[SOLVED] A theoretical regulator question.

Hi everyone,

I have been making a device which requires 5.4V supply. There were such batteries available 20-30 years in our country, they were extremely rare though. There is an addition to the generic scheme, which makes it possible to supply the circuit by 9V. This is a pocket size device so it must be as compact as possible. I've got a case already and battery PP3 doesn't fit into it. I've got a dilemma:
1) Supply the circuit by 2 sequentially connected 2032 battery and forget about any regulators (6V total);
2) Make a regulator and supply its IN by 6V;

The first case solution might kill the entire device (0.6V difference).
The second one probably would not be sufficient to get the regulator properly working (especially when 2032's would low to 2,5V each).
Of course, there is third solution -- to get another case. And here's a problem too: it would be extremely difficult to get a worth one in the place I'm living.

If need I can upload the schemes.

Thank you.

What about putting a diode in series with the load that will drop about 0.6V

MikehZA said:

What about putting a diode in series with the load that will drop about 0.6V

Click to expand...

What kinda diode does it? Is 1n4148 suitable?

Thanks.

The batteries you propose using would suggest that the current flow will be very low so I would think the IN4148 would be fine. Try it with a dummy load to determine the actual voltage drop.
Regards

A diode isn't a regulator. If your circuit is critical and needs 5.4V then battery voltage variations with temperature and battery discharge state combined with diode voltage drop changes with load and temperature make a diode a very bad idea.

Why is your circuit critical that it needs 5.4V? How much current does it need? A single cell with a boost regulator may be one option.

Keith

Grishin states that the battery voltage can drop to 5V so I asume it is not that critical

I will repeat what I have said before on these forums. A regulator maintains a stable output voltage with changes in load, temperature and input voltage. A diode does none of those. If you need a regulator, use a regulator not a diode.

Keith

keith1200rs said:

A diode isn't a regulator. If your circuit is critical and needs 5.4V then battery voltage variations with temperature and battery discharge state combined with diode voltage drop changes with load and temperature make a diode a very bad idea.

Why is your circuit critical that it needs 5.4V? How much current does it need? A single cell with a boost regulator may be one option.

Keith

Click to expand...

A circuit is a general one but that 0.6V would be about 11% higher than its documented voltage. I suppose a new 2032 is a bit higher 3.0V (3.1-3.2 probably) so while the batteries are fresh ones, difference would be even more (15-20%). The circuit is similar to one is in attachment . It needs no more than 2mA.
Note, there is a step-up transformer (T1). The low covers are 2 and 3 winds and high cover is 420 winds, so the coefficient is fairly high. I'm not sure that additional voltage would be OKAY for the circuit.

Grishin states that the battery voltage can drop to 5V so I assume it is not that critical.

Click to expand...

Well, the point was 5V would be okay for the device but it probably would not if I supply a regulator with 5V (which works properly with 9V and gives 5.4V then). It won't get enough voltage to work properly and I don't know how it'd behave.

Thank you.

I agree, that the unregulated DC/DC converter can be a problem, particularly because the sensitivity of the GM tube depends on the HV level. The logic ICs are standard CMOS, I guess, and can manage an even higher supply level.

A LDO regulator, e.g. LP2985-5.0 or -5.3 should do the job. It has only a low mV dropout voltage and a moderate current consumption of 100 - 200 uA at low output current levels. There may be CMOS based LDO with even lower current demand.