Let's made a PC-base metal detector with usb interface !!!

[Aziz]

Forgot the upper limit input voltage behaviour:

[Aziz]

F$!%$!§$& The 78xx/79xx are driving me crazy .

Hi,

I want to have high input voltage possibility at any cost (up to 28V)! But getting closer to the absolute maximum ratings (input voltage), the regulators won't work well. So I can not go with the input voltage up with the 79xx regulator. The DC-DC charge pump can deliver max. -8V (-7V regular). The 79xx has -35V maximum input voltage range. There is also some span necessary to get this regulator work well (lets take max. 27V limit according to recommended operating conditions mentioned in the data sheet).

The max. input voltage would be:
Vinmax = 27V + (-8V) = 19V
That is definitely too low.

Ok, there are other options:
1. Using the LM337 regulator (up to -40V)
2. Discrete solution (up to -40V)

I have choosen the discrete solution. It makes marvelous precise, low power consumption, low noise high-side regulator possible. Have a look into the floating gate voltage (below). I probably can go with the input voltage even higher than 28V.

Aziz

[Aziz]

Hi again,

damn, the 78xx/79xx withstand up to 35V but have a 2V voltage drop. The low-drop types will operate up to 30V or even less. So I have to choose the right part carefully.

I will introduce a 8V (or 8.5V) system voltage and this two times. One for low power part (helper power supply for regulators) and one for stepped down regulation (main power supply). The system voltage will produce all necessary voltage levels (inverters, up-converters, mosfet drivers, etc. ) . With low-drop out regulators, 8V can be regulated with minimum 9V input voltage. I have choosen 8V and not 5 or 6V due to high gate drive voltage and the flexibility of using standard 78L05/79L05 regulators.

So only a span of 8V - 5V = 3V will be heated up.

A comparison:
Input voltage: 15V
Drain current: lets take 200mA @ 5V (electronics part only, coil supply not included)
Buck converter efficiency: 85%

If we use linear regulator, the voltage difference of 15V-5V = 10V will be heated up (10V*0.2A = 2W loss)

With the 8V system voltage:
Loss: (8V-5V)*0.2A = 0.6W
Taking efficiency of 85% into account (buck conversion): 0.6W*1.15 = 0.7 W loss

Now make the calculation, if we have more current drain (400 mA) and 25V input voltage!
Linear regulator loss: 20V*0.4A = 8W
Buck conversion loss: 3V*0.4A*1.15 = 1.4W

That's the difference with high input voltages.
Aziz

[Aziz]

High-side nch-mosfet or pch-mosfet driver?

Hi friends,

regarding the coil power controller:

I have reduced the switch-off time down to 170ns (with p-ch mosfet Si4401DY having 37 nC gate charge). Following dumb calculation: 80ns comparator + 20 ns flip-flop + 70 ns mosfet-driver and mosfet switch-off. It was a discrete high-side p-ch mosfet driver with fast switch-off optimisation (3A switch-off peak current!).

Generally, the high-side n-mos drivers operated in bootstrapping mode won't cut off the switch current instantly and the coil power supply is capacitively coupled to the driver. I have tested the LTC4440 in the circuit simulation and it hasn't the benefit compared to the discrete p-mos solution.

So a high-side p-mos driver is most convenient for this application. Nevertheless, low-side drivers can still be used in the high-side configuration. Indeed, I have several MCP1407 (6A mosfet driver), which I would like to use it. The logic input signal must be level shifted up to the high side and a high-side gate power supply is necessary, if I want to go down to 9V input voltage. Due to the lower input voltage of 9V, the mosfet gate voltage is limitted to 12V or 15V (not anymore 18V). A charge pump inverter will support the switch-on of the p-ch mosfet with negative voltage converter.

The high-side power supply for the mosfet gate driver is critical and must be stable enough (have to float with the input voltage). I will use either the 79L12 (-12V) or a discrete -15V solution or even a low-drop negative regulator. This will ensure same gate voltage drive even the battery voltage diminishes or changes over time.

Input voltage: min. 9V, negative charge pump could deliver -7V: Total span: 9+7=16V. This should be high enough for the 79L12.

Using low input voltage and high gate drive voltage makes an own power supply for the mosfet driver necessary. Fortunatelly, these circuits need not much power and we can use linear regulators. We also need it anyway to regulate the main power supply. Most power draining parts will be efficient down converted.

So I have to change the power supply section completely to be efficient enough and make a wide input voltage range possible (9V .. 28V) . It will very likely the most complex part of the detector.

Aziz

[Aziz]

Voltages and currents (see below):

[Aziz]

Simple Buck Converter Principle

Hi all,

below is the discussed very simple buck converter principle. It has up to 92% peak power efficiency. The mosfet driver and the switching logic (flip-flop, comparator, etc. ) is not shown here. The coil is a typical transformer wound on a ring core ferrite.

I probably will use the LM319 dual voltage comparator as it is fast enough (80 ns), cheap and easy to get. The most delay comes from the mosfet driver and mosfet itself. A switch-off response of 250 ns and less should be possible. Switch-on characteristics is not critical and can even be much higher (1-2µs).

I have not finished the full schematics yet and I am considering to use the buck converter also for the power supply section to maximize the power efficiency further. I hope to get a wide input voltage range (9 .. 28V).

Aziz

[Aziz]

PS: I can get now a power supply rejection ratio of 68 dB at least (< 1mV ripple).

[Aziz]

Hi all,

still working on the coil power regulator. The switching speed is critical and I am trying different configurations to get out better performance. An ultra-fast voltage comparator like the LT1016 (10 ns) would be surely an overkill. The LM311 should do the job as well (200 ns). The mosfet driver has also switch-off delay and fall-times. Finally the mosfet switching times will add to the system delay. The n-mos fets should switch-off faster due to lower gate capacitance.

I did a discrete high side p-mosfet driver, which works well at the expense of more parts. Now I am thinking of changing it to a n-mosfet high side type. There are lots of high side mosfet drivers available. Interestingly not much for p-mosfets.

High side n-mosfet driver (spice simulation results using LTC4440):
The switch-off current is unfortunately not abrupt due to boost capacitor (bootstrapping configuration) of the mosfet driver.
The p-mosfet version shows slightly better results but the higher switching resistance will reduce the overall efficiency.

I probably will try the high side n-mosfet version. Now looking for an available part (LTC4440 is not available to me).

Aziz

[Aziz]

Can you show a schematic of the interesting idea?

-SB

It's too early to show something. It is not finished yet. But the principle is really very easy to understand and to implement.

The capacitors get charged faster now as the magnetic field energy discharge path is decoupled from the regulated coil voltage.


Aziz

[simonbaker]

Hi all,

there is sometime a much simpler solution of a big problem. One sometime don't see the obvious solution. Shame on me.

I have now a very fine solution of the regulation, which will meet hopefully the desired accuracy. Transformer principles will solve my big problem. During charging of the capacitor bank, the transformer stores the magnetic energy. After reaching the desired voltage, the regulator will switch off charging. The stored magnetic energy in the transformer will be discharged on a secondary winding back to the source (back to bypass cap). As the primary and secondary windings are not galvanic coupled, the right polarity can be set to feed the energy back. A snubber circuit is necessary, to damp the minimum energy, which can not be transferred through the diode (due to diodes forward voltage). The secondary windings could be higher than the primary windings to improve the efficiency of the back transforming process.

It now get's very simple.

Aziz

Can you show a schematic of the interesting idea?

-SB

[Aziz]

Hi all,

there is sometime a much simpler solution of a big problem. One sometime don't see the obvious solution. Shame on me.

I have now a very fine solution of the regulation, which will meet hopefully the desired accuracy. Transformer principles will solve my big problem. During charging of the capacitor bank, the transformer stores the magnetic energy. After reaching the desired voltage, the regulator will switch off charging. The stored magnetic energy in the transformer will be discharged on a secondary winding back to the source (back to bypass cap). As the primary and secondary windings are not galvanic coupled, the right polarity can be set to feed the energy back. A snubber circuit is necessary, to damp the minimum energy, which can not be transferred through the diode (due to diodes forward voltage). The secondary windings could be higher than the primary windings to improve the efficiency of the back transforming process.

It now get's very simple.

Aziz

[simonbaker]

Hi simonbaker,



I am already using a synchronized one-cycle buck conversion with faster discharge timing and enhanced efficiency (compared to usual buck converter with diode but without the synchronous rectification). It is of course operated in discontinous mode. No PWM generation and regulator necessary for this. Just a flip-flop, comparator, mosfet-driver, reference voltage, mosfet and a coil. It is already the simplest and cheapest implementation.
Interestingly, the regulation is accurate enough (I got a regulation of <5mV ripple at 10V source ripple). The source voltage would not ripple at such voltage levels of course. The post regulation is probably an overkill but could make the regulator superb (µV regulation).

The chip manufacturer of SMPS regulators don't see the importance of having a synch input pin. The existing regulators with synch input pin and desired specification are unfortunately not available to me. So a discrete solution makes the circuit complexity high.

I need to wire up the solution to see, if it meets my desired specification limits.

Aziz

It's going to be a heck of a platform when you're done Aziz. Great stuff...

-SB

[Aziz]

Hi simonbaker,

What I'm thinking is that for charging the capacitor, just put an inductor in series with the battery and close the switch. Then you have a one-cycle buck/boost sort of: the inductor will store the energy while the current flows into the discharged capacitor, so you don't lose the energy due to I * dV in the battery internal resistance or wire resistance. When you turn off the charging switch, the remaining energy in the inductor goes into the capacitor (need a shunt diode as usual), finishing the charge. So it is just more efficient than charging the cap directly from the battery. A buck-boost regulator does the same thing really, but uses many cycles to do it.

So I'm wondering if a simple series inductor makes a simple solution instead of a whole SMPS.

See what I mean?

Regards,

-SB

I am already using a synchronized one-cycle buck conversion with faster discharge timing and enhanced efficiency (compared to usual buck converter with diode but without the synchronous rectification). It is of course operated in discontinous mode. No PWM generation and regulator necessary for this. Just a flip-flop, comparator, mosfet-driver, reference voltage, mosfet and a coil. It is already the simplest and cheapest implementation.
Interestingly, the regulation is accurate enough (I got a regulation of <5mV ripple at 10V source ripple). The source voltage would not ripple at such voltage levels of course. The post regulation is probably an overkill but could make the regulator superb (µV regulation).

The chip manufacturer of SMPS regulators don't see the importance of having a synch input pin. The existing regulators with synch input pin and desired specification are unfortunately not available to me. So a discrete solution makes the circuit complexity high.

I need to wire up the solution to see, if it meets my desired specification limits.

Aziz

[simonbaker]

If current flows through the inductor, it stores energy. If you don't dissipate or recover the magnetic energy, what would happen at switch-off? If you burn the energy, the efficiency decreases.
Aziz

What I'm thinking is that for charging the capacitor, just put an inductor in series with the battery and close the switch. Then you have a one-cycle buck/boost sort of: the inductor will store the energy while the current flows into the discharged capacitor, so you don't lose the energy due to I * dV in the battery internal resistance or wire resistance. When you turn off the charging switch, the remaining energy in the inductor goes into the capacitor (need a shunt diode as usual), finishing the charge. So it is just more efficient than charging the cap directly from the battery. A buck-boost regulator does the same thing really, but uses many cycles to do it.

So I'm wondering if a simple series inductor makes a simple solution instead of a whole SMPS.

See what I mean?

Regards,

-SB

[Aziz]

..
Do you really need a whole SMPS, or would just a coil between the battery and the capacitor, chosen correctly, do the job pretty well with a single pulse?

Now if you could only recover the energy from the TX pulse...

Regards,

-SB

If current flows through the inductor, it stores energy. If you don't dissipate or recover the magnetic energy, what would happen at switch-off? If you burn the energy, the efficiency decreases.

Recovering energy back from the TX pulse in a metal detector is already patented. Hey, MD is not a SMPS! The energy balance of the transmitted pulse is bad. The efficiency is not high enough as the coil current often saturates (DC losses). The additional circuit adds parasitics to the sensitive receive front-end. It isn't worth to do it. I need to make accurate energy balance calculations to check this again.

There is enough time to charge the capacitor bank between the TX pulses. The regulator must charge the lost energy only - not the whole coil voltage.

A 4700µF capacitor, which is charged up to 12V has the following energy content:
E = 0.5*C*U² = 0.5*4700µF*(12V)² = 0.3384 Joule

Let's assume, we have 1V voltage drop after the TX pulse:
E left = 0.5*4700µF*(11V)² = 0.28435 Joule

So we have lost only 0.3384 - 0.28435 = 0.05405 Joule (54 mJ). 84% of the energy is still in the capacitor.

Aziz