[SOLVED] about the new idea to fulfill the voltage conversion

[Josephchiang]

Hello, everyone

I'm looking for another solution(on chip) which could replace the LDO circuits to supply voltage to digital circuit with simple circuit architecture and very low power consumption. The circuit needs to accomplish the following function:

1. This circuit provides power for digital circuit (digital circuit consumes peak current about 1mA and average current is 1uA). To block the noise which is produced by digital circuit switching ,the circuit need to act as a good isolator.

2. If the circuit is not good enough to block the noise. "Does it is matter if the power is not purely clear (ripple on the LDO output, 0.1~0.3v @ 3V) enough and what happens if it supply the amplifier of crystal oscillator? (32767 KHz)" Does it affect the property of the oscillation circuit ? ex: oscillation start up time or stability

The block diagram is shown like the following figure

 

[BradtheRad]

Do you need to reduce noise on the supply to a sensitive device?

Consider installing a low ohm resistor between it and the noise.

Then add a smoothing capacitor on the isolated side. It will have greater smoothing effect.



If you choose too high a resistance, it will reduce the supply level to the device.

 

[Josephchiang]

Do you need to reduce noise on the supply to a sensitive device?

Consider installing a low ohm resistor between it and the noise.

Then add a smoothing capacitor on the isolated side. It will have greater smoothing effect.



If you choose too high a resistance, it will reduce the supply level to the device.

Thank you for your reply.

But I don't think it is a good idea. The reason is...

First, Power consumption is crossing the resistance. (even though not much)

Second, The 220uF is not available on chip (the major difficulty)

Do you have any idea?



Sorry, English is not my native language. But I will do my best to make it clear .

 

[BradtheRad]

You can use a lesser Farad value of course. Whatever is enough to do the job.

Other ideas:

* a capacitance multiplier

* a shunt regulator (although this may not act quickly enough)

* a separate quiet power supply for the sensitive device.

 

[Josephchiang]

You can use a lesser Farad value of course. Whatever is enough to do the job.

Other ideas:

* a capacitance multiplier

* a shunt regulator (although this may not act quickly enough)

* a separate quiet power supply for the sensitive device.

hello,

I'm glad to see your reply again

I'm confused about your proposed second solution. why is a shunt regulator rather than a series regulator which as shown in the block diagram

Is there any difference between series one and shunt one?

And the other question is "what does happen if it not act quickly enough"

I'm looking for your reply

Best regards,
Joseph

 

[mtwieg]

What's wrong with using an LDO? Any voltage converter is going to require some capacitance for smoothing, and it's hard to get any simpler than a linear regulator.

 

[BradtheRad]

I'm confused about your proposed second solution. why is a shunt regulator rather than a series regulator which as shown in the block diagram

Is there any difference between series one and shunt one?

And the other question is "what does happen if it not act quickly enough"

A shunt regulator acts similarly to a zener diode, but with the addition of a transistor. The zener diode provides bias current to the transistor. The transistor accentuates the zener action.

This is a LDO regulator, as your initial post asks about. It can regulate a supply even when the difference is less than 1 or 2 V. (As you know, series regulator IC's require a couple volts of headroom.)



The series resistor is necessary. A suitable ohm value is between 1 and 10 percent of the load resistance.

Some current is wasted through the transistor.

It may not act quickly enough to filter out all noise and spikes. Therefore a smoothing capacitor might be needed in a strategic location.

 

[Josephchiang]

What's wrong with using an LDO? Any voltage converter is going to require some capacitance for smoothing, and it's hard to get any simpler than a linear regulator.

Thanks, I know that "LDO circuit" is definitely applicable here.

I just confused that have any other good solution to deal with this problem

The reason why LDO is not my first priority is that it's need error amp and bias circuit, hence cause additional power consumption

- - - Updated - - -

A shunt regulator acts similarly to a zener diode, but with the addition of a transistor. The zener diode provides bias current to the transistor. The transistor accentuates the zener action.

This is a LDO regulator, as your initial post asks about. It can regulate a supply even when the difference is less than 1 or 2 V. (As you know, series regulator IC's require a couple volts of headroom.)



The series resistor is necessary. A suitable ohm value is between 1 and 10 percent of the load resistance.

Some current is wasted through the transistor.

It may not act quickly enough to filter out all noise and spikes. Therefore a smoothing capacitor might be needed in a strategic location.

Thanks a lot

It's a useful circuit to deal this the problem

But I think there are something wrong on your statement

"As you know, series regulator IC's require a couple volts of headroom."

Base on my knowledge, the power transistor in the LDO circuit only cause voltage drops of 0.2v is very common rather than a couple
volts

It's that right? I'm not very sure about that

Maybe I just misunderstand your meaning

By the way, the solution with the zener diode may not suitable for fully on chip circuit design because it may not attainable in each

process.

 

[BradtheRad]

But I think there are something wrong on your statement

"As you know, series regulator IC's require a couple volts of headroom."

Base on my knowledge, the power transistor in the LDO circuit only cause voltage drops of 0.2v is very common rather than a couple
volts

It's that right? I'm not very sure about that

I was thinking of common IC's such as 7805, etc.

By the way, the solution with the zener diode may not suitable for fully on chip circuit design because it may not attainable in each

process.

Yes, I suppose it is difficult to manufacture zener values with a sufficiently close tolerance.

I was not sure where you intend to place the LDO regulator, maybe on the pcb where it could be manually trimmed to 3.3 V?

Or in case your plan is to fabricate it on chip, then perhaps you could make 5 diodes in series?

I know little about chip manufacturing.

 

[Josephchiang]

I was thinking of common IC's such as 7805, etc.



Yes, I suppose it is difficult to manufacture zener values with a sufficiently close tolerance.

I was not sure where you intend to place the LDO regulator, maybe on the pcb where it could be manually trimmed to 3.3 V?

Or in case your plan is to fabricate it on chip, then perhaps you could make 5 diodes in series?

I know little about chip manufacturing.

You are a good adviser. Thanks you.