Electrical – Voltage regulator minimum input voltage

voltage-regulator

I have a L78S05V voltage regulator (5V output). The datasheet's maximum ratings suggest the maximum input voltage (35V). I would like to know, is there a minimum input voltage?

I have a circuit that is powered from a 5V power supply. For "forward compatibility" (i.e. if my custom power supply broke) I want to add 5V voltage regulators on the input stage, so I can change the power supply with a 12V one and still have everything powered at 5V.
I can add some jumpers to route the 5V input or 12V input (via voltage regulators), but if the 5V voltage regulator works well with a 5V input I can get rid of the jumpers and work with any input voltage between 5V ad 35V without worrying.

Thanks!

Best Answer

On a linear regulator you can not work with the same voltage at input and output. Basically inside the component there are transitors that are not perfect transistors, because they have a resistance when the current flow through them, and this resistance leads to a loss of voltage in the component. It is called the drop out voltage.

The maximum output voltage you can have with a given component and a given input voltage is : Vout = Vin - Vdropout.

In your datasheet it is wierdly written since Vin is refered at the drop out voltage. The datasheet say the min Vi is 8V, meaning the drop out voltage is actually 3V. Meaning if you put 5V at the input you won't have more than 2V at the output. Then you have to use your jumpers.

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Electronic – Drive voltage regulator Vout when Vin is floating

It is true that the datasheet of the L78S05 does not really give explicit information on this usage. For some other devices, there is a clear limit given in the "Absolute Maximum ratings" table, that indicates the maximum difference between input and output voltage: e.g. for LM317, it says "-0.3V max". When you see this in the datasheet, it clearly means you need a protection diode from the output to the input, so that the input never goes further below the output. But it is not the case here.

But there is a clue given in the datasheet:

The protection diode I was talking about is suggested here "when input is short-circuited, and if there is high capacitance load, and for output voltage grater than 6 volts". The datasheet explains this is because of the base-emitter junction rating.

This actually means there is no real risk of destroying the regulator in your case. The base-emitter junction can tolerate 6V, and in your case there won't be more than 5V. Moreover, the risk only exists when input is short-circuited (because in this case there is a large current flowing back from output to input). In your case, input is just floating, so the current will be null (or maybe just some insignificant leakage).

So you could leave it like that. Now, if you really want belt and suspenders and take care of the case where input is short-circuited instead of just floating, you could add the above protection diode, then you'll be more than safe.

Electronic – Voltage regulator waste too much energy

You are correct in that the linear regulator appears to be using 17mA at 37V so that is a power of:

Vin * Iregulator
37V * 0.017A = 0.629W

But you need to also consider that the linear regulator that you are using is also dissipating power of:

(Vin - Vout) * Iload
(37V - 5V) * 0.031A = 0.992W

Thus you have a total power loss in the linear regulator of:

0.629W + 0.992W = 1.621W !!!

You really need to be using a switching buck regulator for this job. If you do not want to build such a thing you can find a plethora of ready made modules on web sites the like of Amazon or eBay. They do not cost a lot either.

Best Answer

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Electronic – Drive voltage regulator Vout when Vin is floating

Electronic – Voltage regulator waste too much energy

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