Does a switching regulator effect the output of a buck converter with a fixed output voltage of +5 V like the L7805?

[FKamm.2]

I'm trying to design an voltage regulator on a motor control board. I want an output voltage of 5 VDC and the input voltage is 24 VDC. Because of heat production i was thinking about using a switching regulator like the L7983 followed by an L7805. But I'm unsure if the frequency of the switching regulator will effect the output signal of the L7805. I need the 5V for the microcontroller and the current sense amplifier.

Or is there an easier way?

Thank you very much

[Peter BENSCH]

Welcome, @FKamm.2, to the community!

First of all - you have already combined very well if you want to reduce the power dissipation of the linear regulator 7805. However, a microcontroller does not need a linear regulated power supply, the L7983 can supply it directly with the 5V, it normally does not mind if there are a few spikes on it.

But of course it makes sense to filter the voltage supply of the current measurement amplifier, because the mentioned spikes can lead to more or less large distortions of the measurement signal. However, a simple 7805 is not capable of doing this - when it was developed, one could not even imagine that switching regulators could work with the frequencies used today: the L7983 can work with switching frequencies between 200kHz and 2MHz, whereby this frequency then appears on the output voltage as a ripple.

Just look at the Supply Voltage Rejection (SVR) in the data sheet of the L7805: of the 68dB attenuation at 120Hz (table 3), only about 56dB remain at 100kHz according to fig. 31, which can be extrapolated to about 50dB at 200kHz and perhaps 24dB at 2MHz. This shows that such simple linear regulators are rarely suitable for such filtering.

Instead, it is better to route the supply voltage through a good Pi filter (C-L-C; with C~0.1-1µF, L~1...10mH) to supply the analogue measuring circuit.

Does it answer your question

Regards

/Peter

[FKamm.2]

Hello Peter,

Thank you for your answer.

So you mean to take just the L7893 to get the output voltage off 5V and filtering this signal with der Pi-Filter?

My idea was to take a higher Input voltage for the L7805. (Maybe 24 V L7893 10V L7805 5V)

But if i understand your explanation right, it doesnt matter which output voltage i want to generate.

The attenuation will always be the same, because of the depending on the switching frequency.

Another idea is to take two LM317 with a voltage stage between the two regulators of maybe 12V, but the heat production still remains.

obviously i'm pretty unconfident to find the right way between heat reduction and having less noise which will influence the current measurement amplifier.

Regards

Fabian

[Peter BENSCH]

Hi Fabian,

Linear regulators are usually used when no ripple voltage is to occur, the regulation must be realised as inexpensively as possible and the heat loss does not play any role.

As I said, it is not necessary to lower the voltage in two steps, you can regulate directly from 24V to 5V with the L7983. With the Pi filter and the right layout (separate GND from the analogue and digital sections and only connect them at the voltage regulator) you minimise the influence of the ripple on the analogue measuring circuit. Only with the highest demands on precision and very high ADC resolution (>12bit) it makes sense to control the analogue VDD with a special ultra low-noise LDO, which should also have a PSRR (Power Supply Rejection Ration) > 60dB at the appropriate switching frequency - but you will very rarely find this, a Pi filter is an easier and hopefully promising approach.

You can find further useful recommendations for the layout in the data sheet of the L7983 under point 5.7.4, for similar components as in AN1723 in section 6 or in the data sheet of the ST1S10.

Good luck!

If the problem is resolved, please mark this topic as answered by selecting Select as best. This will help other users find that answer faster.

/Peter

[FKamm.2]

HI Peter,

Last question:

Can you help me to understand why one of these two methods is better?

The first would be to take the L7983 and a pi filter. The second would be as the datasheet of the L7805 shows in figure 20 and 21 a

configuration with a resistor and a transistor to reduce the input voltage.

I think the most differences are the heat production and the effeciency. right? And because of the 10-Bit ADC resolution

the solution with the L7983 would be the easier and better one? Or is something wrong with my conclusion?

Sorry by the way. it is my first pcb.

Thank you!

Fabian

[Peter BENSCH]

Every linear controller burns up the differential voltage, the ones in Fig. 20 and 21 only distribute this power additionally to the external transistor. Thus, it is easier to realise the cooling, the efficiency remains the same in total. In this respect, a switching regulator like the L7983 is the most efficient method in terms of energy.

10bit ADC resolution is relatively little, so a Pi filter is easily sufficient. But remember - the layout is extremely important!

Good luck!

/Peter

[FKamm.2]

Hi Peter,

In the datasheet of the L7983 the table 8 is given. We were talking about an PI-Filter (CLC). Table 8 is about some reference applications.

So i have to replace the given LC-filter, an calculate a new CLC-Filter, right?

Thank you for your help,

Fabian

[Peter BENSCH]

No, the table lists the switching frequencies and the necessary component values of the passive components of a normal buck regulator that is to process 24V.

To realise the Pi filter function, an additional LC filter must be connected between the output Vout and the input voltage of the sensitive circuit block. In this case, the output capacitor of the switching regulator forms the input capacitor of the resulting Pi filter.

Regards

/Peter

[FKamm.2]

Hello Peter,

so the result is the circuit diagramm in the picture. Could you please have a look at it?

I want to use the 1000 kHz, therefore i connectet FSW to VCC. And because i want to use the LNM i connected it also to VCC.

Do i have to calculate the whole PI-filter, or just the L4 because C4 = 2.2µF (from the datasheet) and C16 has to be equal?

VOUT will be +5 V and will be connected to the other parts like the INA and the microcontroller.

Do i have to add another capacitor to filter the input voltage?

Thank you very much.

best wishes,

Fabian

[Peter BENSCH]

• You have drawn L3 with 1µF at the input by mistake - this should be a capacitor, otherwise you would have a short circuit.
• VBIAS should not be connected to VCC (3.3V), but to VOUT (which is also mentioned in table 8), because the voltage difference to SWO Threshold (3.2V) is somewhat small.

The output voltage at pin 1 (VOUT) can supply the microcontroller, while the +5V at C16 is fed to your analog measurement circuit.

C4/L4/C16 form the Pi filter. The values of L4 and C16 result from the desired attenuation: for a cut-off frequency of e.g. 100Hz you will need about 100mH and 22µF.

Please choose the saturation current of the inductors L4 and L2 larger than their peak current, because otherwise the inductance collapses and the function of regulator and the filter is no longer guaranteed. C16 should either be a ceramic MLCC only, or a parallel connection of e.g. 22µF tantalum and 0.47..10µF MLCC.

Regards

/Peter