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role of optocoupler for biasing drain of amplifiers
Posted by
john23
on 17 Apr, 2024 16:42
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Hello,In the diagram sown below I have a MOSFET called NDT3055L, the mosfet sends the current into the Qorvo TGA2590-CP drain.
When we test the circuit the source of the mosfet is basically floating although when I will connect the amplifier to source then the QORVO amplifier will introduce a load to the source of the mosfet.
Somehow the optocoupler helps with the transition of the state when we don't have QORVO amplifier load on the source of the MOSFET.
What is the role of the optocoupler in this situation?
I'll be glad to have some analog design logic behind putting this component in this position.
Thanks.
https://www.ti.com/lit/ds/symlink/ucc5304.pdf?ts=1713292114392TGA2590-CP - Qorvo
www.qorvo.com www.qorvo.com
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#1 Reply
Posted by
xvr
on 17 Apr, 2024 17:43
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There is no optocoupler in this schema. UCC5304 is an isolated driver, but not optocoupler.
As for using it here at all - it used to decouple ground of schema from Source of MOSFET. MOSFET controlled by voltage, applied between Source and Gate. Original signal to control MOSFET based to ground of all schema, but MOSFET Source has floating potential relatively ground, so control signal can not be directly passed to MOSFET Gate.
UCC5304 rebased this control signal to MOSFET Source - it used as ground potential on output (isolated) side of UCC5304.
BTW - there is no supply attached to isolated side of UCC5304 in schema. I think something was forgotten
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#2 Reply
Posted by
john23
on 18 Apr, 2024 12:37
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Hello XVR, an updated version is attached of the schematics.
Why do we connect floating ground to a place where we would get the amplifier connected.
The TGA 2590 amplifier will be the load the mosfet source(when we connect them together)
What is the logic of connecting floating ground there?
Thanks.
"it used to decouple ground of schema from Source of MOSFET"
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#3 Reply
Posted by
xvr
on 18 Apr, 2024 12:54
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> an updated version is attached of the schematics.
Floating power still unconnected. You should connect right Vdd of UCxxx to +24V net.
> What is the logic of connecting floating ground there?
Output of UCCxxx should be referenced again Source of MOSFET - this is the way how NOSFET works (it controlled by Gate-Source voltage).
And doesn't matter what else is connected to MOSFET Source.
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#4 Reply
Posted by
moffy
on 18 Apr, 2024 14:11
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I think you need a diode from +24V to Vdd, because the source shifts up to +24V when the MOSFET is on, but +24V exceeds the maximum voltage of the UCC5304, and is also too high for MOSFET gates. You probably need a zener in series with the diode to drop the voltage down to say 15V. I also think you need some moderate value resistor from your floating ground to ground otherwise there will be no current path to charge the driver IC supply unless the opamp can provide that path.
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#5 Reply
Posted by
RFDx
on 18 Apr, 2024 14:56
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Hello XVR, an updated version is attached of the schematics.
Yes, power supply (VDD) for the isolated side still missing, as is the bootstrap diode. The bootstrap capacitor (multiple values in parallel) ist also rather high. Is there a reason for that? How are you charging the bootstrapping capacitor the first time you turn the driver on? Maybe also tell something about what you are trying to achive here, not just how you wanna do it.
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#6 Reply
Posted by
john23
on 18 Apr, 2024 17:08
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Hello ,I am learning the principles of circuit,I only know that its supposed to drive a power amplifier TGA 2590 .
regarding your question two extended schematics photos are attached the node is call VIZ and GIZ and its connected to DB02S2415A component as shown in another picture .
Yes of course the mosfter is opened and closed by the Vgs>Vt condition, however what i have inner logical conflict when on one hand we put "floating" Ground.
But on the other hand we the source is not connected to ground it could be connected to power amplifier and the draing of power amplifier is the load and not plain ground.
So how this virtual ground helps us here?
Thanks.
"Output of UCCxxx should be referenced again Source of MOSFET - this is the way how NOSFET works (it controlled by Gate-Source voltage)"
"BTW - there is no supply attached to isolated side of UCC5304 in schema. I think something was forgotten"
https://www.mouser.co.il/datasheet/2/632/DS_DB02S_D-3106392.pdf
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#7 Reply
Posted by
xvr
on 18 Apr, 2024 17:16
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> But on the other hand we the source is not connected to ground it could be connected to power amplifier and the draing of power amplifier is the load and not plain ground.
We have 2 power sources in schema and 2 different not intersected set of Power/Ground.
One is a whole schema Power source and Ground, and second - your VIZ and GIZ nodes. They also Power and Ground, but different.
So there is no any conflicts - top amplifier connection is a ground in one power domain (isolated) and nothing in another (main).
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#8 Reply
Posted by
moffy
on 18 Apr, 2024 23:12
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There you go, an isolated 15V supply for the MOSFET driver, just what was needed.
Think of the MOSFET as a resistor with one terminal connected to the +24V. When the driver is 'off' the resistance is very high, when the driver is 'on' the resistance is very low, since the driver is referenced to the source of the MOSFET and it now has a floating supply it is happy.
P.S. They could have used a P channel MOSFET and saved quite a bit in complexity and cost.
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#9 Reply
Posted by
xvr
on 19 Apr, 2024 07:57
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Why just not use SSR? Unstead of MOSFET and all circuit that now controll it?
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#10 Reply
Posted by
RFDx
on 19 Apr, 2024 08:00
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regarding your question two extended schematics photos are attached the node is call VIZ and GIZ and its connected to DB02S2415A component as shown in another picture .
Now it makes sense.
Yes of course the mosfter is opened and closed by the Vgs>Vt condition, however what i have inner logical conflict when on one hand we put "floating" Ground.
But on the other hand we the source is not connected to ground it could be connected to power amplifier and the draing of power amplifier is the load and not plain ground.
So how this virtual ground helps us here?
The source of the N-MOSFET has to vary between 0V (MMIC amp off) and 24V (MMIC amp on). In order to do that, the gate of the MOSFET needs to be driven with 0V (for off) or at least 10V (for on), referenced to the level at the source of the MOSFET. If you connect the ground pin (VSS) of the isolated driver to the source terminal of the transistor it becomes a floating ground because it moves up and down (0<->24V, referenced to real ground), depending on the MMIC amp being on or off.
The driver itself is isolated (floating) from the rest of the circuit and is supplied from an isolated voltage source (15 V DC/DC converter) connected between the VDD (floating supply) and VSS (floating ground) pins of the driver. The gate-source Voltage Vgs of the MOSFET alternates between ~0V (amp off) and ~15V (amp on). Referenced to real ground, the gate voltage of the MOSFET will measure ~0V (for amp off) and ~39V (24V+15V) (for amp on).
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#11 Reply
Posted by
john23
on 19 Apr, 2024 18:57
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Hello RDFX,I am used to this of voltages always with reference to some poing gnd.
When we want to mosfet to be ON we have to put at least 10V Vgs.
we have DB02S2415A which gets as input 24V and -12V while the output is Viz and Giz which are refrenced to gnd of the supply of the component.
Then those two voltage are connected to the UCC5304 out and VSS1.
what is the logic of connecting the output of the DB02S2415A to the output of the UCC5304?
Is it a logical error in the schematics or there is some logic for doing this?
THanks.
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#12 Reply
Posted by
xvr
on 19 Apr, 2024 19:02
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This is not output of UCCxxx, this is its power
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#13 Reply
Posted by
john23
on 19 Apr, 2024 19:42
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#14 Reply
Posted by
xvr
on 19 Apr, 2024 20:05
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Page 3. Pin table.
OUT 7 O Output of driver. Connect to the gate of the FET or IGBT.
So, this is output
VDD 8 P Secondary-side power for driver.
VSS 5, 6 P Ground for secondary-side driver.
So, pin 8 is isolated power and pins 5 and 6 - isolated ground
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#15 Reply
Posted by
RFDx
on 19 Apr, 2024 23:07
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When we want to mosfet to be ON we have to put at least 10V Vgs.
That is correct. I have attached a simplified circuit with a similar optocoupler type driver which should make things clearer. The load (think MMIC amp) is shown here as a simple 50 Ohm resistor. When turned on, the gate-source voltage Vgs in this case is 13.4V (37.4-24V). All voltage probes are referenced to real ground. When the load is turned off all voltages and currents are zero.
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#16 Reply
Posted by
john23
on 20 Apr, 2024 05:40
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#17 Reply
Posted by
john23
on 20 Apr, 2024 06:40
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Hello RDFX,could you give intuition regarding the diode in this stucrture?
How it functiones?
Thanks.
"Yes, power supply (VDD) for the isolated side still missing, as is the bootstrap diode. The bootstrap capacitor (multiple values in parallel) ist also rather high. Is there a reason for that? How are you charging the bootstrapping capacitor the first time you turn the driver on? Maybe also tell something about what you are trying to achive here, not just how you wanna do it."
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#18 Reply
Posted by
xvr
on 20 Apr, 2024 08:37
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> i ment what is the exact voltage on the out pin7 and Vss pin 5?
When UCCxxx in turned on state voltage in pin 7 will be the same as on pin 8 (Vcc). You also can measure voltage BETWEEN pins 5 and 7 - it will be isolated side supply voltage.
Measuring any of them relative whole schema ground meanless, because all MOSFET control circuit floating with the Source potential of MOSFET.
So datasheet also lack of this numbers - they have no any physical meaning (except for maximum isolation voltage - difference between any primary net potential and secondary should not exceed it)
In your circuit Source of MOSFET (in turned on state) will bring to +24V
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#19 Reply
Posted by
RFDx
on 20 Apr, 2024 12:17
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Hello RDFX,could you give intuition regarding the diode in this stucrture?
How it functiones?
The bootstrap diode and capacitor are irrelevant because the goal is to turn the MMIC amp continuously (static) on/off. At least I'm guessing that is the case.
I mentioned the diode/capacitor before I knew there is a DC/DC converter to supply the floating driver.
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#20 Reply
Posted by
john23
on 20 Apr, 2024 14:46
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Hello , i was told that i don't need R4u Du1 C4Uon the MOSFET gate. The UCC5304 is an active pull-down device, they wont turn the MOSFET off any quicker.
Could you reccomend me a strtucture will indeed help switching this mosfet i could try and simulate it in LTSPICE.
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#21 Reply
Posted by
xvr
on 20 Apr, 2024 14:51
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> Could you reccomend me a strtucture will indeed help switching this mosfet
How fast you need to switch MOSFET? If you don't need nanoseconds times you can use MUCH simpler circuit
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#22 Reply
Posted by
john23
on 20 Apr, 2024 18:05
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Hello XVR,Its for learning puposes.
Could you reccomend me very fast configuration and normal configuration :-)
So i could simulate them?
Thanks.
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#23 Reply
Posted by
xvr
on 20 Apr, 2024 18:19
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Very fast is a your current one (without R4, R4u, R6u, Du1 and C4u).
Normal is a one where UCC5304 and DB02S2415A replaced by TLP3905 (or by some analog of it)
It's internal LED connected (via resistor) directly to input control circuit and output directly to S-G of MOSFET
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#24 Reply
Posted by
john23
on 22 Apr, 2024 14:36
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Hello XVR i have simulated the driver components as shown in the photos below.
In the datasheet they present a structure shown in the lask link photo, it very similar to my schematics.
Could you give an intuition regarding how the diode and resistor act ?
Why this structure is irrlevant for my mosfet controll of switching?
https://ibb.co/HBq9LTZhttps://ibb.co/64MC3b5https://ibb.co/1KR6vmm