555 Voltage Doubler

[spacecommandant]

I have a circuit which, with 9V, draws around 31mA. I've been using this voltage 'doubler' (image attached) with a NE555P chip to bring the voltage to 15.9V and the current to 75.5mA.
It sounds fine, no issues I can detect, but I don't have much experience using doublers so I'm curious if this looks like a good circuit to consistently use with that amount of current.
I'd likely use an SMD version of a 555 chip based upon availability.
Thanks in advance!

[Clint Eastwood]

As long as the IC doesn't get hot, I suppose it's ok.

[GibsonGM]

If it's not noisy or getting hot or experiencing other problems...and the IC isn't sourcing or sinking greater than 200mA, then it should be ok. 

What I can't easily surmise is how much current it IS sourcing in this kind of arrangement.   

[iainpunk]

as long as the 555 isn't inducing a whine or beep in to your circuit, youre golden.

keep the current the circuit uses below 100mA, as the IC needs to source and sing double that.

cheers

[spacecommandant]

Great, thank you. That's good to hear.

[Rob Strand]

I have a circuit which, with 9V, draws around 31mA

I not sure if you are saying the circuit draws 31mA *input current* with no load?   That would be little high for a charge-pump.

And when you put a load on the circuit draws 75mA *input current* with a load?  But what is the output current at the load? More load current will always cause more input current.

(While not the problem here, R35 values of 1k tend to waste current (say 4.5mA) unnecessarily and IIRC can even make the output saturation voltage increase.  The NE555 alone is probably pulling 6 to 7mA.)

[idy]

I not sure if you are saying the circuit draws 31mA *input current* with no load?   That would be little high for a charge-pump.

And when you put a load on the circuit draws 75mA *input current* with a load?  But what is the output current at the load? More load current will always cause more input current.

I think he means he has an effect board that draws 31ma when drawing power from a 9v supply and 75ma when he bumps his supply up to 15.9v with his charge pump.

[Rob Strand]

I think he means he has an effect board that draws 31ma when drawing power from a 9v supply and 75ma when he bumps his supply up to 15.9v with his charge pump.

That makes more sense but it's still not clear if the 75mA is on the 9V or 15.9V rail, and if 9V you would assume it would have to include the current used by the charge pump.

A) Pedal current current increases proportional to supply voltage and we measure pedal current:
    pedal current =  (15.9/9) *31 = 55mA

B) Pedal current stays same at 15.9V and we measure 9V rail current:
    Power out = 15.9V * 31mA = efficiency * Power in
    Power in =   9V*Current on 9V rail
    efficiency 100%: current on 9V rail =  55mA
    efficiency  70%: current on 9V = 78mA

C) Pedal current increases proportional to supply voltage and we measure 9V rail current.
    Power out = 15.9V * 55mA = efficiency * Power in ; current from (A)
    Power in =   9V*Current on 9V rail
    efficiency 100%: current on 9V rail  = 97mA
    efficiency 70%:   current on 9V rail = 140mA

If measuring 9V we expect something between (B) and (C).
Maybe (B) looks OK for an opamp circuit.
Efficiency is a big assumption and affects the estimate.

For non opamp circuits it's possible the (pedal) current is higher than (A) depending on the way the circuit is biased or if there are zeners.

The 9V current estimate needs to be increased by 10mA for the converter power.   It depends if efficiency includes the 10mA or not.

[spacecommandant]

I have a circuit which, with 9V, draws around 31mA

I not sure if you are saying the circuit draws 31mA *input current* with no load?   That would be little high for a charge-pump.

And when you put a load on the circuit draws 75mA *input current* with a load?  But what is the output current at the load? More load current will always cause more input current.

(While not the problem here, R35 values of 1k tend to waste current (say 4.5mA) unnecessarily and IIRC can even make the output saturation voltage increase.  The NE555 alone is probably pulling 6 to 7mA.)

Sorry, I'll try to be more clear. I meant that if I remove the charge pump entirely and just power the audio circuit with a 9V adapter, I measure 31mA between the DC jack and the circuit.
When I attach the charge pump between the 9V adapter and the audio circuit, I measure 75mA between the DC jack and the charge pump.

Do you mean that R35 should be a lower value?

[Rob Strand]

Sorry, I'll try to be more clear. I meant that if I remove the charge pump entirely and just power the audio circuit with a 9V adapter, I measure 31mA between the DC jack and the circuit.
When I attach the charge pump between the 9V adapter and the audio circuit, I measure 75mA between the DC jack and the charge pump.

OK, got it.  It's probably closer to case B.    When you increase the voltage you are supplying more power to the circuit.  Even if the circuit draws the same current at the higher voltage the circuit still needs more power since Power = Voltage * Current and now the voltage is higher.   That power has to come from somewhere, the battery, since the battery voltage is fixed to 9V it's the current that must increase to meet the required input power.

Do you mean that R35 should be a lower value?

You can save a bit of current by *raising* R35.  The 1k is the lowest value you would use and it's wastes power unnecessarily.  It's best to keep this value above 2.2k and if you want to keep power to a minimum you might up it to 4.7k.     There's a lot of NE555 circuits on the web with crazy R35 values, like 100 ohm!

In your case it's not going to make a big difference.   With 1k it's probably burning up 4.5mA, so if you increased the value to 2.2k it might burn-up about 2mA.   The saving of 2.5mA isn't a make or break decision when you are pulling 75mA.

Where you are likely to save power is by using a more modern charge-pump chip which uses internal MOSFET switches instead of diodes.   Low ESR capacitors can also help.    The idea is to increase efficiency.  You might find the input current drops to 60mA or so. If your circuit is working and you aren't too worried about power consumption you have to ask yourself is saving 15mA worth it?   (You might also find the output voltage goes up a bit, as the diode voltages are removed, and then the current goes up a bit from 60mA because the output voltage is higher!! So you end-up saving less than 15mA.)

[spacecommandant]

Sorry, I'll try to be more clear. I meant that if I remove the charge pump entirely and just power the audio circuit with a 9V adapter, I measure 31mA between the DC jack and the circuit.
When I attach the charge pump between the 9V adapter and the audio circuit, I measure 75mA between the DC jack and the charge pump.

OK, got it.  It's probably closer to case B.    When you increase the voltage you are supplying more power to the circuit.  Even if the circuit draws the same current at the higher voltage the circuit still needs more power since Power = Voltage * Current and now the voltage is higher.   That power has to come from somewhere, the battery, since the battery voltage is fixed to 9V it's the current that must increase to meet the required input power.

Do you mean that R35 should be a lower value?

You can save a bit of current by *raising* R35.  The 1k is the lowest value you would use and it's wastes power unnecessarily.  It's best to keep this value above 2.2k and if you want to keep power to a minimum you might up it to 4.7k.     There's a lot of NE555 circuits on the web with crazy R35 values, like 100 ohm!

In your case it's not going to make a big difference.   With 1k it's probably burning up 4.5mA, so if you increased the value to 2.2k it might burn-up about 2mA.   The saving of 2.5mA isn't a make or break decision when you are pulling 75mA.

Where you are likely to save power is by using a more modern charge-pump chip which uses internal MOSFET switches instead of diodes.   Low ESR capacitors can also help.    The idea is to increase efficiency.  You might find the input current drops to 60mA or so. If your circuit is working and you aren't too worried about power consumption you have to ask yourself is saving 15mA worth it?   (You might also find the output voltage goes up a bit, as the diode voltages are removed, and then the current goes up a bit from 60mA because the output voltage is higher!! So you end-up saving less than 15mA.)

Do you think there are any issues at all with using the NE555 with 75mA?
Also, I'm curious if you are able to estimate its oscillating frequency in this circuit.

Thanks for you insights.

[Rob Strand]

Do you think there are any issues at all with using the NE555 with 75mA?

The NE555 should be able to handle it.   You might find the NE555 is a little warm
but not so hot you can't leave your finger on it.

Also, I'm curious if you are able to estimate its oscillating frequency in this circuit.

If you dig up the datasheet and/or applications notes for the NE555 you will find formulas
to calculate the frequency.   It's the astable oscillator configuration.
In your circuit R1 = 1k,  R2 = 33k, C = 1n.
Period T = (R1 + 2*R2)*C * ln(2) = 46.4us
frequency f = 1/T = 21.5kHz
Making R1 small compared to R2 means the output is nearly a square wave.

Increasing the frequency will decrease the ripple but if the capacitors aren't low ESR types you might find the output voltage drops if you push the frequency up too much.

The frequency should be OK for audio.