Using 7805 Efficiently

[hkBattousai]

I need to use 5VDC in my circuit, so I am planning to design a simple voltage regulator using LM7805 like below:

**broken link removed**

I understand that Vi must be about 3V above the desired output voltage level and below the maximum allowed input voltage level, must be chosen between these two voltage values. What I'm confused is what Vi to supply in order to use LM7805 in the most efficient way.

When we apply an input voltage Vi, does the voltage difference Vx = Vi - Vo appear on LM7805 continuously? If so, it will consume Px = Vx . Ix amount of power and it will be wasted as heat. So we should better supply a Vi voltage which is as low as possible (near 8V for my circuit).

Or, does LM7805 monitor the output voltage level, and pump electrical charge to Co capacitor whenever the voltage level at the output drops down to below a certain level? In this case, applying a high input voltage Vi will not be a much problem as was in the previous case.

How does LM7805 work? Can you please explain me the working principle of LM7805, and specify a Vi value which will make LM7805 work in the most efficient way?

 

[audioguru]

The LM7805 is linear so it throws away at least 2.5V (not 3V) x output current in heat. The amount of input voltage higher than the output is thrown away as heat.

It pumps nothing. If the input voltage is less than about 2.5V (at 1A) more than the output then its output voltage drops.

You need a switching regulator to be more efficient.

 

[colin mac]

The capacitors make the output more stable. The smaller Vin is the more efficient the linear regulator would be . The dropout voltage is given in the datasheet. It's 2V so a minimum input of 7V is needed.
You can get low dropout regulators also such as the LM2940 which have a drop out voltage of 0.5V.

 

[hkBattousai]

So, you say that a 12V to 5V converter circuit implemented with LM7805 would have an efficiency approximately (5V)/(12V) = %42, am I right?

I have this circuit in my mind as an alternative:

**broken link removed**

Comparing with LM7805, which one do you think work better? Which one would be more efficient? If this circuit is OK, what improvements should I do on it, should I add a BJT to the gate of MOSFET for a faster response?

 

[MrAl]

Hi there,

They are both linear regulators so they will both consume the same power. You need a switcher to get better efficiency.

 

[hkBattousai]

Hi there,

They are both linear regulators so they will both consume the same power. You need a switcher to get better efficiency.

But, in my circuit above, the MOSFET is working as a switch, there is no voltage drop on it. Therefore the wasted energy is minimal, isnt it?
I don't understand how they would have same efficiencies?

 

[MrAl]

Hi again,

Well, actually it is working as a linear element not a switch, so the efficiencies will be the same, but even if it were acting as a switch it would not improve the efficiency anyway. Switches by themselves can not improve efficiency because they can not store energy. To create a true switching regulator you need a switch AND at least one energy storage element, and of course the topology has to be right.
Most down converters (or "Buck" as they are often called) require a switch plus an inductor plus an output filter cap.

The simplest buck converters i know of are National's "Simple Switcher" line of products. Just a little more complex than your three terminal regulator.

 

[JimB]

But, in my circuit above, the MOSFET is working as a switch,

Why do you think that?
It is not, it is a simple linear regulator.

there is no voltage drop on it.

What?
If there is no volt drop across the MOSFET, the output would be the same as the input, ie 12volts.

Therefore the wasted energy is minimal, isnt it?

No.

I don't understand how they would have same efficiencies?

Because they are the same circuit configuration.

JimB

 

[hkBattousai]

If there is no volt drop across the MOSFET, the output would be the same as the input, ie 12volts.

The voltage on MOSFET is changing between 12V - 5V = 7V (when it is off - no current on it so it consumes no power) and ≈0V (when it is on - no voltage drop on it so in consumes no power). It stays 'on' until output capacitor is re-charged to 5V, then turns off immediately.

This is how I believe this circuit would work. I'm not sure about it, and since the two of you insist that what I'm saying is wrong, probably you are right. But still I don't understand why this circuit is not efficient. I will appreciate if you explain me in detail.

 

[MrAl]

Hi,

Sorry, that's not the way it works. If you read my previous post that explains more.

 

[RichTheDude]

The pass through device (your FET) is operating in the linear region, not as a switch. In terms of efficiency a linear regulator will approximately give you (assuming the regulators current for its own control circuitry is neglible):

Efficiency (%) = 100 * ( power out / power in)
= 100 * ( Vout * I )/ ( Vin * I )
= Vout/Vin * 100

Heat the regulator will produce (W) = (Vin-Vout) * I

For low voltage drops, I will usually use linear for the sake of simplicity (and with low drops linear can be reasonably efficient).

Some good reading:

https://en.wikipedia.org/wiki/Linear_regulator
https://www.electro-tech-online.com/custompdfs/2011/07/f4.pdf

 

[audioguru]

The zener diode holds the input of the circuit at a constant 5.1V so the output is also a constant 5.1V. The amount of wasted heating is 12V - 5.1V= 6.9V times the current.
If you use an oscillator to switch the input to +12V and 0V then the output will average 5V or 6V depending on the duty cycle.