Reserve Capacitor Query - to prevent injection of noise into supply

[crush]

Hi All,
Have a simple transmitting circuit.**broken link removed**
refer link for circuit and explanation:
http://generous.boy.googlepages.com/transmittingcircuit

Note: Vcc= 10V

Decided to add a reserve capacitor C1 to supply the LED. To prevent the supply from injected with noise when it see's a big gulp of supply being eaten by the LED.

The resistor R1 just provides some resistance, so that the LED sources the supply from the easiest path i.e. from the capacitor.

Question
What will decide the value of C1 and R1? I randomly chose the 47uf since it seemed a standard value.

- after looking throught a bit : delta v= (i *t)/C correct??

i is the current across the LED = 1 A

t = on period = 10us

but what would delta v be? 10V- vdrop across LED = 9.8V??

 

[dknguyen]

What is causing the noise is the rapid change in current demand. Capacitors can react to these changes faster than batteries and power supplies so they will "absorb" the spikes. The power supply can't react to the change in current demand as quickly as the capacitor can so I wouldn't worry too much about what "source" the LED is drawing from if both the cap and power supply are there. I would just stick a capacitor across the power rails so that the capacitor will "absorb" the spikes and reduce noise. In this way, the capacitor should charge up faster than it can discharge too (since there is a resistor in the discharge path, while there is nothing in the charge path).

With how you have it right now, I don't get why you need the 10ohm resistor- all it would do is sap power and make the LED slightly dimmer. I also don't see how it would make that the LED would take power from the capacitor...there is already R1 sitting between the LED and power supply and a short between the cap and LED. The cap is already an easier path. R1 also limits how quickly the capacitor can charge up. It's just not needed.

Size also doesn't matter too much. The LED is probably being switched at low frequency (not MHZ, not even kHz. 30Hz-60Hz is all that is needed to fool the human eye). So you just use a "large" capacitor. 47uF is probably fine, as is 10uF or whatever else.

 

[hjames]

crush doesn't mention it, but this looks like a drive for an IR transmitter - the TSIP4400 is supposed to be a 3mm IR LED with a surge rating of 2A. The extra resistor is probably there to prevent the LED and 10 Ohm resistor from being a fire hazard if some of the logic goes funny.

They way you'd chose R1 probably isn't from the i=CdV/dT equation. Here's some criteria for chosing R1:

1) What's the fastest rate that you want to pulse the LED - if C1 isn't "fully" charged before you start a pulse, you won't be able to do a full brightness pulse. Read up on RC time constants...

2) What's the max current you want to draw out of the power supply - After a pulse, C1 will have dropped in voltage by a bit (use the i=CdVdT equation to figure out how much it dropped), R1 will limit the charging current.

3) What's the maximum amount of power you want the LED to dissipate if the logic decides to turn the transistor on ... for a couple seconds.

 

[Nigel Goodwin]

I wouldn't bother calculating anything! - just look what's used in the millions of IR remote controls - usually 220uF, 470uF or 1000uF. Essentially you only transmit short bursts of IR, so the capacitor provides the power for it, and charges back up while in between bursts.

 

[crush]

The extra resistor is probably there to prevent the LED and 10 Ohm resistor from being a fire hazard if some of the logic goes funny.

i wasnt intending it to be. The way i saw it was the 10 ohm R2 limits the current to the LED to 1 A.
R1 was solely there to add impedance so that instead of the LED sourcing from the supply, it would source from C1 instead- path with least impedance.
Is this wrong?

1) What's the fastest rate that you want to pulse the LED - if C1 isn't "fully" charged before you start a pulse, you won't be able to do a full brightness pulse. Read up on RC time constants....

- the fastest rate i want to pulse the LED is once in evry 10us.
- Cap fully charges up in 5t
- 5t=RC
- hence the largest capacitor i can use is 4.7uf (considring R1=10ohm).
This is much smaller than values of C i have seen around. so just want to confirm, am i right??

2) What's the max current you want to draw out of the power supply - After a pulse, C1 will have dropped in voltage by a bit (use the i=CdVdT equation to figure out how much it dropped), R1 will limit the charging current.

doesnt R1 control the current to 1A? And since i am already choosing C such that it is fully charged (5t) before my next pulse, would mean that there is no drop in voltage in the Cap, correct?

3) What's the maximum amount of power you want the LED to dissipate if the logic decides to turn the transistor on ... for a couple seconds.

I hv changed the LED to a 5mm **broken link removed**
I am thinkin that 1A would fry if it were left on for a few seconds. But at the same time i cannot increase R1 (calculated above to be 10 ohms) because it'll decrease the size of my C1 > ryt?

 

[hjames]

Okay, so first of all, do you really want to pulse this LED for 10uS every 10uS? Depending on how you read that - you want the LED 100% on or just 50% on?! In either case, the LED isn't going to survive. Calculate the number of watts that you want to push into this thing and make sure that the thing will survive. Then make sure the resistors survive, because with those numbers, you need a 5-10Watt resistors...

I brought up the RC time constant thing so that you'd be able to get some minimum value for the capacitor - not the resistor. Then again I didn't think you wanted to have such a high duty cycle - mainly because LEDs won't survive at these duty cycles.

 

[dknguyen]

i wasnt intending it to be. The way i saw it was the 10 ohm R2 limits the current to the LED to 1 A.
R1 was solely there to add impedance so that instead of the LED sourcing from the supply, it would source from C1 instead- path with least impedance.
Is this wrong?

Yes, or unecessarily overthinking it at least. It will draw current from both though varying degrees. What the resistor does is slow the charge up of the capacitor. The goal isn't to "disconnect" the power supply from the LED- spikes are going to happen anyways when the capacitor needs to get charged. The goal is to have the capacitor meet the transient current demand that the power supply is not fast enough to react to.

 

[crush]

I brought up the RC time constant thing so that you'd be able to get some minimum value for the capacitor - not the resistor.

I understood, i thought that is wot i did. I was trying to keep the capcitor as big as possible, hence had to choose a small R.
Instead of the minimum value, from the calculations i posted above it appears that my C1 cannot be larger than 4.7uf. is this correct?

I had the circuit currently workin on 10us pulses (50% duty cyle), without R1 and C1. But noticed the resistor was getting a bit hot. So now have a 1us "on" pulse and 10"us" off pulse as the fastest signal from the transmitter.

The way i have it now:
Vcc = 10v R2=1o ohms. Power = VI = 10V * 1A = 10W.

but the max power dissipation of the LED is 100mW. The LED still works though, even for extended periods like 30 mins!

QUOTE=dknguyen]
The goal isn't to "disconnect" the power supply from the LED- spikes are going to happen anyways when the capacitor needs to get charged. The goal is to have the capacitor meet the transient current demand that the power supply is not fast enough to react to.[/QUOTE]

The goal is to disconnect the power supply. This is because the transients are being seen through the supply at the reciever! Hence my thinking as to adding R1

 

[dknguyen]

You are misunderstanding something I think. You can't disconnect the power supply or it wouldnt run. The only way to do that is to power the LED from it's own battery. When the LED first turns on, it causes a change in power demand for the supply. Supplies can't respond to the change in demand as fast as the demand can occur so it causes dips in the supply voltage (which is the transients that you see). Capacitors, however, can react to this demand. What they do is supply the power while the supply is taking time to react. This allows the supply more time to react so that dips do not occur (or smaller dips, but you can't really eliminate those anyways since nothing can react instantly).

 

[hjames]

First thing's first:

The way i have it now:
Vcc = 10v R2=1o ohms. Power = VI = 10V * 1A = 10W.

but the max power dissipation of the LED is 100mW. The LED still works though, even for extended periods like 30 mins!

Hold on a second there - you haven't fully applied the equations. The forward voltage of the LED is probably ~2V at 1Amp. What's the resulting power dissipated in the LED? I think you've misplaced a decimal point, or your circuit isn't behaving like you think it should.

The spec sheet for the LED indicates that the max dissipation of the LED should never exceed 100mW - if you don't want to cook the LED. At this point, I think the LED that you've been using is probably not it very good health.

 

[Nigel Goodwin]

Perhaps you might like to mention what you're trying to do?.

 

[crush]

Perhaps you might like to mention what you're trying to do?.

Sorry, but i thought it might be too much to write down.
But specifically,

I am transmitting unique ID's:
10us on, 10 us off - ID 1
10us on, 20 us off- ID 2
and so on.. (note have now changed the "on" period to 10 us.

At the moment i am just simply turnin the transmitter on and off with a fixed "on" period and a varying duty cycle. It is controlled by a simple program toggling the pins of a micro which is controlling the mosfet to the transmitter LED.

These transmitted signals are recieved by a circuit and demodulated. the demodulations find out the id based on the delay from one pulse to another. simple system

The Entire system and the concept can be viewed at http://generous.boy.googlepages.com/photodioderecievingcircuit

 

[Nigel Goodwin]

Well that's not really "what you're trying to do", that's "how you're trying to do it" - but presumably it's some kind of remote control or wireless data transfer?.

Why try and reinvent the wheel?, there are many types of existing IR remote control systems - copy one of those. The first thing you need to do is use a carrier frequency - 38KHz or so is commonly used, and allows you to use existing (high performance) IR receiver chips - which will 'wipe the floor' with your simple opamp version.

My PIC tutorials show how to use the Sony SIRC's system, but there are many other system for which you can fine information on the net.

 

[redserpentone]

crush: you did not mention where were you testing, which node?
if you were placing the scope probe at the node of c1, c2, r2 and led yes you are going to see the high floor of the diodes voltage drop approx 0.7 volts which is what I see in the picture of your scope reading. How are you conecting and using your scope, is your coupling dc or ac?

The problem might be your voltageb divider
the potential between crest and floor is 0.2v which is about your voltage drop of your q1 r2(47 hm: )
your circuit works well not what you want though.
As long as your LED is stuck to the bottom rail acting as a slow fuse or circuit on detector all you will see is your LED voltage drop.

Sorry, scrap the cuircuit and use another design.
best of luck, rs1

 

[redserpentone]

I was reading the diagram that you suppled with the pictures of your scope readings , by the way which is totally different from the one above.

cordially, RedSerpentOne

 

[hjames]

I'd go with the 38KHz devices like Nigel mentions - they're pretty much bulletproof - work in daylight, all the good stuff. The only reasons to develop your own are:

1) It's fun - but don't expect to exceed the performance of a module...
2) Need high speed - look at IRDA specs, it is possible to do 16MBit/sec over IR.
3) Need low jitter - some timing related stuff...

Otherwise it's probably easier to just use what every TV for the last 2 decades have used...