TPSI3052-Q1: Spice Model

Vivek Gandham

Part Number: TPSI3052-Q1

Tool/software:

Hi team,

Is there any working Spice model of SSR IC with 15V, from the ti website it is not working spice showing the simulation error.

Can ti help us to solve this issue?

Regards

Vivek

10 months ago

0 Tilden Chen 10 months ago

TI__Expert 6206 points

Hello Vivek,

Thanks for reaching out to our team on E2E. Can you please share your project along with the error message? The TPSI3052-Q1 PSpice model was working for me when I checked a few days ago.

How To Share PSpice Project

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 10 months ago in reply to Tilden Chen

Prodigy 180 points

Yeah, it working in Pspice.

Is it available Simetrix library?

Regards

Vivek

0 Tilden Chen 10 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Thanks for the clarification. Try this file, it was working in March.

0486.TPSI3052-Q1_SIMetrix.zip

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 10 months ago in reply to Tilden Chen

Prodigy 180 points

hi Chen,

This model is generating VDRV 10V output, I am looking for 15Vdc as Vdrv.

Regards

Vivek

0 Tilden Chen 10 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Please try this one.

TPSI3052_SIMETRIX_B1P05.lib

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 10 months ago in reply to Tilden Chen

Prodigy 180 points

Hi Chen,

Its giving pulsed output, even EN given continuous high

On Output side, its giving around 12V not 15V

Regards

Vivek

+1 Tilden Chen 10 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Thanks for the update. The reason you are seeing a pulsed output is due to the resistive load drawing more current than what the TPSI3052-Q1 can supply. VDDH is hitting UVLO and shutting off VDRV. Also, if you are intending to drive a MOSFET in your final solution, resistive load is also not representative of how a MOSFET would behave. Use a capacitor instead.

For example, if you have selected E3M0075120K and want to mimic driving this MOSFET, find the load capacitance (C_VDRV) by taking the total gate charge (Q_G) and dividing by the gate voltage (V_DRV).

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 9 months ago in reply to Tilden Chen

Prodigy 180 points

Thanks chen, its working for me, but why the model is taking 30ms to turn on Vdrv voltage

0 Tilden Chen 9 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Thanks for the update. The model takes 30 ms to turn on VDRV because our device first needs to charge C2 and C3 to sufficient voltage. You can try using C2 = 150 nF and C3 = 450 nF. This should reduce the start time to microseconds range.

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 9 months ago in reply to Tilden Chen

Prodigy 180 points

1. In Pratical it will charge very fast right?

2. In simulation capacitors we need to as per calculation or we can go with your values to run fast?

Regards

Vivek

0 Tilden Chen 9 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Thanks for your reply. If you are using C2 = 3.3 uF and C3 = 10 uF in a real setup, it will still take milliseconds to charge.

Your C2 and C3 capacitance will depend on the MOSFET's total gate charge. If you have selected one, you can use our calculator tool to find suitable capacitance values or I can check that for you with the P/N.

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 9 months ago in reply to Tilden Chen

Prodigy 180 points

For me it's charging in uSec, with 576nC as per the below circuit

0 Tilden Chen 9 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Thanks for the update. To model a MOSFET with Qg = 576 nC, we would want to use a C_LOAD = 38.4 nF (instead of the 3.67 nF).

Would recommend using C2 = 1.5 uF and C3 = 4.7 uF here.

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

0 Vivek Gandham 9 months ago in reply to Tilden Chen

Prodigy 180 points

Thanks for the support, in turn off gate pulse the delay form EN to VDRV is depends on these capacitors?

0 Tilden Chen 9 months ago in reply to Vivek Gandham

TI__Expert 6206 points

Hello Vivek,

Thanks for the update. The EN-to-VDRV delay does not depend on C2 and C3, when EN goes low the driver (VDRV) is pulled down to VSSS. So the EN-to-VDRV delay will be the sum of the time for the capacitive load (C_VDRV) to discharge plus the TPSI3052-Q1's propagation delay (microseconds range). In the datasheet, t_{HL_VDRV} assumes C_VDRV = 100 pF, which would be equivalent to MOSFET with 1.5 nC total gate charge. Since that RC discharge time is near 500 ps, the TPSI3052-Q1's internal propagation delay is mainly contributing to the 2.5 µs spec.

Best regards,
Tilden Chen

Solid State Relays | Applications Engineer

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