Replacing 1N4001 Diode With Capacitor to Prevent Reverse Flow and Boost Voltage?

I am trying to figure out how to compensate for voltage drop in my power section due to diode drop. There are 2 diodes in the design and both are 1N4001, the first helps stabilize and regulate the voltage of an LM7809 to 9 volts but I am reading 8.91 volts on the meter.

The second diode is possibly redundant, but I thought it was a good idea to have it based on other designs I have seen. After that diode the voltage is down to 8.40v.

This what I am wanting to know...

Would it be possible to replace the second diode with an electrolytic capacitor since this diode is mainly there to make sure there is no reverse power flow?
If so how do I calculate the cap size needed to make up the .09 volts that are lost.

It seems like the Electrolytic would be able to boost the voltage and stop any reverse power flow by it's design.

Please go easy on me, I am self trained and know I have a lot to learn still. This forum seems like a place that encourages questions and learning, so I have put it out to you for help.

Thanks,

Danny

Hi Danny,
It would be a lot easier to answer your questions with a circuit schematic, can you post it if you have one?

Lets look into your first problem, the low output voltage of the 7809. If you look at the lm7809 datasheet you will notice that the typical output voltage is 9V but that the minimum is 8.64V so it isn't strange at all that your output is 8.91V. You must remember that with most voltage regulators the output voltage will never be exactly the value you want, there is always a minimum and maximum which you can look up in the datasheet.

Short answer to your question: A voltage cannot be boosted with a capacitor, if you dont want that voltage drop, remove the diode. I need to see the schematic before i can tell you exactly what to do, i need to know where that diode is placed.

I agree with Boi Okken, a circuit diagram would be helpful.

I'd would assume that one diode is reverse biased from the 12V to the 9V and the other is attached from the 9V to ground on the output. I would assume that a 0.1uF capacitor on the output would be sufficient for stability. 0.33uF on the input will keep that 12V steady as well, which will help on the output as well. The capacitors will only really maintain a steady output, they won't boost the voltage.

What is the specifications of your circuit? What tolerance would you like to have on your output? You will always have to deal with tolerances, so understanding what your circuit can handle is helpful. There's a higher tolerance version of the same part: LM7809A, which ranges from 8.82V up to 9.16V. Another option you have of increasing the voltage output, which is in the datasheet by attaching a voltage divider on the output connected to the ground pin of the device. There's even an equation there you can plug your values into. See Figure 12, Circuit for increasing output voltage. In effect it behaves like a DC offset. Also from the notes in Figure 11, you can only boost up to 2V lower than the input voltage, which makes sense.

I agree lets have a circuit first Otherwise no one has a clue as to what your on about ??

OK, sorry for the delay, I was having PDF issues. Here is the schematic.
Thanks for all the help!

Danny

BTW - I know I spelled Referred incorrectly, whoops!

Here is the Schematic to look at.

damn, its one of those days... i accidentally hit the vote down on your answer. very sorry, is there a way to fix that?

I guess that if you want a diode there to be you can use a schottky diode to lower the voltage drop. If you dont want the voltage drop i would suggest to just remove the diode all together

I will vote it up to compensate

In your opinions do I even need that diode? If not I'll yank it out of there. Do you see anything in the schematic that is obviously wrong that I have missed?

Thank you.

Here's the schematic

Here's the schematic.

Hi Danny

Have you checked your battery in the DVM ? Silly question but I’ve known this to give false readings
What is Vin to the LM7809 ?
Have you tried increasing the voltage to the input of the LM7809?

Too low a voltage on the input to this device gives this effect typically you want at least 12 to 15 volts dc in

What is the purpose of D2 ?? Is it not to ensure that the input voltage remains higher than the output of the LM7809

What would happen if the input voltage dropped below 9 volts ?? Interesting !!

Why do you need an input filter and why isnt the input filter R1 ,C1 , C2 connected as follows C2 -> R1 - > C1

The other point I noticed was your current supplying the leds Why are you using a 3K3 resistor with a 390R in series The led with these values would never ullimnate assuming your current supply even if it were 9 volts

What is your forward voltage drop across D4 and D4B


If you disconnect R2 and R2B does the voltage = 9volts or at worst 8V95

Why do you need D3 ?

I can only see this as an attempt to drop the output voltage under load conditions by 0.6 volts or possibly to prevent another part of the circuit which may rise above 9volts damaging the regulator

Where does + V Bridge connect to have you disconnected that part of the circuit ?

What happens if you disconnect pin 3 to the TLE2426 ? Maybe its drawing more than the specified operational current or close too 800ma there about

Have you disconnected the output leg of the LM7809 and performed a current check to the rest of your circuit to determine how much current is being consumed by the circuit ?

Is the regulator getting hot ??
Does the current into the device exceed 1 amp ?
What is the maximum current of V IN ?

In other words does your supplyinto the Reg deliver more than the 1 amp ?

Need those answers first

Mark Harrington

www.harrington.force9.co.uk

The usefulness of the diode totally depends on the situation, the diode can be safely removed if there wont be a higher voltage at the input.

*I am running this circuit from a 12v 100mA DC adapter. Most circuits that will run from this will average a draw of 20 mA or less.*

What is the purpose of D2 ?? Is it not to ensure that the input voltage remains higher than the output of the LM7809
*Yes it is to ensure proper input output levels and supposedly help to stabilize and regulate the circuit even more. Now that I look at it and think I am wondering if D2 is actually needed since the input source is a 12 volt adapter and will always have the voltage headroom for the LM7809 to operate correctly.*

Why do you need an input filter and why isnt the input filter R1 ,C1 , C2 connected as follows C2 -> R1 – > C1

*The input filter is a hum eliminator to help kill any possible noise coming from an adapter. This power section is for musical instruments and effect pedals and this design is supposedly "tried and true" or so they told me… That is why I have them going R1 -> C1 -> C2. D1 protects against reverse polarity in case someone plugs in the wrong adapter. R1 C1 & C2 are the filter that reduces hum from the line. What would be the difference switching them to C2 -> R1 -> C1 from how I have it?*

The other point I noticed was your current supplying the leds Why are you using a 3K3 resistor with a 390R in series The led with these values would never illuminate assuming your current supply even if it were 9 volts

*I used an online calculator to re-crunch these numbers and the 3k3 is incorrect, it should be a 2.74k instead. Here is what the calculator gave me… 9 V Input, 3.6 volt drop across LED, LED current 2 ma = 2.74k resistor. 2 mA across the LED does light it up pretty well actually, and for being onstage in dark situations that 2 mA of current makes the LED more subdued and less glaring. I had originally tried the LED at 20 mA and found it to be a bit distracting on stage. I added the 10uf Electrolytic and the 390r resistor to help combat "Pedal Pop" from the LED turning on when the unit is activated. Having this configuration helps the LED come on and go off gradually instead of instantly. The 390r resistor limits the current in the LED when C1 dumps its charge as the switch is toggled.*

What is your forward voltage drop across D4 and D4B
*Voltage drop across D4 and D4-B are both at 3.6v, theLED's are blue ultraviolet*

If you disconnect R2 and R2B does the voltage = 9volts or at worst 8V95
*With this current schematic configuration, removing R2 and R2-B would make the voltage hitting the LED's would be 8.91 v*

Why do you need D3 ?
*Good question… I originally thought I needed it to protect the regulator but I am starting to think it was overkill for the small current and voltage present. D3 is really the biggest problem in the schematic and it is the original reason I posted this question because I started thinking it might be unnecessary.*

I can only see this as an attempt to drop the output voltage under load conditions by 0.6 volts or possibly to prevent another part of the circuit which may rise above 9volts damaging the regulator
*It was intended to protect the voltage regulator*

Where does + V Bridge connect to have you disconnected that part of the circuit ?
*V+ Bridge gives the full voltage from the LM7809 to power IC's that do not use the split rail for power (chips like TL074's LM13700's etc) so they are getting 9v of power. This power section is designed to be a drop in to power many different synthesizer and effect pedal designs. I wanted a unified plug and play power module to place into any schematic I come up with.*

What happens if you disconnect pin 3 to the TLE2426 ? Maybe its drawing more than the specified operational current or close too 800ma there about
*The design in the rail splitter section was the only good example of using the IC with the noise reduction offered in the 8 pin configuration that I could find. Disconnecting pin 3 would stop the current from entering the IC wouldn't it? Here is where I located that circuit design http://tangentsoft.net/elec/vgrounds.html*

Have you disconnected the output leg of the LM7809 and performed a current check to the rest of your circuit to determine how much current is being consumed by the circuit ?
*Please explain this question to me a little more, I don't fully understand how to do this check. If i disconnect the output from the LM7809 wouldn't that kill all power to the rest of the circuit? Please be kind, remember I am still learning as I go.*

Is the regulator getting hot ??
*No it does not get hot.*


Does the current into the device exceed 1 amp ?
*No, the DC adapter only pushes 100 mA and most of the circuits this will be applied to run in the area of 20 mA.*


What is the maximum current of V IN ?
*12.6 volt DC 100 mA from the adapter*

In other words does your supply into the Reg deliver more than the 1 amp ?
*No, 100 mA is the limit on the adapters I am using.*

I hope that I have answered everything clearly and correctly. All of your help is very appreciated.

Hi Danny

Have you disconnected the output leg of the LM7809 and performed a current check to the rest of your circuit to determine how much current is being consumed by the circuit ?

Explanation for you here
See attached drawing as to how to
I'll will explain what I’m doing
What we want find out is what part of the circuit is drawing current either the supply into the LM7809 is insufficient in terms of current because the load on the LM7809 is higher than 100 ma "Possible at this stage " and we are seeing the regulator not behaving properly because there is insufficient current to start with

The reason why I say this is because you are telling me "V+ Bridge gives the full voltage from the LM7809 to power IC’s"

I take it that you mean some type of audio output IC if that’s the case you need at least 100 ma in idle mode i.e no output no signal into the power IC

So what we do here is unsolder the output leg of the LM7809 and we connect a dvm , Milliamps range one meter lead to output leg and then the other to where the output leg which connects to the rest of the circuit

If the reading is lets say 80 ma or somewhere there about’s we know that the input from the adapter is insufficient i.e we should be supplying 500ma as opposed to 100 ma I suspect very strongly that is where you problem lies

As a general guide this capacitor should be rated at a minimum of 1000uF for each amp of current drawn that’s on the output side at the moment you only have 247uf present on the input no where near what they recommend

What we need to do is sum all the currents drawn on the 9volt rail and see whether we are reading the same on the DVM

I.e. How much current is the led network drawing so we have as an example D4 + D4b which gives us lets say 2ma as you have told me that’s 4ma total so far

Next we need to add what the TLE2426 might consume

If we look at the datasheet this suggests maximum Output current, 80ma now add 4ma and we are now at what 84ma (Not looking to good !! )

Don’t forget that for our regulator to function then it must consume some current even if disconnected on the output side to the rest of the circuit

So lets say 4 ma for the Reg in idle mode (Not connected to the rest of the circuit but being supplied by your DC Adaptor

Now we are up to 88ma Doesn’t leave a lot for any other devices does it not to mention possible power amp circuitry which you are telling me is connected to V+ Bridge

So I’m now looking at what exactly is being drawn

You cant draw more current than current into the Reg say your circuit draws 120ma and your Dc Adaptor only supplies 100ma you will only ever get slightly below 90ma at best from the Regulator which means the voltage into the regulator is going to drop causing the output to drop

This looks as though from what you are telling me is exactly what I think is happening

So lets find out just how much the total current sums to then we can ascertain if the DC adaptor you are using is in fact too small Knowing how these Adaptors are built that wouldn’t actually surprise me

I looked at your schematics and my first question is do you need that compensation or are compensating because you read that it should be 9V?
As i read your schematic the diod is a decoupling. Used to eliminate noise.
Looked a fast look at the tle2426 and it should work own to 4 V so there should not be any problem.
I dont remember if the LM7809 has a spec of +-5% or if its +- 10% but 8.91 V would say 5% :-)

So do you have a problem or are you creating one?
But if you really need that compensation you could use an diod from gnd on LM7809 to your real ground.

Vin---lm78---Vout
|
V
-
|
GND

Since the LM7809 have a voltage drop of about 3-4 V you should check that you have 9.7+ 4= 13.7Volt or more.
Good luck

That’s what I said all along Insufficient voltage in and not enough current to start with .i.e 100ma at 12v6 He needs at least 500ma and more at 15v0 and above up to 26 volts max for this to function correctly Not forgetting its a 1 amp device with inbuilt current protection capable of delivering 1v5 amps max with a Quiescent Current of 8ma i.e no load conditions Under load the device will drop voltage to 8v6 if approaching 1 amp load current but since he is only supplying 100 ma to start with then it wont regulate properly if peak current on switch on is 100ma or above

Thanks to all for the help, pointers, advice and knowledge! It all works now!