Filippo Salviati (bishop)
Catholic prelate
From Wikipedia, the free encyclopedia
Filippo Salviati (1578–1634) was a Catholic prelate who served as Bishop of Sansepolcro (1619–1634).[1][2][3][4]
ChurchCatholic Church
DioceseDiocese of Sansepolcro
In office1619–1634
PredecessorGiovanni dei Gualtieri
Most Reverend Filippo Salviati | |
|---|---|
| Bishop of Sansepolcro | |
| Church | Catholic Church |
| Diocese | Diocese of Sansepolcro |
| In office | 1619–1634 |
| Predecessor | Giovanni dei Gualtieri |
| Successor | Zanobi de' Medici |
| Orders | |
| Consecration | 18 August 1619 by Ottavio Bandini |
| Personal details | |
| Born | 1578 |
| Died | 1634 (age 56) |
Biography
Filippo Salviati was born in Florence, Italy in 1578.[2]
He had been Provost of the cathedral Chapter of Prato.[5]
On 12 August 1619, he was appointed Bishop of Sansepolcro by Pope Paul V.[1][2] On 18 August 1619, he was consecrated bishop by Ottavio Bandini, Cardinal-Priest of San Lorenzo in Lucina, with Francesco Sacrati, Titular Archbishop of Damascus, and Horace Capponi, Bishop Emeritus of Carpentras, serving as co-consecrators.[2]
In Fall 1629, Bishop Salviati conducted a formal visitation of the ecclesiastical institutions in his diocese.[6]
He died in 1634.[2]
