[sergey23 0]

Hello , I am confused with the components, there is a regulators and there is DC-Dc both can create voltage for my PMA5-83-2W+ sequencer
I can do this l3505 DC-DC and i can do it with regulator like LM7805 or LT1761.

What is the key difference between DC-DC and a regulator with performance?
https://www.sparkfun.com/datasheets/Components/LM7805.pdf
https://www.minicircuits.com/pdfs/PMA5-83-2W+.pdf
https://www.analog.com/media/en/technical-documentation/data-sheets/1761sff.pdf
https://www.qorvo.com/products/d/da007774
https://www.analog.com/en/products/lt3505.html

[Yuri7751 25]

1 hour ago, sergey23 said:

What is the key difference between DC-DC and a regulator with performance?

The key difference is the way they control the amount of energy delivered to the load. A linear regulator (like 7805 or LT1761) has a transistor inbuilt which  operates in a linear mode dissipating the excess of energy.

If the input voltage is say 12V, output is 5V and current 0.1A the dissipated power is (12-5)*0.1= 0.7W. And all that power will heat the IC. The efficiency (Pout/Pin) in the case is about 41% only.

In case of what you call DC-DC (step-down converter or switching regulator) the transistor inside the IC operates in the switching mode (on/off). The energy is proportional to the duty cycle. To get the DC at the output a LC filter is essential. The efficiency may exceed 90% even at high load currents. But at the cost of higher noise and more complicated design (calculations, part selection, PCB artwork etc.).

In your case taking into account very low current (4.5mA) needed and your obvious lack of experience (in switching regulators design at least) I'd recommend using a linear regulator.

Maybe 7805 is not the best choice because you should note the following:

To provide such a sequence you may need a regulator with enable/shut-down pin such as LT1761 for example.

[sergey23 0]

Hello, Suppose I have a general bias of +15V gnd -15V.
As you reccomended for  5V  Vg2 i will use LT1761.
What device you reccomend to use for -2V ?

What device you recommend for 12V step?

Thanks.

Изменено 18 июня пользователем sergey23

[Yuri7751 25]

6 hours ago, sergey23 said:

What device you reccomend to use for -2V ?

Well you have provided qorvo example of such a circuit yourself. LTC1261 can do the job. You should carefully study the datasheet to understand how to adjust the negative bias to -2V. See Fig.6.

Maybe in will be necessary to adjust the voltage in two steps :  -2V at first and some higher voltage (adjustable?) after the 12V is on.

Btw it is the third variant of step-down regulators - switched capacitor inverter. 

LTC1261-Switched Capacitor Regulated Voltage Inverter (analog.com)

As to the 12V you can use LT3505 as in qorvo example or you can go to Power Designer (ti.com) enter your data (15V input, 12V 0.4A output) and it will propose you more than 300 variants calculated and simulated in Webench 🙂 

Any variant requires your own efforts - you should carefully study datasheets and application examples, PCB artwork requirements, etc. And any basic knowledge books/papers on switching regulators too. Remember that this power amplifier can produce smoke very easily in case you make a mistake in powering it 😉 Good luck.

 

[sergey23 0]

Hello, I have implemented the basic functionality of LT1751-5 datshet shown below. I get perfect 5V when SHDN=Vin pure 15V DC.
.
If i have to use SHDN pin then i need to have Vshdn=Vin as shown in the photo below.
I have tried to put 15V pulse so there will be a time period where Vhsdn=Vin and still i get Vout almost zero.

Where did i go wrong in switching on  the regulator with SHDN and get it 5V?

Thanks.

https://eu.mouser.com/datasheet/2/609/1761sff-3124194.pdf
 

Изменено 19 июня пользователем sergey23

[Yuri7751 25]

6 hours ago, sergey23 said:

Where did i go wrong in switching on  the regulator with SHDN and get it 5V?

Nothing was wrong except timing 🙂 It is not that fast:

 

[sergey23 0]

Hello Yuri ,I have tried to upgrade the circuit so i could connect INPUT to SHDN when pulse is 3.3V  and to discharge the output when pulse is 0V.
I see the error in the circuit below where whr mosfet discharges when pulse is 3.3V so i tried to reverse the opamp inputs to make the mosfet be closed .

But the bigger problem is that SHDN not equal Vin when i open the M2.there is 2V gap between them.
How do you reccomend to solve this issue?

Thanks.
 

 

opamp_mosfet_switching.asc

[Yuri7751 25]

 

1. If you look at the page 11 charts you will notice that the threshold voltage of SHDN is about 0.7V. So you don't need to apply 15V there to turn it on.

2. I don't understand why you need to discharge anything. After you shut down RF and  Vdd (12V) nothing bad can happen to your amplifier no matter how fast Vg1 and Vg2 decrease naturally. 

Try to make it simple. Avoid unnecessary parts.

There is a mistake in your circuit - you turn the IC on by high S level and M2 and at the same time you short the output by M1.U2 or M1 are heating and smoking 🙂

[sergey23 0]

Hello Yuri , for the -2V i used LTC1261L ,I dont see that an LDO the less noisy version in the datasheet below.
somehow the 5V regulator managed to charge and discharge the output , but in this case i did the schemtics as shown below.
For G1 I need -2V and 4mA according to the datasheet shown below.
But as you can see in the plot i get 400mA peaks of current.

How can i improve this -2V circuit so i will meet the required demands on the datasheet for Vg1?
LTSPICE and relay simbols of the relay compnent are attached.
DC_Dc_relay_m2p4.asc

https://eu.mouser.com/datasheet/2/609/1261lfa-3124642.pdf

Well you have provided qorvo example of such a circuit yourself. LTC1261 can do the job. You should carefully study the datasheet to understand how to adjust the negative bias to -2V. See Fig.6.

Maybe in will be necessary to adjust the voltage in two steps :  -2V at first and some higher voltage (adjustable?) after the 12V is on.

VO14642_CT.asy VO14642_CT_V2.txt

[sergey23 0]

UPDATE :

Hello , I put 0.5UF as a load with no thought at all.
4.5mA is the maximal rating so i prever 2mA so it wont burn 
So given -2V and 2mA then load resistance is 2/2mA =1Kohm
suppose my gate capacitance is 100nf(more logical then 0.5u) and i put series resitance as 1Kohm as shown in the photo below.

I get a very bad Vout signal (its jumping a lot)
but the good news that the current on the load capacitor is around 1-2mA.
How can i fix the voltage signal?
https://eu.mouser.com/datasheet/2/609/1261lfa-3124642.pdf

Изменено Суббота в 09:40 пользователем sergey23

[Yuri7751 25]

On 6/21/2024 at 4:47 PM, sergey23 said:

I dont see that an LDO the less noisy version in the datasheet below

1. You don't have any LDO in your circuit. As I (and the datasheet) said LT1261 is a switching capacitor regulator. It has inherent ripples and to get a 'good', clean DC you need to filter the output (see pp12,13 of the datasheet). And you should worry about the output voltage not the C1 current.

2. It is a DC-DC inverter. When I say DC I do mean it. It is a DC not a pulse generator. You can't expect that fast operation from it. If you really need that short pulses like in your model (200uS) the approach should be different - you first generate DC and don't touch it any more 🙂 Don't turn it on and off that fast. Instead put a pulse forming circuit (switches) between the DC-DC and the load.

For example:

[sergey23 0]

Hello Yuri, I dont care at all about PULSE rise time ,it could be even 0.1sec.
I am worried about burning Vg1 connection.


From the datasheet we have two hints shown in the photos below.
The basic caracteristic says its -0.8V with 15 micro amp which meeans that the resistance at Vg1 is almost highZ.

So we start with -2V and increase till we reach  15uA.
What load capacitor you reccomend to use? 

my simluation below says 250mA where as the Vg1 can handle only 4 as i shown in previos post.

I was told "This current will be ONLY in circle comprised output capacitor and discharge SSR"  this current will not reach G1, I cant see how its simple I=CdV/Dt
if you could clarify regarding :

1.what load to use to represent the gate G1.

2.gven the simulation below how exactly these 250mA will not touch G1?

Does it make any sense to you?

https://www.minicircuits.com/pdfs/PMA5-83-2W+.pdf

Изменено Понедельник в 16:09 пользователем sergey23

[Yuri7751 25]

9 hours ago, sergey23 said:

I dont care at all about PULSE rise time ,it could be even 0.1sec.

Did I say anything about rise time? You showed the waveform with 200uS pulse and complained about the "very bad Vout signal". I showed you how to fix it. That's all. if you don't need it just forget.

9 hours ago, sergey23 said:

I was told "This current will be ONLY in circle comprised output capacitor and discharge SSR"  this current will not reach G1

That's right.

9 hours ago, sergey23 said:

What load capacitor you reccomend to use?

Any. This capacitor doesn't effect the current flowing into the G1 pin. This capacitor is a filter capacitor - it reduces ripples. Your simulation of C1 current is senseless. Forget.

It happened that when people wish to limit a current they put a resistor in series with the protected circuit, not a capacitor in parallel. I don't know why. Maybe because of the Ohm's law 😃

Your target is to keep the G1 voltage in the range of -0.2V - -2V. Actually 4mA is just a reference value. Ok if you still worry that much about this current put a resistor in series (between the capacitor and G1 input). 1.3k guarantees the current will never ever exceed 4mA even at -5V at the LT1261 output. I don't know how it  affect the amplifier operation (as no internal structure shown) but you can sleep well 🙂 Later you can safely remove the resistor.