Experimental work | Working with PS 9505
11:29 AM
PS9505 is an optocoupler with the capability to produce large output current up to 2.5 A. It can be used to replace a traditional tag team of optocoupler and a gate driver. Its salient features are as follows.
Rg = Vg/Ig
where Vg is the gate voltage applied and Ig is the gate current required to switch the device.
Ig= Qg/ts
where Qg is the total gate charge and ts is the switching time
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| Pin configuration (click to zoom) |
- VCC-VEE can be supplied up to 30 V max.
- The output follows the input hence it is non inverting.
- Max operating frequency is 50 kHz.
- Inherent under voltage lock out with VCC-VEE < (9.5 to 12.5V)
The circuit connection is as follows
- Pin 2 and 3 takes the input through a resistor normally in the range of 200-500 ohm.
- Pin 1 and 4 are not connected . They should be grounded or left open.
- Pin 8 gets the VCC while pin 5 is grounded.
- Pin 8 also needs a bypass capacitor with value greater than 0.1 uF.
- Pin 6 and 7 are shorted externally and output is connected with them with a series resistance RG.
The value of RG should be selected carefully as it has a great impact of the working of PS9505. If RG is large then it may lead to higher switching dissipation. Selecting a small RG leads to larger voltage variations. Therefore the selection of RG needs calculations as presented below.
- RG must be selected such that the ic never exceeds the maximum current rating i,e 2.5 A.
- RG must be selected so that the ic power dissipation remains bounded within 178mW
- RG will be selected based on its calculation from input a well as from the output side.
- RG value is based on the calculation from input side aand from the output side. The value calculated from the input side is designated RG and the one calculated from the output side is termed as Rg.
- Fine tuning of Rg for optimum results.
RG calculation from input side
According to the application note of PS9505 the following equation
RG > {VCC-VEE-VOLM}/ IOpeak
where
VCC = Biasing voltage
VEE = 0 if grounded
VOL = Low level output voltage
IOpeak = Peak output current i.e 2.5 A
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| VOL vs IOL characteristics (Click to zoom) |
In my circuit VCC= 18V, VOL is provided in the application note and is selected as 3.5 V , hence
RG > (18-0-3.5)/2.5 = 5.8 ohm
RG > (18-0-3.5)/2.5 = 5.8 ohm
RG > 5.8 ohm as calculated from the input side
RG calculation from output side
To calculate Rg from the output side first we need to ransack the datasheet of the device connected at the load. For example let us consider the mosfet IPA65R650CE . The information required from its datasheet is the total gate charge. Now comes the calculation of Rg from the output sideRg = Vg/Ig
where Vg is the gate voltage applied and Ig is the gate current required to switch the device.
Ig= Qg/ts
where Qg is the total gate charge and ts is the switching time
Now using the datasheet of the above mentioned mofet Ig= (23nC x 100kHz) = 0.0023
Therefore Rg = 18/0.0023 = 7826 ohm
According to the application note of this ic , following condition must be true .
RG calculated from input side < Rg calculated from output sideIf this is not true then you must find some other optocoupler which can drive a larger current.
Fine tuning RG
In order to tune the RG we need to find the power consumption of the device. The power consumption consists of two parts.
- Power consumption on the input side (PD) = Vf x If x duty cycle
- Power consumption on the output side (PO)= Po(circuit)+Po(switching)
In this regard Po(circuit) is the consumption of the photo detector and Po(switching) is the power consumed by the output current.
Calculation of PD= 16mA x 1.8 x 0.5 = 14.4mW
Calculation of Po(circuit)= Ic x (VCC-VEE) =
where Ic is the current supplied to the photo detector and typically it is 2mA and maximum 3 mA as provided in the application note.
Calculation of Po(switching) = Esw (RG,QG)xFsw
where Esw (RG,QG) is the per cycle power consumption when charging the mosfet and fsw is the switching frequency (60 Hz in my case). The Esw (RG,QG) can be calculated based on the graph presented in application note. Based on 5.8 ohm RG calculated above the Esw is found to be 0.5uJ.
Calculation of PO = Ic x (VCC-VEE) + Esw (RG,QG)xFsw = 3mA x(18-0) + 0.5uJ x 60 = 54 mW
This PO is less than the maximum power rating of this ic i.e 178 mW.
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| Switching loss per cycle ( click to zoom) |
Therefore, based on the switching loss graph presented in the application note, the RG should be between 6 ohm - 40 ohm.
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