schematic, three are shown in series.
But I have also labelled the battery voltage as 9~12V in the schematic.:)

This is what I have said.

eblc1388 said:
The resistor values for the LEDs are not shown but please refers to previous post about the number of different color LEDs you can place in series with the battery voltage you intended to use.
 
Last edited:

SgtWookie

Joined Jul 17, 2007
22,230
The 1kohm resistor will be put on the vIn or Vout of the 7805? Where in eblc1388's drawing would it go?
It would go on the Vout of the 7805 to ground (battery -).
Look at the datasheet for a 7805 regulator. You will need a small capacitor on the output of the regulator; if the wire to the battery is more than 6" long, you will also need a capacitor on the input.
 

Thread Starter

critiera119

Joined Nov 21, 2008
66
It would go on the Vout of the 7805 to ground (battery -).
Look at the datasheet for a 7805 regulator. You will need a small capacitor on the output of the regulator; if the wire to the battery is more than 6" long, you will also need a capacitor on the input.
Understood. Thank you.
 

Thread Starter

critiera119

Joined Nov 21, 2008
66
eblc1388, I just want to make you sure that you also concur on the usage of the 7805 over the 78L05 with the .33uf at the in and 0.1uf at the out and also a 1k ohm resistor on the out? If using the 7805 with those caps and that resistor, do any other changes to the circuit need to be made?


And also, to make this easier for me..I am going to use just one white LED on each of the three channels with a 220ohm resistor on each...
 
Last edited:

eblc1388

Joined Nov 28, 2008
1,542
eblc1388, I just want to make you sure that you also concur on the usage of the 7805 over the 78L05 with the .33uf at the in and 0.1uf at the out and also a 1k ohm resistor on the out? If using the 7805 with those caps and that resistor, do any other changes to the circuit need to be made?
The 78L05 is just a smaller size 7805 for reduced current output. You can use a 7805 without problem.

I'm not aware that a 1K resistor is needed on the output or can't find this mentioned in the device datasheet. For LM317, a minimum load is needed but I see no reason to use one on 7805.

The input and output capacitors requirement are taken care of in my original circuit and you'll fine.
 

SgtWookie

Joined Jul 17, 2007
22,230
I'm not aware that a 1K resistor is needed on the output or can't find this mentioned in the device datasheet. For LM317, a minimum load is needed but I see no reason to use one on 7805.
You really have to look at the output specifications.
Vo (Output Voltage) and ΔVo (Load Regulation) is usually given when Io=5mA to 1A. This implies that a minimum 5mA load is required to achieve guaranteed regulation.

Since our OP is considering using a 7805 regulator, a minimum 5mA output load is required. I feel that it is likely that his control circuit is CMOS, and as such may have a very low current draw at times - which may cause the 7805 output to become unregulated and possibly damagingly high.

Since the output is 5v and R=E/I, R=5v/5mA=1k Ohms to ensure guaranteed regulation. It may prove to be unnecessary once the current draw is determined. I suggest to err on the side of caution, rather than risk destroying the logic circuit.

References:
Fairchild LMxx/LMxxA 3-Terminal 1A Positive Voltage Regulators datasheet
ST Microelectronics 7800 Series Regulators datasheet
Panasonic AN78xx Series Voltage Regulators datasheet
 

eblc1388

Joined Nov 28, 2008
1,542
I feel that it is likely that his control circuit is CMOS, and as such may have a very low current draw at times - which may cause the 7805 output to become unregulated and possibly damagingly high.
I would suggest you or anyone do an experiment, which would take only 30 seconds at the most, using a 9V battery, a 7805 and a DVM, nothing else.

Post back the output voltage reading on the DVM.
 

Thread Starter

critiera119

Joined Nov 21, 2008
66
And also, if you could address these question marks and also which caps should be electrolytic ( I am guessing at least the ones going in and out of the 7805) and which should be non-polar such as ceramic.
 

SgtWookie

Joined Jul 17, 2007
22,230
I would suggest you or anyone do an experiment, which would take only 30 seconds at the most, using a 9V battery, a 7805 and a DVM, nothing else.

Post back the output voltage reading on the DVM.
I don't wish to appear argumentative; I'm merely passing along conclusions based upon my interpretation of the information contained in the datasheets that I've perused.

All of the 78xx datasheets thus far have indicated that Vo and ΔVo are specified with 5mA < Io < 1A, or 5mA <= Io <= 1A.

National Semiconductor's LM340/78xx series datasheet Vo/ΔVo are specified in the same manner as the others:
http://www.national.com/mpf/LM/LM340.html
Datasheet direct link: http://www.national.com/ds/LM/LM340.pdf

Since the Io in our OP's circuit may fall well below the collective datasheets' minimum specifications, correct output voltage regulation should not be assumed - unless additional load is applied to make up for the shortfall.

I can't predict whether the output may be low, high, or even the possibly of oscillations. In the latter case, a DVM would not be sufficient test equipment; one would need an oscilloscope.

I do happen to have ten 7805's made by several manufacturers, but the results of such a woefully small statistical sample would be irrelevant in light of the millions of these regulators that have been sold by a large number of manufacturers over the years.

There must be a reason that all of the various manufacturers' own datasheet specifications begin at Io >= 5mA.
 

eblc1388

Joined Nov 28, 2008
1,542
I do happen to have ten 7805's made by several manufacturers, but the results of such a woefully small statistical sample would be irrelevant in light of the millions of these regulators that have been sold by a large number of manufacturers over the years.

There must be a reason that all of the various manufacturers' own datasheet specifications begin at Io >= 5mA.
I have an open mind. I can't argue with actual results nor statement quoted in a datasheet. I can't remember seeing any mentioning of a minimum loading requirement.

Can you state or give reference on where(document & page) there is mentioning of a minimum load requirement on 7805 regulator? There is one on LM117 if I remember correctly.

You are actually in a better position than many of us here because you have 7805 from different manufacturers. A test performed with products from different manufacturers would be quite helpful towards this argument.

Why not just do the test and put me and others out of our mysteries. If you like, you can use an CRO and fit Cin and Cout to the 7805 too. I just want to know if the output voltage changes so much as you have mentioned when there no load or when there is a 5K connected at the output.

Of course I have done the experiment myself before asking you to. I would respect any results you have obtained and I would offer you a public apology for doubting your judgment if the results you have got supported your previous argument of "become unregulated and possibly damagingly high".

There must be a reason that all of the various manufacturers' own datasheet specifications begin at Io >= 5mA.
Of course there is and you should know why.
 

SgtWookie

Joined Jul 17, 2007
22,230
7805 test results:



Four different manufacturers. I forget who uE is offhand; it was the only new one of the bunch, the remainder were salvaged from various equipment. All were at least a decade old.

Interesting to note that under the conditions of this test, only the Motorola parts exhibited tendencies to oscillate without caps. I'm sure that given a different set of test parameters (such as power supply lead length, Vin, temp, etc.) the others could also be made to oscillate.

The reason that the output is specified beginning at 5mA is for the same reason that the LM117/LM317 specifications require a minimum 10mA load for guaranteed regulation. In the 78xx series regulators, most manufacturers use 240 to 250 Ohms for R1 (see an LM117/LM317 datasheet). The internal R1 resistance would result in only a 5mA internal load, requiring an additional 5mA external load for guaranteed regulation.

The informal test I conducted is by no means conclusive; it was performed under static load and no-load conditions under ideal temperature conditions (24°C).

However, from the photograph posted it appears that our OP tested their 7805 regulator without using input or output caps. I strongly suspect that theirs is oscillating like my Motorola samples.
 

Attachments

SgtWookie

Joined Jul 17, 2007
22,230
Can either of you address my first post on this page 4? I really want to get this working.
OK, I already mentioned about the input and output caps for the 7805. I'm pretty sure you didn't have any when you tested your regulator, and that it was oscillating at around 7MHz to 12MHz. Adding caps to the input and output of the regulator like I mentioned above should take care of that problem.

As far as the output voltage from pin 3 of the 555, I'll have to guesstimate that you'll see somewhere around 1.3v to 2.3v less than the battery voltage, depending upon the load. Split the difference, and call it battery voltage less 1.8v.
 

eblc1388

Joined Nov 28, 2008
1,542
The reason that the output is specified beginning at 5mA is for the same reason that the LM117/LM317 specifications require a minimum 10mA load for guaranteed regulation. In the 78xx series regulators, most manufacturers use 240 to 250 Ohms for R1 (see an LM117/LM317 datasheet). The internal R1 resistance would result in only a 5mA internal load, requiring an additional 5mA external load for guaranteed regulation.
Thanks for providing a set of informative results. Your results, apart from indicating there is no "big" output voltage changes with or without a 1K (5mA) output load, also made clear that output capacitors(or even input capacitor) must be used to avoid high frequency oscillation.

I do not concur to your reasoning of why regulation figure is often quoted for load current starting from 5mA and up in the datasheet.

I envisage the reason for manufacturers doing so is to just give better specification figure in the datasheet, instead of presenting an inferior regulation figure at 0.5mA which perhaps could be times worse.

Our contention is whether a 5mA load is needed for correct output voltage and not what is the best connection with 7805. If I were to use a 7805 to build a power supply for a preamp, I for sure will follow your advice of adding this loading to enable my 7805 to perform at its best.

I'm sorry for making you do the test and let's view it as being an engineering exchange with no lasting hard feelings whatsoever.

Coming back to the existing problem of our OP, if your examine the photo of OP's post, you will notice that there should be a zener diode to regulate the supply voltage but it was replaced simply by a 12K resistor instead. The voltage required is about 5V and may be a zener diode will also do.

And also, if you could address these question marks and also which caps should be electrolytic ( I am guessing at least the ones going in and out of the 7805) and which should be non-polar such as ceramic.
@critiera119:

Sorry we are caught up in the moment and have ignored your problem.

For capacitor type, one has to look at the symbol and often it will tell you what type of capacitor is required.

If there a "+" marked and there are extra hatching lines inside the capacitor symbol, then it will be an electrolytic. In this case, most often the working voltage of the capacitor is provided. One need to get a capacitor with that rating or higher.

For non-polar capacitor, the voltage rating is often not marked because manufacturer usually manufacturer them to a minimum of 25V or 50V and this is suitable for low voltage application with supply voltage of a few volts. However, if the application involves high voltages, then the voltage requirement will also be marked along with the capacitor value.

For absolute certainty, best performance and sometimes safety purposes, the type of capacitor will also be shown. e.g. Type "X" capacitor for connection across AC lines.

I hope this answer your question.
 

eblc1388

Joined Nov 28, 2008
1,542
You have made several mistakes in the wiring up. Here they are:

1. the connection of the 555 Vcc is wrong. it should be directly to the battery +V instead of to the 7805 5V output

2. I don't see a jumper from pin#4 of 555 to pin#8

3. input capacitor voltage rating at 7805's input is too low. It should be at least 25V, not 100uF 10V as battery is already 16V.

4. output capacitor of 0.1uF ceramic is missing. this should be placed next or very close to the 7805 5V output pin in order to prevent oscillation. (You are definitely having oscillation so voltage measured on your DVM is only 4.57V and not close to 5V)

5. 120K resistor correct marking is Brown-Red-Yellow. Check resistance of yours with DVM.

 

Attachments

Last edited:
Top