Simple tremolo?

Started by xorophone, September 23, 2016, 04:20:12 PM

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xorophone

Hello! I'm looking for a very simple tremolo schematic which uses a 555 or 556 timer. I just want to have a simple square wave because I'm actually not building a tremolo pedal. I want to implement the circuit in a function generator to be able to hear effects such as delay and reverb easier.

I've been looking at this one: http://www.diystompboxes.com/smfforum/index.php?topic=87322.760
Which value should the LDR be? There's only 2-20k Ohm available at my local electronics store, so of course I want to keep it in that range if it's possible.

Thanks!

Edit: Also, can the 1n4001 diode be replaced with a 1n4148? I've got some 1n5817 and 1n34a too, but I've never seen anyone use them this way before. Which one should I choose?

Transmogrifox

The value of the LDR all depends what you're doing with it.  This LDR works against the output impedance of whatever is feeding it -- in this case presumably a guitar.

You can use any value LDR if you are driving this with a low impedance source.  Then you can place a resistor between the input to the LDR and tap output between LDR and resistor.  Then the resistor can be sized to work with whatever LDR you have...even a pot if you want to have adjustable modulation depth.

Now if all you want is a square wave oscillator then you don't even need the LDR.  I'm assuming you want the square wave oscillator to turn something on and off, though.

So...if all you want is signal on/off you can probably just connect a BJT collector to the signal path, emitter to ground, and the 2.2k resistor (driving the LED) to the base, instead of an LED.

It will produce clicks as it cycles on and off (not a very nice effect) but as a test generator it may be a simple way to do what you need.
trans·mog·ri·fy
tr.v. trans·mog·ri·fied, trans·mog·ri·fy·ing, trans·mog·ri·fies To change into a different shape or form, especially one that is fantastic or bizarre.

xorophone

Quote from: Transmogrifox on September 23, 2016, 06:19:52 PM
The value of the LDR all depends what you're doing with it.  This LDR works against the output impedance of whatever is feeding it -- in this case presumably a guitar.

You can use any value LDR if you are driving this with a low impedance source.  Then you can place a resistor between the input to the LDR and tap output between LDR and resistor.  Then the resistor can be sized to work with whatever LDR you have...even a pot if you want to have adjustable modulation depth.

Now if all you want is a square wave oscillator then you don't even need the LDR.  I'm assuming you want the square wave oscillator to turn something on and off, though.

So...if all you want is signal on/off you can probably just connect a BJT collector to the signal path, emitter to ground, and the 2.2k resistor (driving the LED) to the base, instead of an LED.

It will produce clicks as it cycles on and off (not a very nice effect) but as a test generator it may be a simple way to do what you need.

Thank you for the explanation, Transmogrifox.

I think the BJT method you explained should be enough for my build. Should I use an NPN or PNP transistor? Right now I've got MPSA13, MPF102 and 2n3904, but i'm guessing only MPSA13 and 2n3904 are BJT transistors (Sorry but I don't know much at all about transistors.) Both of them are NPN too, but if I need another one I can of course buy it.

Do you know if the 1n4001 diode can be replaced with a 1n4148, 1n5817 or 1n34a diode? I'm guessing the 1n4148 would be my best bet.

GibsonGM

The 1N4001 going to ground there, that's a protection diode. It will "clamp" if you apply the wrong polarity voltage (install the battery backwards). It does not conduct in normal operation, as you can see by where the cathode (bar) is pointing  (toward positive).   

You CAN use a 1N913/4148, but their ability to carry current is much less than a power diode, and so they would probably burn out - when a polarity issue is going on, the diode IS conducting in a little loop there, and must be able to dissipate the power involved...so, the 5817 Schottky SEEMS to me to be the better choice, if you can't get a 1N400X series diode.
In the future, I'd go with those, they will give you a better guarantee...

To find out more about parts, go to Google etc., and type in "1N5817 data sheet", for example...then you can read about your parts!  I made a library for myself of PDF's of the common 'things', to refer to...
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xorophone

Quote from: GibsonGM on September 23, 2016, 08:24:26 PM
The 1N4001 going to ground there, that's a protection diode. It will "clamp" if you apply the wrong polarity voltage (install the battery backwards). It does not conduct in normal operation, as you can see by where the cathode (bar) is pointing  (toward positive).   

You CAN use a 1N913/4148, but their ability to carry current is much less than a power diode, and so they would probably burn out - when a polarity issue is going on, the diode IS conducting in a little loop there, and must be able to dissipate the power involved...so, the 5817 Schottky SEEMS to me to be the better choice, if you can't get a 1N400X series diode.
In the future, I'd go with those, they will give you a better guarantee...

To find out more about parts, go to Google etc., and type in "1N5817 data sheet", for example...then you can read about your parts!  I made a library for myself of PDF's of the common 'things', to refer to...

Thank you so much! I think I'll buy some 1n4001 diodes. Just wanted to see if the ones I already had would work. That PDF-thing sounds like a very good idea, so I'll have to do that for myself sometime too.

Just to make sure my other question doesn't get lost in the thread; Should I use an NPN or PNP transistor for the switching? Which transistor would work the best?

Kipper4

Welcome Anton

"So...if all you want is signal on/off you can probably just connect a BJT collector to the signal path, emitter to ground, and the 2.2k resistor (driving the LED) to the base, instead of an LED.

It will produce clicks as it cycles on and off (not a very nice effect) but as a test generator it may be a simple way to do what you need."

This suggests an NPN to me.

Mpsa13 is a Darlington transistor
2N3904 is an NPN bjt
MPF102 is a jfet

They are not the same animals.

Check out the 2N3904 data sheet for the transistor pinout
Save to your new pinout and data sheet folder.

Ma throats as dry as an overcooked kipper.


Smoke me a Kipper. I'll be back for breakfast.

Grey Paper.
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Transmogrifox

Yes, NPN is what I had in mind. 
1N4148 is better than none at all.

As mentioned the 1N4001 is just reverse-polarity protection.  The circuit will work fine without it if you're not worried about accidentally reverse-connecting power to it.    The reason for putting a power diode like that is to provide a kind of reverse supply protection without losing the headroom from the diode drop during normal operation.  You can use the 1N4148 in series with the battery like this, substituting 1N4148 for the 1N5817:

And D1 is not necessary in your situation but you could substitute a 1N4148 here too if you are concerned about getting nasty reverse voltage spikes >100V (thinking ESD), although a series resistor like the 100 ohm above and a small (10nF) capacitor across the IC power pins is a better way to absorb that.  I really can't think of a reasonable situation where D1 would ever do anything useful

Also if you are only using a battery then the 1N4148 will take it long enough for you to notice something isn't right.  A 500mA wall-wart supply would probably damage a 1N4148 if this circuit was reverse-connected.

If you use this in a situation where you know you won't accidentally connect power backwards you can forget about the diode.  It's just nice insurance that will save your 555 IC if you get polarity backwards for a couple moments.
trans·mog·ri·fy
tr.v. trans·mog·ri·fied, trans·mog·ri·fy·ing, trans·mog·ri·fies To change into a different shape or form, especially one that is fantastic or bizarre.

xorophone

Quote from: Kipper4 on September 23, 2016, 09:57:01 PM
Welcome Anton

"So...if all you want is signal on/off you can probably just connect a BJT collector to the signal path, emitter to ground, and the 2.2k resistor (driving the LED) to the base, instead of an LED.

It will produce clicks as it cycles on and off (not a very nice effect) but as a test generator it may be a simple way to do what you need."

This suggests an NPN to me.

Mpsa13 is a Darlington transistor
2N3904 is an NPN bjt
MPF102 is a jfet

They are not the same animals.

Check out the 2N3904 data sheet for the transistor pinout
Save to your new pinout and data sheet folder.

Perfect! Thank you so much!

Quote from: Transmogrifox on September 23, 2016, 10:45:20 PM
Yes, NPN is what I had in mind. 
1N4148 is better than none at all.

As mentioned the 1N4001 is just reverse-polarity protection.  The circuit will work fine without it if you're not worried about accidentally reverse-connecting power to it.    The reason for putting a power diode like that is to provide a kind of reverse supply protection without losing the headroom from the diode drop during normal operation.  You can use the 1N4148 in series with the battery like this, substituting 1N4148 for the 1N5817:

And D1 is not necessary in your situation but you could substitute a 1N4148 here too if you are concerned about getting nasty reverse voltage spikes >100V (thinking ESD), although a series resistor like the 100 ohm above and a small (10nF) capacitor across the IC power pins is a better way to absorb that.  I really can't think of a reasonable situation where D1 would ever do anything useful

Also if you are only using a battery then the 1N4148 will take it long enough for you to notice something isn't right.  A 500mA wall-wart supply would probably damage a 1N4148 if this circuit was reverse-connected.

If you use this in a situation where you know you won't accidentally connect power backwards you can forget about the diode.  It's just nice insurance that will save your 555 IC if you get polarity backwards for a couple moments.

Thanks for the great explanation! I'll look further in to it and see if I'll include the diodes.

GiovannyS10

My first tremolo was like this. Only a vactrol liked to a 555 IC with a pot to control the frequency and other to control the led bright - its control the effect deep. If you want some explanation, can PM me, but i think the guys said you enough  :icon_mrgreen:

That's all, Folks!

"Are you on drugs?"
-ARSE, Duck.

www.instagram.com/allecto

xorophone

Quote from: GiovannyS10 on September 24, 2016, 05:49:31 PM
My first tremolo was like this. Only a vactrol liked to a 555 IC with a pot to control the frequency and other to control the led bright - its control the effect deep. If you want some explanation, can PM me, but i think the guys said you enough  :icon_mrgreen:

Nice! :) Does the circuit work well? Any value or something you think I should change to make it better?

GiovannyS10

Yes, worked very well... I will send you a PM showing it.
That's all, Folks!

"Are you on drugs?"
-ARSE, Duck.

www.instagram.com/allecto

xorophone

Quote from: Transmogrifox on September 23, 2016, 06:19:52 PM
Now if all you want is a square wave oscillator then you don't even need the LDR.  I'm assuming you want the square wave oscillator to turn something on and off, though.

So...if all you want is signal on/off you can probably just connect a BJT collector to the signal path, emitter to ground, and the 2.2k resistor (driving the LED) to the base, instead of an LED.

It will produce clicks as it cycles on and off (not a very nice effect) but as a test generator it may be a simple way to do what you need.

Hmm.. I just realised that I'm going to be turning the 9v signal on and off, so shouldn't I be able to just connect the 555 pin 3 (output) to the rest of my circuit and skip the transistor?

Transmogrifox

If you're just turning the 9V on and off then forget all about the LDR or transistor, LED's etc.  Just drive directly off the 555 chip.  The 555 is already turning 9V on and off for you.

Leaving the one LED+resistor in circuit might be nice so you can see the LFO is working.
trans·mog·ri·fy
tr.v. trans·mog·ri·fied, trans·mog·ri·fy·ing, trans·mog·ri·fies To change into a different shape or form, especially one that is fantastic or bizarre.

xorophone

Quote from: Transmogrifox on September 26, 2016, 01:37:11 PM
If you're just turning the 9V on and off then forget all about the LDR or transistor, LED's etc.  Just drive directly off the 555 chip.  The 555 is already turning 9V on and off for you.

Leaving the one LED+resistor in circuit might be nice so you can see the LFO is working.

I'll try that. :)

For future reference; when I'm using a transistor as a switch, does the emitter have to go to ground or can I use it as an output?



This is how I'm guessing you told me to wire it, but this would ground the 9v going in to the 555 chip too and kill the circuit, right? So can I instead just use the emitter as the square output? If that doesn't work, I'm guessing I should add a diode or something to separate the power going in to the 555 chip and the power that's being modulated by the transistor. Is that right?

I'm sorry for all these questions. I'm just trying to learn as much as possible.
You're all very helpful. :)

balkanizeyou

yes, this arrangement would fry the transistor within seconds (probably less, depending on the power supply).
The idea was like this:



As with everything in electronics, the behaviour always depends on what exactly are you trying to switch. In the case of the circuit on the picture, as you noted, the input cannot be a voltage source for example. This should work fine when the "signal in" has a rather high output impedance, because connecting it directly to, say, an opamp output wouldn't make the opamp very happy. It would be easier to help if you showed us what exactly you're trying to switch on and off.

If you just want to stick the square wave somewhere, you don't need the transistor. Using an emitter follower (if you wanted to do that you would need an emitter resistor) would make sense if your square wave source was a high impedance source - the emitter follower would lower the output resistance which would make driving loads easier. But the 555 can supply 100mA of current without breaking a sweat, so in most cases using a transistor here would be redundant (but, as ususal it depends on what exactly you're trying to switch).

xorophone

#15
Quote from: balkanizeyou on September 26, 2016, 06:54:26 PM
yes, this arrangement would fry the transistor within seconds (probably less, depending on the power supply).
The idea was like this:



As with everything in electronics, the behaviour always depends on what exactly are you trying to switch. In the case of the circuit on the picture, as you noted, the input cannot be a voltage source for example. This should work fine when the "signal in" has a rather high output impedance, because connecting it directly to, say, an opamp output wouldn't make the opamp very happy. It would be easier to help if you showed us what exactly you're trying to switch on and off.

If you just want to stick the square wave somewhere, you don't need the transistor. Using an emitter follower (if you wanted to do that you would need an emitter resistor) would make sense if your square wave source was a high impedance source - the emitter follower would lower the output resistance which would make driving loads easier. But the 555 can supply 100mA of current without breaking a sweat, so in most cases using a transistor here would be redundant (but, as ususal it depends on what exactly you're trying to switch).

In this case I'm going to be switching the power on and off on this circuit: http://www.circuitstoday.com/simple-function-generator-circuit . I'm assuming this should work fine with the 555 output without using a transistor.

Ah, emitter follower. I'll have to remember that.

Edit: The reason I'm planning on modulating the op amps power supply is because the outputs split to square, triangle and sine and I don't want to mix them. From what I've understood this should work fine.